Alder - Lectures on Integration of Several Variables (Cálculo)

M��

Lectures on Integration of Several Variables�

Second Semester ��

Dr� Michael D� Alder

Department of Mathematics

The University of Western Australia

Nedlands W�A� �� AUSTRALIA

Contents

� Introduction �

�� What The Course is About � � � � � � � � � � � � � � � � �

�� Ideas versus Techniques � � � � � � � � � � � � � � � � � � �

�� Reminder � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Summary � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� Integrating Functions ��

�� Integration Over Rectangles � � � � � � � � � � � � � � � � ��

�� Integration Over bits of disks � � � � � � � � � � � � � � ��

�� Over wilder regions� but not too wild � � � � � � � � � � � ��

�� General Ideas and Examples � � � � � � � � � � � � ��

�� Algorithm � � � � � � � � � � � � � � � � � � � � � ��

�� Exercises � � � � � � � � � � � � � � � � � � � � � � � ��

�� Applications and Subversive Thoughts � � � � � � ��

�

� CONTENTS

�� Change of Variables � � � � � � � � � � � � � � � � � � � � � ��

�� Three Variables � � � � � � � � � � � � � � � � � � � � � � � �

�� Swindle � � � � � � � � � � � � � � � � � � � � � � � ��

�� More Exercises � � � � � � � � � � � � � � � � � � � ��

�� Higher Things � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Embeddings � � � � � � � � � � � � � � � � � � � � � ��

�� Parametrisation � � � � � � � � � � � � � � � � � � � �

�� Integration over Curves and Surfaces � � � � � � � � � � � �

�� Exercises � � � � � � � � � � � � � � � � � � � � � � � ��

�� Doing it Piecewise � � � � � � � � � � � � � � � � � ��

� ��Forms and ��forms on R� ��

�� Introduction� A New Idea � � � � � � � � � � � � � � � � � ��

�� Advice� Learning This Stu� � � � � � � � � � � � � ��

�� Notation � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Exercises � � � � � � � � � � � � � � � � � � � � � � � ��

�� Surprises� Closed and Exact Forms � � � � � � � � � � � � ��

�� forms � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Holes � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Connections � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� The exterior derivative and k�forms � � � � � � � � � � � � ��

CONTENTS �

�� Change of Variables � � � � � � � � � � � � � � � � � � � � ��

�� forms � � � � � � � � � � � � � � � � � � � � � � � ��

�� forms � � � � � � � � � � � � � � � � � � � � � � � ��

�� Some simple rules � � � � � � � � � � � � � � � � � � ��

�� What is a di�erential form� � � � � � � � � � � � � ��

� Forms on R� ��

�� Some examples of Exterior Derivatives � � � � � � � � � � ��

�� Closed and Exact k�forms � � � � � � � � � � � � � � � � � ��

�� Stokes� Theorem � � � � � � � � � � � � � � � � � � � � � � ��

�� History � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Integrating ��forms and ��forms in R� � � � � � � � ��

�� Stokes� Theorem� Modern Dress � � � � � � � � � � ��

�� Stoke�s Theorem� Classical Dress � � � � � � � � � ��

�� Gauss � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Tying Up Stokes � � � � � � � � � � � � � � � � � � ��

�� Integration of forms over manifolds � � � � � � � � � � � � ��

�� A Summary � � � � � � � � � � � � � � � � � � � � � ��

�� A note on i� j� k � � � � � � � � � � � � � � � � � � ��

� Partial Di�erential Equations ��

� CONTENTS

�� Introduction � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� The Di�usion Equation � � � � � � � � � � � � � � � � � � � ��

�� Intuitive � � � � � � � � � � � � � � � � � � � � � � � ��

�� Saying it in Algebra � � � � � � � � � � � � � � � � � �

�� Laplace�s Equation � � � � � � � � � � � � � � � � � � � � � �

�� The Dirichlet Problem for Laplaces Equation � � � � � � � ��

�� Laplace on Disks � � � � � � � � � � � � � � � � � � � � � � ��

�� Solving the Heat Equation � � � � � � � � � � � � � � � � � ��

�� And in Conclusion��

CONTENTS �

Preface

These notes� whether you get them on paper or o� the Web� are for theM�� course in Mathematics at the University of Western Australia�They are intended for second year students of Mechanical� Civil and En�vironmental Engineering� They are not intended for Pure Mathemati�cians� Physicists� or Electrical Engineering students who do a di�erentunit� Anybody can read them of course� but if you �nd mistakes donot be too surprised� They were given in second semester in �� and�� I was guided by Grant Keady who pointed out at least one awfulerror and numerous minor ones in the �rst version� I am also gratefulto Malcolm Hood and Nev Fowkes for useful perspectives� None of theabove� however� should be blamed for the �nal form�

The approach was required to be intuitive and geometric rather thanformal or precise� and is� Do not� therefore� be surprised or shockedat the sloppiness or scru�ness of the exposition� It is supposed to bethat way� On the other hand� if you have trouble seeing why Stoke�sTheorem or PDEs are worth studying� I would like to think that thistext will give you some grounds for thinking they might be�

I have never been able to �nd Engineering text books interesting� norhave I the slightest taste for ritual� so bashing through the standardtexts was� for me� only something I did after I had found out usuallyfrom reading something quite di�erent or talking to a colleague thatthe subject was fascinating and a triumphant demonstration of thepower of the human mind� I think this is terrible� To take somethingquite amazing and wonderful� and then turn it into boring drudgeryought to be a hanging crime� Yet the e�ort involved in following thetechnicalities is considerable� If you see the material in terms of passingexaminations� then it is� let us face it� just one shitload of integrals�Why anybody should want to pass exams that much is a mystery to me�there being so many much more enjoyable ways to pass the time� butmany do� or think they do at the beginning of the year� I don�t know ifthis helps you� but it helped me a lot to feel� as I was sweating throughthe material� that I was treading again a path blazed by some of thecleverest men the world has seen� and that the horrible complexity of it

CONTENTS

is there because the world is a di�cult and complicated place� Beatingeven a small part of it into submission� understanding clearly how itworks so that you can make bits of it do what you want� is horriblydi�cult� Being muddled and confused is the natural state of man andwoman� Getting slightly less muddled and confused by exploring theworld by pen and paper and doing ghastly sums is not easy� but atthe end you can be awestruck by the things that the human mind can�gure out� It is this pleasure in making sense of the world that drovethe men who invented this stu� to do the work they did usually aftera hard day at their proper job� To learn the skills in order to pass anexamination and to miss the point of having them is to concentrate onthe �nger and to ignore what it is pointing at�

Heinlein remarked that to be ignorant of mathematics was to be lessthan fully human� such a person is just a house�trained monkey� Toconfuse mathematics with knowing how to do scads of integrals is tobe less than human too� I guess there are a lot of part�human beingsaround�

Chapter �

Introduction

�� What The Course is About

In �rst year mathematics� you learnt about Calculus and Linear Alge�bra� The Calculus was about di�erentiating and integrating functionsf � R �� R � such as fx � x�� The linear algebra dealt with linearmaps f � R n �� R

m represented by matrices� On the other hand� linearmaps are very special� and you didn�t do much calculus with them�

In Second year� you get to do more Linear Algebra� and also moreCalculus� In this course we do di�erential and integral calculus forfunctions f � R n �� R

m� Mainly Integral Calculus�

If you think about the sort of things done in mathematics� you candivide it into two types� you could call them geometry and algebra� orpictures and symbol strings� or ideas and techniques� For instance� indoing integration of a function such as fx � x�� you have the symbols�

Z �

�x�dx

and the picture of �gure ��

�

� CHAPTER �� INTRODUCTION

Figure �� y � x�

The picture gives you an idea of what the symbols mean� the inte�gral

R �� x

�dx is the shaded area under the curve� The de�nition of theintegral tells you that you have to draw little rectangles and add upthe area of the rectangles in order to get an approximate idea� thenyou go back and do it with smaller� narrower rectangles to get a betteridea� and in the limit as you calculate the area of thinner and thinnerrectangles� you get the actual area and the value of the integral�

The picture is useful� it tells you what integration is about from theconceptual point of view� but it doesn�t give you much of an idea ofhow to actually calculate the integral in a reasonable time� Fortunately�there is a very useful theorem which tells us how to do the job� we �rst�nd an antiderivative of fx � x�� which is easy� x�� will do nicely�then evaluate this at each end of the interval and take the di�erence�So we get �

� as the answer with very little e�ort� The less e�ort wehave to go to� the better� so the theorem is very handy� Combined withother theorems telling us how to di�erentiate polynomials and otherfunctions� we get a lot of hard sums done very easily� which is a GoodThing� So Good Theorems save us a lot of Work�

The main theorem we use to do the sums here is the FundamentalTheorem of Calculus which tells us that Di�erentiation and Integration

�� IDEAS VERSUS TECHNIQUES �

are inverse operations� and that to evaluate the integral of a functionf over an interval� we look at another function� F � the antiderivativeof f � over the boundary of the interval� This generalises in very usefulways to higher dimensions� and is called Stokes� Theorem� We shall nottry to prove this theorem� but we shall be trying to understand it anduse it� There are some specialised version of Stokes� Theorem calledGreen�s Theorem and Gauss� Divergence Theorem which we shall alsodo�

This is the main point of the course� The only other thing we dois rather tacked on the end� it generalises the Ordinary Di�erentialEquations you did in First Year� We generalise to what are calledPartial Di�erential Equations� This is a huge subject and I am goingto spend a few lectures on it at the end of the course� The good newsis that you will learn how to solve quite hard problems about heat�ow and di�usion� again at fairly small cost in terms of the amount ofsu�ering you have to do�

�� Ideas versus Techniques

It is traditional in teaching Engineers to train them to do the sums byshowing them the techniques� and to hope that they will take it forgranted that the sums are worth doing� Also to take it for granted thatthey have terri�c memories� Most of the books that cover EngineeringMathematics are based on this assumption� and it makes them ratherless fun to read than� say� the Sermons of the Bishop of Bath and Wells�

I shall spend rather more time than is conventional drawing picturesand explaining the ideas in geometric terms� Someone had to inventthe techniques� and it is easier to remember them if you can see howthe inventor thought about them� So there will be a lot of pictures inthis course� It is most important that you have some mental image ormodel or interpretation of the strings of symbols�

There may be more than one� It is worth pointing out that there

�� CHAPTER �� INTRODUCTION

are often several di�erent models or ways of thinking about a piece ofmathematics� For example�

R �� x

�dx can be thought of as the area abovethe x�axis and under the graph of fx � x� between � and �� or it canbe thought of as the average value of y over the range of possible xvalues� multiplied by the length of the interval� Both are useful waysto think about integrals� Don�t go looking for the one right way ofthinking about something� there may be lots�

At the same time� the bottom line in Mathematics is being able to dothe sums� and the best way to do this and guarantee that you willpass the exam is to practise doing a lot of examples until the detailsburn themselves into your brain� I will give a lot of examples� In theend however� it is you who have to do the skull�sweat which gets youfeeling con�dent about how to tackle the problems that can come up�

Everything in the course is useful� There isn�t anything which is put injust to make Engineers su�er� Cling to that assurance� because thereis going to be a lot of su�ering� It is good for you� it will make youmore spiritual and noble� more thoughtful and intelligent� Just likeyour lecturers�

�� Reminder

When you have a function of one variable� f � R �� R � you are usedto drawing the graph of the function and thinking of the function asjust its graph� This can be carried a little way forward� if you have afunction f � R � �� R then it is still possible to think of the graph� Inthe one dimensional case� the input to the function is a single numberand the output is a single number� so you can imagine that you walkalong to the input number along the real line� then you do some quicksums on your calculator to work out the value of fx� then you lift apole up to that height and make a mark on the sky� Then go to anotherpoint on the line and lift up your pole again� Keep doing this for allpossible real numbers than can be input� and the track of the top ofthe pole for each output is the graph�

�� REMINDER ��

X

Y

Figure �� fx� y � x� � y�

Now if you have a function f � R � �� R � it is exactly the same exceptthat now you have two numbers� say x and y as input� which we canthink of as a point in the plane� So you go to some point in the plane�x� y� and then you calculate fx� y and lift your pole up that highbecause the output is a single number� As you go all over the plane�the top of the pole forms a surface over the plane or under it if thevalue of fx� y is negative� As an example� f � R

� �� R given byfx� y � x� � y� has graph a parabolic bowl� like a radio telescopepointing straight up� as in �gure ��

If you have a function of three variables� f � R � �� R � then you wanderabout a huge room� in three dimensions� and you get a single numberas output� It is not possible to visualise the graph of this� because it issitting in four dimensions� You can imagine that each point is assigneda colour� yellow for positive� red for high positive� green for zero orsmall negative and blue for very negative� This would be a useful wayto think of the function� but it isn�t exactly a graph�

If you have a function f � R �� R� then it is possible� though a little

strange� to think of its graph� You have to imagine that you wanderalong a line and at each point on the line� t say� you calculate ft�which is a pair of numbers or a point in the plane� So you will get some


0

1

2

3

45

6

78

9

10

11

12

Figure �� x� y � ft

sort of a curve in three dimensions�

It is probably easier to think of a map f � R �� R� as being a bug

wandering about the plane in time� The input� t is the time� andthe value of ft is where the bug is at that time� You can draw thetrajectory of the bug on some graph paper� and put tick marks on tolabel the times� I have sketched this in �gure ��

It would be a big mistake to think that this is the only way to think ofa function f � R �� R

�� but it is useful� It is looking at a parametrisedimage of the function� NOT the graph of the function� There areother ways of looking at functions besides graphs� If you are muddledabout this distinction� stop and think about it now by making up someexamples� Otherwise you will stay muddled� which is a uncomfortableand b makes doing the sums much harder� When you are muddledyou can write down some really daft things without knowing they aredaft�

For maps f � R �� R�� it is easy to generalise and imagine a bug �ying

about in three dimensions� Replace the bug by a spaceship and youhave a typical problem in dynamics�

Suppose we have some water �owing through a rectangular region� At

�� REMINDER ��

each point inside the rectangle� the water is �owing in some directionat some speed� We can imagine putting a little arrow at each pointof the rectangle� pointing along the direction of the �ow� the lengthproportional to the speed� This is almost possible� the air��ow inwind�tunnels is sometimes indicated by putting threads into the tunnel�Tracer dyes can be released into water and a video of the shape ofthe dye drops can show how the stu� is stretched� Certainly� a goodcomputer simulation can draw lots of little arrows� if not all of them�Since a little arrow is also a vector in the same space� we can usefunctions f � R � �� R

� and f � R � �� R� to represent vector �elds in

R� and R

� respectively� The input to the function is where you are� andthe output of the function is where the head of the little arrow is at thatpoint� relative to the tail� which is on the point� Such an assignmentof little arrows� one to each point of the space� is called a vector �eldon that space� and we show a bit of it in �gure �� Conversely� oneway of thinking of a map f � R

� �� R� is as a vector �eld on R

��Again� there are other ways� and the more ways of visualising whatyou are doing the better� and we shall come across some of them inthe course� One way of thinking about functions from R

� to R� is to

imagine them picking up a piece of graph paper and stretching anddistorting it and then hammering it �at into another piece of graphpaper� Similarly� a function from R

� to R� can be used to describe

taking a piece of graph paper and bending it and putting it in a threedimensional room� There are lots of visualisation methods� You will bynow have had some experience of using SymbolicMathematics packagesto assist in this�

We shall not have much use in this course for maps f � Rn �� R

m

except in these cases� i�e� when n�m are �� or �� although you wouldbe mistaken if you thought that these are the only �real� cases� SpeechRecognition by computer� for example� is a hard problem� and a mansaying �hippopotamus� into a microphone is producing a trajectory ina space of �lterbanks of dimension �� or � � Recognising the wordamounts to matching it with other such trajectories� And there are azillion other cases where high dimensional spaces and maps betweenthem are important� But we have to leave something for third year�


Figure �� fx� y � u� v

�� Summary

This course is about integration and di�erentiation applied to multi�dimensional things such as vector �elds and functions f � R n �� R

m�mainly in the case where n�m � �� The central result is Stokes�Theorem� which generalises the Fundamental Theorem of Calculus andwill tell us how to do some thoroughly nasty sums with the minimumof stress and pain� As a free extra� we shall apply the Fourier theory of�rst semester to the di�usion equation� allowing us to solve some simplePDEs Partial Di�erential Equations governing the �ow of heat andthe di�usion of substances� No theorems will be proved� but the ideaswill be emphasised so that you can hope to �gure out how to solve newproblems�

This is old fashioned eighteenth and nineteenth century mathematics�and I shall present it in an old fashioned eighteenth and nineteenthcentury manner� with levels of rigour so low as to shock a Pure Math�ematician of this century� I shall do this because while rigour has animportant place� it doesn�t have much meaning for mechanical and civilengineers in general� If you feel uneasy about the ideas because theyseem to you unclear and confusing� you may have mathematical in�

�� SUMMARY ��

stincts and are probably in the wrong course� Please speak up if youfeel a need for more precision and want to know what functions andvector �elds really are� If you don�t� I shall assume my sloppiness meetswith your approval� When my conscience gets the better of me and Iwince at something particularly outrageous I have just written� I shallput in a footnote� to try to justify myself�

There is one important di�erence of style between �rst year Mathemat�ics at least as you probably got away with doing it and this course�In �rst year Mathematics the key thing is to identify the problem type�pull out an algebraic manipulation for solving it� and hit the problemuntil you kill it� The lecturer probably went on about ideas� but yougot away without knowing what he was talking about� In this course�you will need to link up the geometry with the algebra� Instead ofgiving you closed problems where you only have to do some patternrecognition� I shall give you open problems with lots of possible solu�tions� In particular� you will be given curves and surfaces and requiredto express them in terms of functions� For example� if I tell you thatthe unit sphere in R

� is the set of points��B�

xyz

�CA � R

� � x� � y� � z� � �

��

I shall ask you to express this parametrically� that is� as the image ofa square� This is geometrically like taking a square of plasticene andwrapping it so that it covers the surface of a beach�ball� If I had askedyou to do a similar thing for the unit circle in R

�� namely to express itparametrically as the image of a line segment� the answer

x � cost� y � sint� t � ��

would have been acceptable�

I shall do this with lots of di�erent surfaces and curves� I shall give lotsof examples� but you will need to practise drawing graphs of surfaces inorder to get a feel for it� This sort of thing is important for engineers

�like this�

� CHAPTER �� INTRODUCTION

who deal with shapes of objects which may be complicated� and mayneed to be able to express these shapes algebraically�

This course is oriented to standard problems and methods for solvingthem� it slithers around many of the subtleties needed for proper gen�eralisation to higher dimensions� Altogether it is pretty disgusting� butengineers seem to like being disgusting� and I aim to please the cus�tomers� If the customers want to pass the examination� all they haveto do is take the assignments seriously and give them their best shots�

Chapter �

Integrating Functions

�� Integration Over Rectangles

Given a function of two variables� f � R � �� R � we visualise a surfacesitting over the plane� the graph of the function� as a simple generalisa�tion of functions of a single variable� If I take a rectangle in the planeand call it U �

U �

� xy

�� R

� � � � x � �� y � �

�

and if I look at the region over the rectangle and under the graph� thenit is a three dimensional chunk of space and can reasonably be assumedto have a volume� If fx� y � � for all points inside U � then it is areasonable generalisation of the one dimensional case to say that thevolume is given by Z

Ufx� y dA

where dA is some little bit of area� just as dx is a little bit of x� �

�In the crude� barbarous days of old� people used to use �x to denote �a little

��

�� CHAPTER �� INTEGRATING FUNCTIONS

Figure ��RU fx� y

The de�nition of the one dimensional integral is in terms of limits ofsums of areas of little rectangles� the dx telling us something aboutthe limit of little line segments as the bases for the rectangles� For thetwo dimensional case� we can chop up the rectangle in �gure �� intolittle squares� erect a box on each square reaching up to the surface ofthe graph� add up the volume of the boxes� and then do it again withsmaller squares� In the limit� the integral would give us the volume� iflife has any meaning� �

bit of x�� and dx was used to mean an in�nitesimal bit of x� where an in�nitesimalwas a number so small that it was impossible to tell it from zero� but it wasn�t zero�Isaac Newton who invented the things wasn�t very happy about them� and BishopBerkeley had a dig at them� calling them �the ghosts of departed quantities�� He wasnarked because Robert Hooke� a friend of Newton �insofar as Newton had friendsand a declared atheist �a burningatthestake crime in those days had sneered atthe doctrines of the church� particularly the idea of a threeinone god� which hethought daft� It took over two centuries for mathematicians to work out what thedx really means� Mind you� most of them were having too much fun doing mathsthe scru�y way to care� We shall do mathematics in the old fashioned piratical way�with a rude gesture and a heynonnynuts to all those serious people who insiston knowing what they are talking about� A vague impression is good enough forscru�y buggers like us� provided we get the right answers to the sums� and most ofyou who do this course will never �nd out what dx or dA really means�

�Recall that the expressionR b

af�x dx is code for lim�x��

Pf�xi�x� where

the xi are taken one in each interval of length �x partitioning the interval �a� b �

�� INTEGRATION OVER RECTANGLES ��

X

Y

Figure ��RU x

� � y�

So at the level of ideas� it is easy to understand that it is possible tointegrate functions of two variables over rectangles in the plane�

Just to get things nailed down� it would be nice to have a rough ideaof the value of the integral of �gure �� This we could write asZ

Ux� � y� dA

where U is the rectangle with vertices �� Youcould build the shape shown with plasticene� The lowest corner hasheight zero and the highest has height �� The volumetherefore must be between � and �� To get a better estimate� although

The extension to integrals of functions of two variables� f�x� y is obvious� Wechop up the base rectangle U into little squares of some small area �A with nooverlap and every point of U in one of the little squares� then we take an �x� yi ineach little square� and multiply each of the f�x� yi by �A� and add them all up�P

f�x� yi �A� Now take the limit as �A� �� This is the de�nition of the integralfor the case where f is a reasonable� wellbehaved� lawabiding� decent function thatdoes not do drugs� It is a little trickier to �x up the de�nition for sleazy functionswhere you cannot immediately tell if they do drugs or not� Of course� the bit herethat throws doubt into the staunchest heart is the bit about taking limits� It soundsOK if you say it fast� but taking a limit is sometimes easier to talk about than do�Like taking a mistress� I daresay that�s easier for some than others�


not a very good one� the line in U along y � � has a minimum heightof � and a maximumheight of �� The volume of the bit over the square� � x � �� y � � is not bigger than �� and the volume over thesquare � � x � �� y � � is at least � and less than �� So the volumelies between � and �� You might think this is a pretty rough estimate�but to be able to say

� �ZUx� � y� dA � �

is at least a start� Crude estimates such as this are quite useful thingsto be able to do quickly in your head to stop you writing somethingreally daft� If you use smarter methods and come out with �� or�� you know you blundered� It also beats into your skull whatthe things mean� which makes it all a bit more interesting�

All the same� the work involved in getting the answer to four places ofdecimals by this method would seem to be excessive� There has to bea better way� and there is�

To see a smart way to do this� look at the area of the slice alongthe x�axis� where y � �� We already know that this is just the onedimensional integral Z �

�x� dx �

�

�

Similarly the slice at y � � is just the one dimensional integral

Z �

�x� � � dx � �

�

�

and the slice at y � � has area given by the one dimensional integral

Z �

�x� � � dx � �

�

�

Indeed� at each value of y� the area of the slice is given by

Z �

�x� � y� dx � y� �

�

�


Now if we think of each little slice having thickness dy� we get that thevolume given by Z

Ux� � y� dA

is just Z �

�y� �

�

� dy �

��

�

which looks a reasonable number�

Now this idea is not very di�cult� We have in fact a nice useful theorem�a particular instance of a more general theorem� due to Fubini� whichI shall state here in a special case�

Theorem �� Fubini� If U is a rectangle in the plane given by

U �nx� y � R

� � a � x � b� c � y � do

then ZUfx� y dA �

Z y�d

y�c

�Z x�b

x�afx� y dx

�dy

usually written Z d

c

Z b

afx� y dx dy

providing both of the integrals on the right hand side of the equationexist�

In other words� we can reduce integrals of functions of two variablesinto a sequence of two integrations in each variable separately� Thiscertainly works when the functions are nice and can be integrated� andwhen we have to integrate only over rectangles� As we shall see later�it works for integrals over nastier regions of the domain space also�

You may not feel it is easy to love a nice friendly theorem� but if youhad to evaluate many volumes from �rst principles� you�d feel very welldisposed to Fubini�s Theorem� You get the idea that every new theoremmeans extra work� so you usually hate them� but you try solving the


problems without them� and you�d really appreciate what cuddly thingstheorems are�

The enquiring mind will notice that we could just as well have donethe integral by reversing the order� doing the y integral �rst� so thatwe integrate along slices at di�erent values of x� and we had better getthe same answer� There are cases when integrating in one way is veryhard� while integrating the other way is easy� So if the integration looksparticularly foul� try reversing the order and see if it looks any better�

At this point� you need to get stuck into some practice sums� I shalldo the �rst one for you as a worked example�

Exercise �� For each of the functions below� for the rectanglespeci�ed� obtain a crude estimate of the integral of the function overthe rectangle by sketching the graph and using some common sense�Then compute the integral using Fubini�s Theorem�

��RUx� y� dA�U � fx� y � � � x � �� y � �g

��RU xy dA� U � fx� y � � � x � �� y � �g

�RU sinxy dA�U � fx� y � � � x � �� y � ��g

�RU e

�x�y� dA�U � fx� y � � � x � �� y � �g��

RU

y�x�y�� dA�U � fx� y � � � x �� y � �g

Example �� x� y� � � on the square �� Sincethe area of the square is �� the integral also lies between � and �� Sincethe graph of x � y� lies below the graph of �x � y which cuts thecuboid on the unit square of height � into two equal halves� the integralmust be less than �� Since the graph of x� y� lies above the graph ofx � y on the half of the square having x � y � �� the value of theintegral is greater than �� That is� �

�� I � �� where I is the value

of the integral� That gives some bounds on the integral� To do it byFubini�s Theorem


Figure ��R��x� y� dA

ZUx� y� dA�U � fx� y � � � x � �� y � �g

�Z x��

x��Z y��

y��x� y� dy dx

�Z x��

x��

x� y�

�

�y��

y��

dx

�Z x��

x��

x� ��

�� x�

�dx

�Z x��

x��x� � x�

�

� dx

�

x�

��x�

��x

�

�x��

x��

��

��

�

��

�

�

��


X

Y

Figure ��RU x

� � y�

�� Integration Over �bits of� disks

The next step is to ask if the domain of integration U has to be arectangle� and the answer is obviously not� if U were a disk� for instance�we could still chop it up into little squares� although some of themmightnot be wholly inside the disk� Still� the contribution to the volume ofthe bits outside the disk would get smaller as the squares got smallerand smaller� So it seems reasonable that it would work for domainsof integration that are more complicated than rectangles� or for thatmatter� disks�

There is an obvious way of trying to do integration over disks� and thatis to change to polar coordinates� Let�s go back to the mickey�mouselevel example Z

Ux� � y� dA

where now

U �nx� y � R

� � x� � y� � �o

This is shown in �gure ��

It is good practice to study simple examples to death� until you un�

�� INTEGRATION OVER �BITS OF� DISKS ��

derstand exactly what you are doing� Generalising to harder cases isthen much less e�ort� Mathematicians are very good at saving them�selves e�ort� A mathematician is someone who when confronted witha problem that would take him an hour� will cheerfully spend a weeklooking for the quick way� �

The height is � on the bounding circle� and zero in the middle� If we takeany angle � and draw a radius along that that angle� the function is justthe function fr � r� along that line� If we wanted a rough answer�we could take the perimeter of the disk over which we are integrating�which has length �� and multiply it by the area of the section� which is�� to get ��

�� This is obviously an over estimate� because it is the average

area multiplied by the greatest length of the arc� and if we wanted abetter estimate� we would have to divide the disk into concentric rings�and multiply the area by the perimeter for each ring� This is posible�and you should try to get an estimate of the required integral this way�

The alternative� which involves a bit of thought but much less comput�ing� is to have a polar version of Fubini�s Theorem� In the case of thecartesian coordinate system� the dA turned into dx dy� The questionarises� what ought dA to become when we work in polars� and the an�swer� obtained from contemplating �gure �� is r dr d�� The blob is atpolar coordinates r� �� and the little shaded region is approximately ofsize �r� r�� Taking limits� the product �r� r�� becomes r dr d��

We therefore obtain�

ZUx� � y� dA � U �

nx� y � R

� � x� � y� � �o

�Z ��

��Z r��

r��r� r dr d�

�Z ��

��

r�

�

��

d�

�This is often cost e�ective� If he can �nd a method which will reduce it so itonly takes ten minutes� and then he tells a thousand other people� that is one weekdown but one thousand times �fty minutes saved� I am not sure how long this is� Iam still looking for a quick way of �guring it out�

� CHAPTER �� INTEGRATING FUNCTIONS

θ

Δ

r

r

Δθ

Δ

θ

Figure �� A � �r r��

�Z ��

��

� d�

��

�

Z ��

��d�

��

��

��

�

Exercise �� For each of the functions below� for the disk or partof disk speci�ed� obtain a crude estimate of the integral of the functionover the region by sketching the graph and using some common sense�Then compute the integral using the polar coordinate version of Fubini�sTheorem�

��RUx� y� dA� U � fx� y � x� � y� � �� x � �g

��RU xy dA� U � fx� y � x� � y� � �g

�RU x

� � y� dA� U � fx� y � x� �� y � �� g

�RU e

� �x��y�� dA� U � R

�

�� OVER WILDER REGIONS BUT NOT TOO WILD ��

�� By considering the last exercise� the symmetry of the function andthe fact that ex

��y� � ex�ey

�� evaluate the integral

ZR

e�x�

�� dx

where �� is any strictly positive number� Explain in a short sen�tence why anybody should care about this integral�

�� Over wilder regions� but not too wild

�� General Ideas and Examples

It is reasonably good news that we can integrate functions of two vari�ables over rectangles and disks� Before we go out and celebrate however�it is worth pointing out that there are other shapes� The de�nition ofthe integral for functions f � R � �� R makes sense for more than justrectangles and disks� Take� for example� the triangle formed by thex�axis� the y�axis and the line x � y � �� as shown in �gure �� Ifthe function fx� y � x� � y� were to be integrated over this region�and call it U again� it is easy to make crude estimates� it is clearlyless than �� for instance�� Equally clearly�� the value of the integralis more than �� Chopping the base triangle� U into little trianglesinstead of squares would seem to be a reasonable thing to do� and that�sextremely easy� So there really is an integral� and there has to be asmarter way to evaluate it than taking zillions of little squares�

If we divide the triangle U into thin strips parallel to the x�axis� of width�y� and if we integrate f over a line segment at position y between �and ��

Rfx� y dx� we note that what has changed is that the limits

of the integral depend on where you are for your y value� whereas this

�The word �clearly� is intended to make you feel stupid until you have �guredout why this claim is true�

�Ha�


X

Y

y yΔ

Figure �� Integrating over a triangular region

didn�t change in the case of rectangles� Or to write it out more cleanly�ZUx� � y� dA� U � fx� y � x � �� y � �� x� y � �g

�Z y��

y��

�Z x��y

x��x� � y� dx

�dy

�Z �

�

x�

�� y�x

�x��y

x��

dy

�Z �

�

�� y�

�� y�� y dy

��

�

Z �

�� y � y� � �y� dy

��

�

�y � �

�y� � �y� � y�

��

��

�

��

��

�

��

This example sets the general principle although a certain amount ofcommon sense is required to actually evaluate some integrals over nastyshapes� If necessary� we can chop the nasty shaped regions in R

� into a


X

Y

Figure �� Integrating over a slightly wilder region

set of smaller regions� each of which is relatively nice� and simply addup the answers� It is obvious from the de�nition of the integral thatthis will give the right answer�

Example ��

Problem De�ne U to be the region between x � �� x � �� the x�axisand the graph of y � x�� First shade the area so we can get an idea ofwhat it looks like� Now integrate the function fx� y � x��y� over theregion U �

Solution The region is shown shaded in �gure �� and you have toimagine a parabolic bowl coming out of the page at you� The smallestvalue of x� � y� on the region is � at �� and the largest is �� at��

The area of the region is � � � The integral

I �ZUx� � y� dA

clearly satis�es� � I � ��


and a thoughtful guess and some arithmetic suggests we can say

�� I � ��

Now doing it properly� we have to work out the right limits so we starto� by supposing we will be taking the area of horizontal slices at �xedy by integrating x �rst� If y � �� then the left hand side of such a slicewill be �� and the right hand side will be �� If y � �� the right handside will still be � but the left hand side will be

py� We must divide

this into two bits� to get

I �ZUx� � y� dA� U �

nx� y � � � x � �� y � x�

o

�Z y��

y��

�Z x��

x��x� � y� dx

�dy �

Z y��

y��

�Z x��

x�pyx� � y� dx

�dy

�Z �

�

x�

�� y�x

�x��

x��

dy �Z �

�

x�

�� y�x

�x��

x�py

dy

�Z �

�

�� y��

�

�� y�

�dy �

Z �

�

� � �y��

y��

�� y��

�dy

�

� y

��

�y�

�

��

�

�y � y� �

�y��

��

�y��

�

��

��

��

��

��

��

�

��

��

Just to do the thing to death� I shall also do it by reversing the order ofintegration� This gives us

I �Z x��

x��

�Z y�x�

y��x� � y� dy

�dx

�Z �

�

x�y �

y�

�

�y�x�y��

dx


�Z �

�x� �

x

� dx

�

x�

��x�

��

��

� ��

��

��

��

�

��

�

��

��

��

This shows you a more or less typical case� the others are all the samewith a few changes to the limits and the functions� Practice with theexercises below� Notice that this integral was slightly easier the secondway than the �rst� If you get lots of practice at doing these� you canchoose the easier way by instinct in a lot of cases� To give a nice� easyrule for doing all cases is a bit complicated� but approach it as follows�

�� Algorithm

To evaluateRU fx� y dA for a region U bounded by curves given as

graphs of functions�

�� Take out any symmetries of U and f � that is� if the functionyou have to integrate is symmetric and so is U � you only need tointegrate over part of U �

�� Write down the integral as

Z y��

y��

�Z x��

x�fx� y dx

�dy

�� Draw the region U in the plane and draw a horizontal line at someconstant value of y which goes from left to right across U �


�� The leftmost value of x at that value of y is given by some functionof y� replace � by the appropriate function of y so as to get thecorrect left hand end of the line segment cutting across U �

�� Similarly� the rightmost end of the line segment cuts U in somevalue of x which depends on the height y� write down the generalexpression for where the right hand end of the line segment acrossU meets the rightmost boundary of U as a function of y and put equal to this function of y�

� This works provided the horizontal line at height y cuts theboundary of U in two places only� if this isn�t true� either in�terchange x and y and try again� or chop up U into two bitsby a suitably chosen vertical line and try the above for each bitseparately�

�� Now put equal to the lowest possible value which y has any�where in U � and � equal to the highest value� and you are readyto integrate� � � have to be numbers since the answer to thewhole thing is a number�

�� It may well be that the functions �� are de�ned piecewise� inthis case� split up the integral into the di�erent pieces�

�� Try writing out the integral with x� y reversed to see if it looksany easier�

�� Try a polar coordinate form if it still looks bad�

�� If all else fails� chop up U into little bits each of which looks easy�and do them separately�

�� Exercises

Exercise �� For each of the following descriptions of a region Uin R

�� sketch the region� then write down a double integral which wouldgive the area of the region if evaluated� Reverse the order of integration


Figure �� Problem Solution

and do it again� Finally� select the easier looking of the integrals andevaluate it to obtain the area of U �

�� U � fx� y � R� � �� x � �� x� � � � y � �� x�g

�� U � fx� y � R� � x� � � � y � x�g

� U � fx� y � R� � �� x � �� x� � x � y � x� � x� �g

� U � fx� y � R� � x� � y � cosxg

�� U � fx� y � R� � jxj � �� jyj � �� x� � y� � �g

�� U � fx� y � R� � jxj � �� jyj � �� x�y� � �g

To give you a start� I do the �rst one for you�

Example �� U � fx� y � R� � �� x � �� x�� y � ��x�g

The sketch of the region U is given in �gure ��

Area �Z y��

y��

�Z x��

x�� dx

�dy


Then

� ��

for � � � y � �� q� � y� � �

q� � y�

for � � y � �� q�� y� � �

q�� y�

Area �Z �

��

Z �p��y

�p��y� dx dy �

Z �

�

Z �p��y

�p��y� dx dy

This does it the �rst way�

To do it with x and y reversed� we do vertical sections

Area �Z ��

��

Z ��x�

x�� dy dx

�Z �

�� x�� x� � � dx

� �

x� x�

�

��

��

�

This does it the second way and gives the answer� which we could havedone with our eyes closed trying to get to sleep o� nights�

�� Applications and Subversive Thoughts

The above material has supposed that you want� for some mysteriousreason� to integrate some function de�ned on a region in R

�� You nowhave a very fair idea of how to do it for nice easy functions and forreasonably nice� easy regions� Most of the functions and regions youmay have to deal with in adult life are less nice� less easy� and not


particularly reasonable� but you can usually approximate them withthe functions and regions you know about�

The question arises� who cares� If you are a good little student of thecivil sort� you just do it in order to conform with the expectations ofproperly constituted authority� to pass your examinations and to collectthe maximum number of brownie points� This is good preparation foradult life in Australia� Or maybe you do it because you are youngenough to want to test out your brain against every problem that comesalong� to develop intellectual skills� This is somewhat uncommon� Themost reasonable assumption is that you only want to do it so as to besure of passing the examination� after which you will forget it�

It would� however� be nice to believe that problems do arise in the realworld which require us to be able to use these hard won skills�

Even if there aren�t� we can make up some which possibly might arise insome people�s lives� So as to prepare you for a cheap simulation of reallife� here are a few exercises where the problem is to turn the string ofwords into proper mathematics� then solve the mathematical problem�Amuse yourselves by pretending that you care about the matters raisedin the next set of exercises�

Exercise ��

�� A bath has a horizontal section consisting of a rectangle with semi�circular ends� as in �gure ��

The radius of the semicircles is � unit� and the length of the bathaltogether is units� The bath has vertical sides� Water is pouredinto the bath� and then treacle is added and stirred into the water�The density of the treacle is given by the formula � � �e�h� whereh is the height measured from the bottom of the bath� The mixturehas depth of � unit� How much treacle is in the bath�

�We can only hope that your lives will be more interesting than most of thosewhich give rise to the examples�


2 units

4 units

Figure �� A bath�

�� The same experiment as above is conducted with yet more treacle�and with a bath having the same base as the other one� but withsides that slope up so that at a height of h units� the linear di�mensions have been multiplied by h� �� The second bath is �lledto a depth of � units with the same treacle�water solution� Howmuch treacle is in the new bath�

� To hell with treacle� The atmosphere on the planet Earth �� hasdensity at sea level of exactly ��gms�ml� It falls o� exponen�tially with altitude so that at a height of ten kilometres above sealevel� the density has halved� A sleeping grendler in a raft at seais approximately elliptical in shape with a major axis of � me�tres� and a minor axis of � metre� What is the weight of the airpressing down on the grendler�

� A mass of some unlikely substance in R� has density given by xy

and is in the shape of a triangle with vertices at �� and�� Find the centre of gravity of the object�

�� Darth Vader and Luke Skywalker are �ghting with laser sabres�Each sabre emits a laser beam of circular section having radiusone unit� length proportional to the wielders virility and frequencyproportional to the user�s moral stature� The beams momentarily

�� CHANGE OF VARIABLES ��

b

a

Figure �� An Ellipse�

intersect along the long axes when these axes are at right angles�What is the volume of the region inside both light sabre beams�

�� Obtain a formula� preferably the right one� for the volume of aright circular cone of base radius r and height h�

�� Change of Variables

The ellipse in �gure �� has equation x�

a�� y�

b�� The lengths of

the semi�axes are easily seen to be a and b respectively� I claim thatthe region inside the ellipse has area �ab� This is very easy to see asfollows�

�� The unit circle encloses area �� well known fact� try proving it�

�� If you stretch R� by the linear map g � R

� �� R� which takes

x�y to ax� by� then it takes the unit square to the rectanglewith vertices �� a� �� b� a� b with area ab� It also takesthe unit circle to the ellipse and the origin to the origin�


�� Note that the matrix for the map g is just a �� b

�

�� The area stretching factor of this map for any square is ab� thedeterminant of the above matrix� i�e� if you take a square of sideu�v� it gets taken to a rectangle of side au�bv which is ab timesthe area of the original square�

�� If you cover the unit disk with little squares� each of area �A�the sum of the areas of the squares gets closer to � as the squaresget smaller�

� The stretched squares have areas ab�A� and the sum of thoseapproximates the area enclosed by the ellipse� equalling it in thelimit�

�� My Case Rests�

The same argument obviously applies to any shaped region U in theplane� if g � R

� �� R� is any linear map� then we need to calculate

the absolute value of the determinant of the linear map� and this is the�area stretching factor� of g� This is the whole point of the determinantof a square matrix� it tells you the area� volume� whatever of the imageof the unit square� cube� or n�dimensional thingy� multiplied by �� ifthere has been a re�ection done by the map� Since all little squares getstretched by the same amount� jdetgj� any region U with an area Uwill have the region turned into a new region� gU� with area gU �jdetgj U� I suppose this ought to be called a theorem� but it�stoo late� the argument I gave you is the proof� The n�dimensionalgeneralisation works the way you would expect it to�

Exercise ��

Write down a � � � matrix� calculate the image of the vertices of theunit square �� Draw the result on a sheet of


graph paper� and by counting little squares� estimate the area of theparallelogram� Now work out the orientation of the map by looking tosee if the direction of the smallest angle from the image of �� to theimage of �� is clockwise or anti�clockwise� If clockwise� multiply thearea by �� Now calculate the determinant of the matrix� Is this amiracle or what�

The same thing works if g is an a�ne function� that is to say if it islinear with a shift of the origin� We can forget about the shift� whichhas no e�ect on the area� and still just take the determinant of thelinear part of the function�

We can push this even further� suppose the function g � R � �� R� is

not a�ne but is di�erentiable at every point� and is one�one� I shallwrite this as the function or map

g � R � �� R�

xy

��

g�x� yg�x� y

�

Now at the point x� y in R�� the derivative is given by the matrix� g�

xg�y

g�x

g�y

�

This matrix of partial derivatives is the standard representation of thederivative� and at each point x� y� it is a de�nite linear map� In factyou should think of the matrix as a variable linear map which dependson where you are� Just as the derivative for functions f � R �� R

gives you a linear map with a straight line as graph� the tangent tothe graph of f� so the matrix above tells you which linear map �ts fbest near a point x� y of the domain of f ��

Strictly speaking� I ought to talk about the linear part of the best �tting a�ne

function because the tangent map should go through the point where it is a best �t�and we ignore this� moving the point of contact back to the origin� We can alwayswork out where to put it separately� and so we only bother about the linear part�


This matrix then� with numbers obtained from putting in values for xand y� is precisely the linear part of the a�ne map best approximatingg in the vicinity of x� y� Consequently� although the amount of areastretch is no longer constant but varies from place to place in U � at eachpoint we can say how much area stretch is being applied by g� it is theabsolute value of the determinant of the matrix of partial derivativesof g� This determinant is otherwise known as the Jacobian of the map�

If we integrate the mount of area stretch over the domain� we get thearea of the image� as for the linear case� It is worth stating this as atheorem�

Theorem �� If g � R � �� R� is a di�erentiable function which is

one�one on a region U � and if we use Jgx� y to denote the jacobianof g at the point x� y � U � and if X is the area of a subset X R

��then

gU �ZUjJgx� yj dA

I said we wouldn�t be proving any theorems� but the argument I havejust given you essentially is the proof� I hope you have followed all thiscarefully� because it is kinda cool when it falls into place� If not� readit again� very� very slowly until you get an orgasm� Better yet� do thenext exercise�

Exercise ��

We explore the map f � R � �� R� given by fx� y � x�� y�� in the

neighbourhood of the point �� Take two sheets of graph paper� Callthe �rst sheet �domain� and the second �range�� Draw the axes and thepoint �� on both sheets� since f takes �� to ��

Now draw a baby square with its bottom left�hand corner at �� ofside say �

�� as in the diagram �gure �� on the domain sheet� Shade

in the square and label it �baby�� We want to �rst work out the imageof baby�


Domain

baby

f(baby)

Range

Figure �� Some square stretching�

Do this by looking to see what happens to the components when werestrict x to be � and � � y � �

�� and likewise for the other four sides of

the square� It is easy to see that we get a square back� label it f�baby��

Count the number of little squares of the graph paper inside f�baby��Now divide by the number of little squares inside baby� This is ourestimate for the amount of area stretch which f does at �� If youwant a better estimate� do it with a smaller baby�

Now compute the derivative matrix of f at the point �� Now writedown its determinant� You get a number which is not too far di�er�ent from the estimate obtained from baby� You should be able to con�vince yourself that repeating with smaller and smaller babies will getyou closer and closer to the value you got from taking the determinantof the derivative�

We can use this idea to calculate areas of nasty shapes� if we can workout the area of a nice shape and �nd a di�erentiable function whichtakes the nice shape to the nasty one�

Example �� In the case of g being the linear map which stretchesthe unit circle into the ellipse� we have g�x� y � ax� g�x� y � by� so


Figure �� A deformed square�

the derivative has matrix

g�x

g�y

g�x

g�y

��

a �� b

�

and its determinant is ab� and a � �� b � �� so gU � area of ellipse�

RU ab dA � ab

RU dA � ab U � �ab� as advertised�

Example �� De�ne g � R � �� R� by

g

xy

��

uv

��

x�� y�

y�� x�

�

This takes the square fx� y � �� x � �� y � �g into theregion shown in �gure ��

There are two ways to compute the area enclosed by the deformedsquare� the simplest is to work out the size of the bites taken out� Themore exotic one is to integrate the jacobian of g over the original re�gion of size � units� As is evident to the meanest intellect�� the matrix

� clearly�


representing the derivative is

ux

uy

vx

vy

��

��y�

�xy

xy ��x�

�

�

The determinant of this is �� x�� y�� x�y� and integrating

this over the interior of the square we started from� we obtain

�

�

Z �

��

Z �

�� x� � y� � �x�y� dx dy

��

�

Z �

��

x�

x�

�� y�x� x�y�

�x��

x��dy

��

�

Z �

��

��

�

�� y� � y��

�� y� � y�

�dy

��

�

Z �

��

�

�dy

��

��y��

��

�

It is easy to see that the size of the bite taken out of each face of thesquare is �

� � there are four bites� and the square had area � before it wasmangled� or bitten� whichever way you want to see it� So both methodsgive the same result�

Example �� Take the map P � R � �� R� given by

P

r�

��

xy

��

r cos�r sin�

�

The reason for calling it P is obvious P for Polar Coordinate trans�form� Now look at what it does to the rectangle � � � � �� r � ��

The picture for this map is �gure ��


r

θ

X

Y

Figure �� A deformed rectangle�

The problem is to compute the area of the unit disk� The derivative isgiven by

xr

x�

yr

y�

��

cos� �r sin�sin� r cos�

�

and the determinant of this is r� from which we deduce that the areaenclosed by the transformed rectangle is given by

Z ��

�

Z r��

r��r dr d�

which works out as �� which is what we claimed earlier to be the areaof the unit disk�

Note that the �stretching factor� by which a little region �r �� getsstretched is just r� This was illustrated by �gure �� The geometricpicture in this case agrees with the algebra nicely� �

This is good clean fun� keeps us o� the streets� and enables us to cal�culate areas of some bizarre regions of R � with little fuss� which we cantake to be a Good Thing� Now the calculation of an area of a region can

�It had damn well better�


be thought of as calculating the integral of the characteristic functionor indicator function of the region� the function which takes the value� for points inside the region and zero for those outside� Seen from thispoint of view� it is not too surprising that we can carry the above pro�cess for calculating areas of regions to a process for calculating integralsover regions� If we want to evaluateZ

Ux� � y� dA

where U is the unit disk� we take the map P of the last example� andfollow it by fx� y � x� � y� to get the function f � P r� � � r�

which we integrate over the rectangle � � � � �� r � �� We mustnot however neglect the area stretch� which was r dr d�� So we get

ZUx� � y� dA �

Z ��

�

Z r��

r��r� r dr d�

which we already knew�

The general rule can be remembered from this particular example�

Theorem �� Change of Variable� If g � R � �� R� is a di�er�

entiable� one�one function on a region U � R�� and if f � R � �� R is

an integrable function de�ned over gU� then

Zg�U�

f �ZUf � g jJgj

I have left out some dAs in the above statement� you can put themback in if it makes you feel better� but in this form I can change the �in R

� to any positive integer n and it stays true��

Some books do the general theorem �rst and then do the polar trans�form as an example� I felt it would be useful for you to have some handson experience �rst and then you could see the point of the theorem�

��And in any case� we are extremely hazy about what the dA means anyway�right�


Some exercises will develop your mental muscles and burn the applica�tion of the theorem and hence what it means into your brains�

Exercise ��

�� The region V � fx� y � R� � �� x � �� sinx � � � y �

sinx � �g Find the area of V by the usual means� Now �nd afunction g � R � �� R

� sending u� v to x� y so that g takes theregion U � fu� v � R

� � �� u � �� v � �g to the regionV � i�e� V � gU� Recalculate the area using the last theorem�Which method is easier�

�� The region V R� is bounded by the curves x � t cost� y �

t sint for � � t � �� x � �t cost� y � �t sint� and thepart of the x�axis between �� and �� Sketch the region and� bychoosing a suitable map g � U �� V � where U is the rectangle� � s � �� t � �� nd its area�

� Let U � �� and gu� v � u�� v�� The image gU isjust the region �� Integrating jJgu� vj � �juvj fromu � �� to u � �� and likewise for v gives an answer di�erentfrom the area of the unit square� What went wrong�

� Take U to be the region enclosed by the unit square� �� and invent a map g � R � �� R

� which takes U to some interestingshape� Find the area of the shape� Integrate the function fx� y �x� � y� over the shape� Check your answer by estimating thevolume over the shape and under the surface by chopping the shapeup into rectangles�

�� Three Variables

What we have done so far is all about integrals over regions in R�� It is

not hard to extend the ideas to R�� The idea of an integral over a region

�� THREE VARIABLES ��

in R� should not cause us much concern� Suppose I take a plastic bag�blow into it and close the end so as to de�ne a three dimensional regionof the room you are sitting in� If I then ask you to integrate a functionof the three coordinates of the room over the inside of the bag� it iseasy to see that the idea of chopping the interior of the bag up intolittle boxes� �nding the value of the function at some point in each box�multiplying the value of the function by the volume of the little box�and then adding up all the numbers� is conceptually straightforwardenough� although possibly time consuming� So the idea of the integralmakes sense� at least over sensible regions such as might live insideplastic bags� and for reasonable well�behaved functions� which are theonly sort you know about�

The warm� comforting glow that you get from re�ecting that if all elsefails you can go back to little boxes may not survive doing many sums�but given that it could be done with litle boxes� doing it more intelli�gently by a three dimensional version of Fubini�s Theorem is the nextstep� and a pretty obvious one to take� Likewise� taking weird regionsin R

� and �nding ways of mapping nice simple regions of R� to them

by di�erentiable maps� then calculating integrals over the weird regionsby the change of variables theorem can be done in three dimensions aseasily as in two� Actually� they can also be done in twenty seven dimen�sions� although you will be pleased to hear that that isn�t examinable�

Rather than bore you with more explanation� I shall give you someexamples so you can get the hang of the applications� then you canbash out the following exercises so as to burn the principles into yourbrains� The examples introduce you to two kinds of polar coordinatesystem� cylindrical and spherical� both having their uses in R

��

Example �� Problem

Compute the volume enclosed by the unit sphere in R�� Hence compute

the volume of a ball of radius R�

Solution �


X

Y

Z

r

φ

θ

Figure �� Spherical Polars

The area of a horizontal section of height z is �� z�� so the volumeof the ball is

��Z z��

z�� z� dz �

��

�

If the unit ball is mapped to the ball of radius R by the linear mapwith matrix RI� where I is the three by three identity matrix� then theJacobian is R� and so the volume enclosed by the sphere of radius R is��R

��

Solution �

The diagram �gure �� shows the transformation �map� from R� �

��

�� where R

� denotes the non�negative real numbers� toR�� which takes the cuboid to the unit ball� We have� as the diagram

clearly shows

SP � R � � ��

�� R

�

�� r��

��

�� x � r cos � cos �y � r cos� sin �z � r sin�

��


The Jacobian of this� often written��x� y� z�r� ��

��is easily veri�ed to be��

cos� cos � �r cos � sin � �r sin� cos �cos� sin � r cos � cos � �r sin� sin �

sin� � r cos �

��which after evaluating the determinant comes to just

r� cos �

Thus the volume enclosed by the unit sphere in R� is

Z ��

��

Z ��

��

Z r��

r��r� cos� dr d� d�

��

�

Z ��

��

Z ��

�cos� d� d�

��

�

Z ��

�� cos� d�

��

��

The same argument as last time gives the general formula ��R

� for thevolume enclosed by a sphere of radius R�

Solution �

The Cylindrical Polar coordinate system is indicated in �gure �� andis speci�ed by the map from R

� � �� R given by

CP � R � � �� R �� R��

�� r�z

��

�� x � r cos �y � r sin �z � z

��


X

Y

Z

r

θ

z

Figure �� Cylindrical Polars

The Jacobian of this� written ��x� y� z�r� �� z

��is easily veri�ed to be r� The volume enclosed by the sphere is easilyseen to be

Z z��

z��

Z ��

��

Z r�p��z�

r��r dr d� dz

��

�

Z �

��

Z ��

�� z� d� dz

� �Z �

�� z� dz

��

��

Three out of three sounds convincing�

Check these out to see if you can follow the argument and that you �ndit plausible� Note that the number of basic ideas here is fairly small�Burn them into your brain by doing lots of exercises�


�� Swindle

You may or may not have noticed that I have been getting away with asmall swindle� It is to do with the boundaries of the regions in R

� andR� which have been used for various transforms� particularly the polar

maps�

When I write �� you note that the left hand end of theinterval of points between � and �� is in the set� but the right handend is not� This interval is going to get wrapped around the unit circle�and I keep explaining that maps have to be one�one� Which it won�tbe if � and �� get sent to the same point�

But if you look at the polar map� you see that when r � � we have that�� x is sent to the same point for any x� which appears to screw thingsup somewhat� If the map is not one�one� then the theorem on change ofvariables seems to go bung� Yet I have cheerfully used it several timesin the last example� shouldn�t I feel bad about this�

The answer is yes� I do feel bad� but only a little bit�

I can almost �x things up by removing the � end of the interval� toget �� except that this now leaves the point at the origin inR� uncovered� If I write the set �� f�� g� that does it�

although now I have to specify the maps a little more carefully�

All this is fairly fussy� because the area of lines is zero anyway� exceptin some pathological cases where it is almost impossible to write downthe functions� Similarly the volume of a surface such as a sphere iszero�

Note that I must be careful to distinguish between the volume of asphere and the volume enclosed by a sphere in R

�� This is being ratherfussy� it is normal to talk about the area of a parallelogram� so theparallelogram must be the set of points inside the parallel lines� If Italk about the area of a square� it is slightly more of a problem to saywhether when I say �square� I mean the region inside or the bounding


lines� When I say circle� again� I might mean either� English languageis scru�y here� because it seldommatters muchwhich I mean� and whenit does� I can usually guess from context whether you mean the circle orthe disk bounded by the circle� In order to make our lives less troubled�I de�ne the unit ball in R

n to be the set fx � Rn � kxk � �g� and write

Bn for it� The boundary is the set fx � Rn � kxk � �g and is called

the unit sphere in Rn� written Sn�� In the case where the dimension

n is two� I shall use the term �disk� instead of �ball�� and the ��sphere�S� is usually called a circle� Then the volume of the unit sphere in R

�

is zero� but the volume of the unit ball is �� So when I say �circle� I

mean S� or something which is S� scaled and shifted� And it has area�� This sounds a bit silly� but saves you from saying what you don�treally mean�

Later� it will be important to be able to have a name for the boundaryof a set in R

n such as a ball� I shall use �U to refer to the boundary ofthe set U � In this notation� �Bn � Sn��

From now on� I shall slither over di�culties caused by sets of measurezero such as curves in R

� or curves and surfaces in R�� Measure� is

the generic word used to mean length for one dimensional sets� area fortwo dimensional sets� volume for three dimensional sets� and we just ranout of names� Intuitively this is reasonable� and it won�t get you intotrouble so long as you stick to low dimensions and safe� well�behaved�respectable functions and sets��

�� More Exercises

Exercise ��

�� Compute the Jacobians for the spherical and cylindrical transfor�

��My mathematical conscience gives me severe twinges about this disrespect forformal analysis� and the fact that this course doesn�t actually contain much in theway of de�nitions or theorems gives me a real pain in cold weather� Still� eighteenthcentury it was gonna be� and eighteenth �and early nineteenth it will be�


mations�

�� The mass of a line segment �a� b�� where the density along the seg�ment is given by a function �x� is de�ned to be M �

R ba �x dx�

The centroid or centre of mass or centre of gravity �x of a linesegment �a� b�� where the density along the segment is given by

the same function �x� is de�ned to be

R b

ax ��x� dx

M� The cen�

troid for a two or three dimensional object U with mass density�x� y� �x� y� z respectively is a two or three dimensional vec�

tor

�x�y

��

�� x

�y�z

�� respectively� where �y � �

M

RU y�x� y dA� or

�y � �M

RU y�x� y� z dV � or �z � �

M

RU z�x� y� z dV respectively�

where M �RU � in every case�

Find the centroid of the mass enclosed by a right regular cone withheight h and base radius r on the assumption of uniform density�Find the centroid of the conical surface bounding this solid�

� The moment of inertia of a solid object U about an axis L� whenthe solid has �mass� density given by �x� y� z� is de�ned to be thevalue of

RpLx� y� z��x� y� z dV � where pLx� y� z is the per�

pendicular distance of x� y� z from the axis L� Obvious variantsexist for one and two dimensional objects�

Calculate moments of inertia of

�a� a uniform density bar along the X axis of length � unit andnegligible thickness� about the Z axis�

�b� a bar of the same size but where the density at x� �� isgiven by x�� Where is the centroid of the bar�

�c� The solid cone of the last question about an axis through thevertex and orthogonal to the axis of the cone�

�d� a solid ball of radius � and centre �� having uniformdensity� about the Z axis�

�e� A solid disk of uniform density and radius r about an axisthrough its centre and orthogonal to its plane�


Rr

Figure �� A Torus

�f� a circle of radius r of uniform density about an axis throughits centre and orthogonal to the plane of the circle�

�g� A �Wedding ring� or solid torus� of uniform density and ma�jor radius R� minor radius r� as in �� about the axisthrough the centre of the major circle and orthogonal to itsplane�

� Make up a shape of your own in R� and assign it some density

function� then calculate its centroid and moments of inertia aboutsome choice of axes�

�� Higher Things

�� Embeddings

Casting your mind back to the various transformations which stretchedarea and volume� recall that I told you� among other things� how tocalculate areas of weird shapes� you toook a non�weird shape such asa rectangle� you deformed it until it �tted on top of the weird shape

�� HIGHER THINGS ��

exactly� and then you integrated the �amount of stretch� that the defor�mation transformation did over the rectangle or whatever� Of course�the weird and non�weird shapes were all sitting in R

�� There is a threedimensional version� but there the nice shapes and the weird shapeswere all sitting in R

�� In all cases� the nice shapes and the weird�stretched shapes� were all sitting in the same space�

But they don�t have to be� You can see that taking a rectangle� deform�ing it and putting the result in R

�� providing you don�t have the shapeintersecting itself� will not be too di�erent� If we take the cylinder in R

�

consisting of the set fx� y� z � R� � x� � y� � �� z � �g� then we

could take the rectangle fu� v � R� � � � u � �� v � �g and map

it by g to the cylinder by sending u� v to x � cosu� y � sinu� z � vin a nice one�one and di�erentiable manner� Now we ought to be ableto calculate the area of the cylinder by the above method� calculatethe amount of �area�stretch� that g does at u� v and integrate over therectangle� This must involve looking at the matrix of partial derivativesof g� and looking to see how much area�stretch is accomplished by thislinear map�

If g � R � �� R� is a linear map� it has a matrix representation� relative

to the standard basis� which looks like

��a db ec f

��

for some numbers a � � � f � The map g takes the basis vector

��

�to

the �rst column� and the basis vector

��

�to the second column of the

matrix� The unit square gets taken to the parallelogram with vertices�

��

��

�� abc

��

�� def

��

�� a� db� ec� f

��

and the area stretch of this linear map is the area of the parallelogram�


(a,b,c)

(d,e,f)

Figure �� A stretched and twisted square in R�

There is no problem about the parallelogram having an area� the factthat it is in R

� doesn�t make a di�erence� See �gure ��

The area of a parallelogram is given by the length of one side multipliedby the length of the other side multiplied by the absolute value of thesine of the angle between them� This makes perfect sense� incidentally�if we take a parallelogram in R

n for any n � �� The angle between themcan be worked out using the dot product� at least up to sign� which inany case we don�t care about�

In R�� if we take the cross product of the two column vectors� we get a

third vector� the length of which is precisely the required area� So inthe case of the above map g� the area stretch is simply

�� abc

��

�� def

��

If g is not linear but is di�erentiable� then we get the area stretch of g


at the point u� v � R� by taking the value of��

��

g�ug�ug�u

��

��

g�vg�vg�v

��

Then this function of u and v� can be integrated over the domain U ofg to give the area of the surface gU�

Example � �� Let U be the rectangle �� in R�� as above�

and let gu� v be given by giu� v � cosu� g�u� v � sinu� g�u� v �v� Then gU is the cylinder of radius � and height �� It�s area is

ZU

��

g�ug�ug�u

��

��

g�vg�vg�v

�� dA �

Z v��

v��

Z u��

u��

�� sinu

cosu�

��

��

�� du dv

This is

Z v��

v��

Z u��

u��

��

cosusinu

�

�� du dv �

Z v��

v��

Z u��

u�� du dv � ��

The area stretch is just �� which feels right to anyone who has bent asheet of wrapping paper around a bottle�

The next exercise is for those of you feeeling particularly hairy chested�

Exercise � ��

De�ne V R� by

V �

��

��xyzw

�� R

� � x� � y� � �� z� � w� � �

��


Write V as the image by a one�one di�erentiable map g � R � �� R� of

the square �� Verify that the map is one�one and onto�Hence compute the area of V � Can you decribe the object V � whichconsists of a two dimensional surface sitting in four dimensions�

When we take some curve� surface or more general set U � and mapit continuously by a one�one map into R

n� we say that we are embed�ding the object by the map� Usually the map will be di�erentiable aswell� We also need the inverse map from the image back to U to becontinuous� This means that the map of the last exercise is not anembedding� the inverse map would rip the object apart at the pointswhich are images of the boundary points of the square� Everywhereelse the map is an embedding�

It is worth pointing out that the linear map putting the unit squareinto R

� has image a parallelogram except when the matrix de�ning themap is degenerate� when it collapses to a line segment or a point� andthat there is a normal vector in R

� which is orthogonal to the plane ofthe parallelogram� The normal vector is not entirely unique� since itsnegative is also a normal vector pointing in the opposite direction�

When the embedding of the unit square into R� is di�erentiable and not

just linear� the unit normal will point in di�erent directions dependingon where you stand on the embedded surface� It is intuitively clearthat if you were to walk over the surface� at each point there wouldbe an �up� normal to the surface at that point� and a �down� normalopposite to the �up� normal� If gs� t is the point at which you stand�g being the embedding� imagine keeping the t constant and runningthrough nearby values of s� you would get a curve through the pointat which you stand� If you take the partial derivative with respect tos� you would get� at the point s� t� a vector tangent to the curve� andhence tangent to the surface� We often write this as g

s� and it is short

for the tangent vector ��

g�sg�sg�s

��


Similarly� if we kept s constant and changed t in the neighbourhoodof s� t� you get another curve� also passing through s� t and hencelying in the surface� with a tangent vector g

t� which is also tangent to

the surface� Provided the two vectors are linearly independent� i�e� therank of the derivative is two� then we have two independent vectors inR� tangent to the surface at the point s� t� The cross product� g

s� g

t

is normal to the surface at that point� And we have already seen thatthe area stretch of the map is the norm of this vector�

Exercise � ��

By choosing a simple parametrisation of at least part of the unit sphere�calculate the normal vector at the north pole of the unit sphere in R

��Do you get the answer you expect�

Example � ��

Problem

Compute the area of the unit sphere S��

Solution �

Let D� be the unit disk� We shall lift up the disk at each point until itbecomes the northern hemisphere of S�� If we call a point of the diskr� � then it has to be lifted up to a height of

p�� r� in order to be on

the sphere� We therefore get the map from the rectangle

��

into R� which �rst goes into polars and then �xes up the height

gr� � �

�B�

x � r cos �y � r sin �

z �p� � r�

�CA

To get the area of the northern hemisphere� we need only to integratethe � area stretching factor� over the rectangle ��


The �area stretching factor� is the norm of the vector

�g

�r� �g

��

which is ��B�

cos �sin ��rp��r�

�CA �

�B��r sin �r cos �

�

�CA��

that is ��

�BB�

�r� cos �p��r�

�r� sin �p��r�r

�CCA��

which after a small amount of algebra becomes

rp�� r�

This� it is worth noting di�ers from the simple r stretching factor whenwe go to a �at disk�

The area of the northern hemisphere of S� is therefore

Z ��

��

Z r��

r��

rp�� r�

drd�

This is just

��Z �

�

rp� � r�

dr

The integral for r is easily solved by the substitution r � sin�� dr �cos � d� to give

��Z �

�

rp� � r�

dr � ��Z ��

��

sin�

cos �cos� d�

�� HIGHER THINGS �

Figure �� Another way of wrapping a rectangle onto a hemisphere

� ��

It follows that the area of the unit sphere is ��

Solution �

The particular way of plastering a rectangle onto the sphere� or at leasthalf of it� is not the only possibility� Another way of doing it is to take arectangle of height � and length �� and make a cylinder of it which �tsaround the equator of the northern hemisphere� as in �gure �� Thenwe need to take each parallel of latitude on the cylinder and shrink it intowards the sphere until it has been squashed onto the hemisphere� It iseasy to visulaise the map g � �� R

� which does this� andnot much harder to write it down algebraically

g�u� v �

�B� x � cos v

y � sin vz � u

�CA

takes the cylinder into the position shown in the �gure� We have tomultiply the x� y component by a factor which depends on z � u to


bring it onto the surface of the sphere

gu� v �

�B� x �

p� � u� cos v

y �p�� u� sin vz � u

�CA

has the required image in R�� By taking the rectangle

��

we could cover the whole sphere in one go�

We now have a second map which takes a nice� easy rectangle in R�

and plasters it over the sphere� or a half of it� whichever� In order tocompute the area of the sphere� all we have to do is to integrate the�area stretching factor� over the rectangle� As before� this is the normof the vector

�g

�u� �g

�v

which is ��

�BB�

�up��u� cos v�up��u�a sin v

�

�CCA �

�B� �p�� u� sin vp

�� u� cos v�

�CA��

�

��B� �p�� u� cos v

�p�� u� sin v�u

�CA��

Which is just �� This is a bit surprising at �rst glance� it means thatthe squashing of the cylinder by the horizontal shrinking of circles oflatitude around the cylinder into circles of latitude on the sphere doesnot change the area� It also tells us immediately that the area of thesphere is �� the result we got doing it the other way�

�� HIGHER THINGS �

�� Parametrisation

The job of expressing a surface V as the image of some rectangle bya di�erentiable map is called a parametrisation of V � This is the poshway of saying that we take a rectangle and use a map to plaster it overthe surface� the last example gave two di�erent parametrisations of the��sphere�

It is a two dimensional version of parametrising a curve by writing itout as the image of a smooth function from some interval of R � Forexample�

g � �� R�

� �� cos�sin�

�

parametrises the unit circle� It is all about wrapping string aroundcurves� just as parametrising surfaces is all about putting wrappingpaper over them� The string is made of chewing gum� because it hasto be stretched in general� likewise the wrapping paper��

All this stu� about stretching surfaces by maps should not really comeas a surprise to you� because you have been doing the correspondingthing for one dimension for yonks without knowing what you were do�ing� In the last example� the �rst solution required you to integrate

Z �

�

rp�� r�

dr

and I used the substitution r � sin� to do it� The dr then turnedinto cos � d�� which is exactly the stretching factor� only now it is thelength stretching factor� because we are one dimension down� So youhave been doing all the stu� about taking nice shapes to nasty shapesbefore� but last time you did it so as to make the integrals easier� Andyou did it in only one dimension ��

��Yeuk��The question� �Is it better to have a geometric idea of what you are doing or


Exercise � ��

�� Sketch the spiral ramp described by the map

g � �� R�

s� t �� s costs sint

t

��

Show that the map is a di�erentiable embedding and deduce thearea of the ramp�

�� Drop down a dimension to do the next problem �nd the lengthof the curve of the parabola y � x� between �� and ��

� Show that the map

g � U �� R�

xy

��

��

xy

q� � x� � y�

��

with domain the interior of the unit circle �x� � y� � �� in R�� is

di�erentiable� one�one and onto the northern hemisphere of theunit sphere in R

�� Deduce the area of a sphere of radius R�

just to learn the recipes�� deserves a bit of thought� If there are not too manyrecipes� and if you are only going to use these recipes day after day� there isn�t a lotof point in seeing more deeply into why they work� You are just a handle turningrobot� and the hell with all the geometry� If you have a terri�c memory and novisual imagination� then again� it is easier to forget about the underlying geometry�On the other hand� it is horribly dull to just know that this is the right way but

not have any good idea as to why� and anybody with a bit of imagination �nds thatsort of thing intolerable� And if you are going to be given a humungous great list ofrecipes� all derived from the same basic idea� then you can cut down on the amountyou have to memorise by a tremendous amount�Mathematicians are people with bad memories but good imaginations� They do

not make good robots or good slaves for doing lots of handleturning� They like toknow why the machinery works� not to treat it like a black box that does miraclesfor incomprehensible reasons� They do not go in for mindless ritual much�Di�erent people get bored by di�erent things� Suit yourselves� of course�

�� INTEGRATION OVER CURVES AND SURFACES �

� A car drives at uniform speed around the square� starting at ��and going to �� then to �� then to �� and �nally back to�� The temperature at the point x� y is x� � y�� Estimatethe average temperature on the journey by taking a �nite sam�ple of points� Get bounds for your estimate� Find the averagetemperature on the journey by calculus�

�� Let

f � R � �� R�� xyz

�� x� � y� � z�

give the temperature at each point of R �� Let V R� be the cubical

box with vertices a�� a�� a�� where each ai is either � or �� Whatis the average temperature on the surface of the box�

�� Using the same temperature function as in the last exercise� calcu�late the mean temperature on the spiral ramp of the �rst exercise�

Observe that if I give you a curve in R� by describing it as the graph of a

function g � �a� b� �� R � then there is a particularly obvious parametri�sation� write � � �a� b� �� R

� and de�ne �t � t� gt� Note that ifthe function g has y � gx this gives the right answer� and if x � gywe can just take �t � gt� t� This is rather too obvious to spendtime on� so I won�t�

�� Integration over Curves and Surfaces

Take a friendly little function of two variables� f � think of its graph asthe surface of a rather badly cooked cake� and slice the cake along theX axis with a sharp knife� as in �gure �� I have taken out some ofthe cake in the positive quadrant to show the section�

CHAPTER �� INTEGRATING FUNCTIONS

Figure �� Section over the X axis of fx�y

The result is a real valued function of one variable only� y being set tozero� It is quite clear that this function can be integrated in the usualway� and if we think of the part of the X axis over which we cut thecake as being a particular curve� call it �� We could refer to this asZ

fx� y ds

where dsmeasures the distance along the curve and is� in this particularcase� the same as dx�

Now suppose you cut the cake along the line y � x� holding your knifevertical so that the tip travels from �� to �� Again� if the functionf is polite� law�abiding� kind to animals and various other somewhatvague terms connoting approval� we could expect the integral to exist�Chopping things up into little rectangles and adding up areas is stillan intelligible thing to do� Limits of the process ought� in a properlyconstructed universe� to exist� Whether God� or whoever is responsiblefor the Universe� is as nice a guy as Riemann was is open to doubt��but remember� this only has to work for nice easy functions we canwrite down without using too much ink or paper�

��Riemann would never� never have invented the Ichneumon Wasp� Even I� whoam a right bastard� wouldn�t have thought of anything as nasty as that� What thatmakes God isn�t clear� Ine�able� I guess�


Figure �� Section over the unit circle of fx�y

Now suppose I cut the cake by going around the unit circle as in�gure �� All the outside bit of cake is thrown away or mailed toBangladesh�

It is still easy to believe that the integral of the function around thecurve still makes sense at the conceptual level� The side of the cake hasan area� if the Universe is only half way sensible� And going laboriouslyaround the circle in little steps� chopping the cake over the step� takingthe height of fx� y for some point x� y in the step� multiplying bythe step size� adding up the numbers� then doing it all over again withsmaller steps� and again� and again�� and then taking the limit of thisprocess� Yes� you wouldn�t want to do it this way� but you could if youhad lots of paper� a stack of pens and nothing much on television�

Exercise �� If fx� y � x� � y� and � is the unit circle� whatwould you expect

R f ds to be�

Example ��

Problem

If fx� y � x� � y� and � is the unit circle�


Figure �� Graph over the unit circle of x� � y�

�� Get a crude estimate of bounds forR f ds in this case�

�� Parametrise the unit circle and hence obtain a value for the inte�gral�

� Sketch the graph of f over the unit circle�

Solution

At �x��y�� the function f takes the value �� likewise at the points�� At �p

�� p

� it takes the value �

� � Likewise at the

other three midway points� It is clear that when x� � y� � �� we havex� � y� � � � �x�y� and hence �since � � x�y� � �

�for points on the

circle� we have �� x� � y� � � on the unit circle� Hence we have

� �Zx� � y� � ��

If we write x � cos �� y � sin � then fx� y becomes

f � �� sin� � � cos� �


where � � �� R� is the obvious parametrisation of the unit circle�

It is easy to see that this can be simpli�ed to

f � �� sin��

�

and the desired answer is then

Z ��

�� sin��

�d�

which is easily seen to be ��

The graph can be drawn by taking the graph of � � sin�� between �

and �� and wrapping it around the cylinder with base the unit circle�A rather clumsily drawn version is shown in �gure �� Don�t shootthe artist� he�s doing his best�

It is worth going over the above example to see the idea used to solvethe problem� Basically� we took a line and wrapped it around the circle�looked at what was over the circle� then unwrapped it to get an ordinaryintegral of one variable� You can imagine doing this with a real cakeand a bit of wrapping paper� Wrap a paper cylinder around the cakethat remains when we have thrown out every bit of cake outside theunit circle� Make sure that is higher than the cake at every point tostart o� with� then trim it down until it exactly �ts the side of thecake� Now unwrap the paper and measure its area in the usual way�This idea will clearly work for other curves� apart from circles� withonly minor modi�cation�

One thought that crosses the re�ectivemind is� how far does this dependon the function � that traces the curve� in this case a circle� Suppose�for instance� instead of taking the map

� � �� R�

� �� x � cos �y � sin �

�


which undoubtedly covers the curve� we had taken the map

� � �� R�

� �� x � cos ��y � sin ��

�

We see that this takes a shorter line segment and stretches it as it wrapsit around the circle� It winds up going once around the circle as before�but we must take account of the length stretching factor�

What is the length stretching factor for the maps � and �� We havealready looked at this in the two dimensional case of area stretching fora surface in R

�� and now we are doing it for length in R�� and the idea

is similar� We look at the derivative of the map at each point� whichgives a linear problem� and then we look at linear maps from R intoR�� Well� the derivative of a vector valued function is just the vector

of derivatives� And if g � R �� R� is a linear map� it is represented

by the matrix

ab

�for some values of the real numbers a and b� This

is the image of the standard basis vector in R � which is the number ��

The amount of stretch this does is just the length of the vector

ab

��

or alternatively the distance of the point

ab

�from the origin� written��

ab

�� which is simplypa� � b��

The length stretching factor for non�linear or non�a�ne maps will notgenerally be constant but will change according to the value of �� Itwill however be de�ned so long as the map has a derivative� and is justthe norm of the derivative� In old fashioned language� if xt� yt is

the parametrised curve� the functionqdxdt� � dy

dt� is the multiplica�

tive term which needs to go into the formula to describe the lengthstretching factor of the map�

Exercise ��

�� INTEGRATION OVER CURVES AND SURFACES ��

�� Calculate the length stretching factors for the maps � and � de��ned above�

�� Usually the stretching factor isn�t constant� What is the length

stretching factor of the map ��

��

�for � � � � �� Sketch

the curve�

� A curve in R� is given as the graph of a function yx�

c �

� xy

�� y � yx

�

Find an expression for the length of the curve between

�

y�

�

and

�

y�

�

If you think of this as just another change of variables problem� thenyou can see that we need only have the parametrisation one�one andonto� The only other problem is the direction of the curve� since if we goin the other direction we get a negative answer� So part of specifyingthe curve is to say which way around you go along it� Under theseconditions�

If � � �a� b� �� Rn is a smoothly parametrised curve in R

n and iff � R n �� R is a well behaved integrable function� then we de�ne

R f

by�

Zf �

Z b

af � ��k ��k d�

The result will not change if we change the parametrisation to another�as long as both are one�one and onto the same underlying set� and alsoin the same direction� This is because the answer has to depend on thecurve regarded as a set of points and a direction� and not on the rate at


which a little bug traces out the curve� which is what parametrisationamounts to� Reversing the direction will change the sign of the result�exactly as it does for ordinary one dimensional integrals�

Integrating functions around closed curves� where the end of the curveand the beginning of the curve are the same point� happens a lot� Inthis case� there is a special integral sign which is used in old fashionedbooks� It is I

cf d�

and it means nothing very special� just a reminder that when you goaround the curve you are supposed to get back to your starting place�It is probably done to throw terror �� into the hearts of the uninformed�

In order to convince you that this is all easy and geometric and to getyou to remember it all� do the following exercises and make sure youcan read and understand the text book section on this subject and dothe problems�

�� Exercises

Exercise ��

�� The curve coshx � ex�e�x

� may be parametrised in the obviousway by sending x to x� coshx� Find the length of the curvebetween x � � and x � ��

�� Integrate the function f � R � �� R given by fx� y � y sinhxalong the curve of the last exercise�

� Find the length of the curve given by the polar equation r � sin �for � � � � �� Do this by �rst sketching the curve and obtainingbounds for the length� next by writing the composite map which

��Mediaeval wizards wore black robes with stars all over them for much the samereason� as is well known to anybody who is up to date on the Wizard of Id� Thereis a lot of this sort of thing in Mathematics�

�� INTEGRATION OVER CURVES AND SURFACES ��

takes the interval �� into the �� rspace� then follows it by thepolar map P � Compute the net length stretching factor by usingthe chain rule� and integrate this over the interval ��

� An asteroid travels in an elliptical orbit described by the formulax�

a�� y�

b�� where a � �� b � � measured in Astronomical

Units � A�U�� The temperature �in degrees absolute� is givenapproximately by ��

d�� where d is the distance from the sun� which

is at �� What is the average temperature along the orbit�Why might an inhabitant of the asteroid feel that you were over�estimating the average�

�� All the asteroids and planets are collected and hammered into aDyson Sphere� a humungous sphere centred on the sun� and oneA�U� in radius� Using the same assumptions as for the last ques�tion� it is obvious that the temperature on the sphere is a pleasant��oK� and the human beings� together with what few other ter�restrial species haven�t been eaten or turned into handbags� liveon the inside of the sphere� Unfortunately a Dyson Sphere isn�tstable� As a result of putting the school entrance age up to �� andconsequently forgetting how to do integrals� the society responsi�ble for setting up the Sphere loses the technological ability to keepthe sphere centred on the sun� After a while� the sun is actuallyhalf way to the edge of the sphere� What is the average temper�ature on the sphere� Make a crude estimate �rst� then try to doit by integration� Does this incline you to want to be able to dointegrals� �

�� Make up an example of an integral over a curve in R� and evaluate

it�

�� Make up an integral over a surface in R� and evaluate it�

��It should� at the very least� make you wonder what will happen if our societyloses the skills for running a moderately complicated technological industry� andmaybe make you wonder if we aren�t half way there� At least somebody ought toknow how to do integrals� Which is why you are doing this course� right�


�� Doing it Piecewise

Suppose that instead of asking you to study integrals along nice smoothcurves� I had asteroids travelling around in square orbits� This may beunrealistic� but there are serious problems where the path is not smooth�

This is no big deal� you can probably see that we can simply splitthe path into bits� do each bit separately� and then add the answerstogether� After all� the whole idea of integration consists of doingexactly that in the limit� so it can�t do any harm to do some non�smooth curves piecewise�

Exercise ��

A surface is made up of a hemispherical bowl centred on �� andhaving radius �� so that it touches the origin� Another hemisphericalbowl of radius �

�has the same centre but is the other way up� so that it

touches �� The region between the boundaries is the third partof the surface and consists of the set

�x� y� z � R

� ��

�� x� � y� � �� z � �

�

Sketch the surface� If the temperature at the point x� y� z is given byx� � y� � z�� nd the average temperature over the surface�

Chapter �

��Forms and ��forms on R

�

�� Introduction� A New Idea

We have been integrating real valued functions of several variables sofar� and the big di�erence between the one variable case and the twoor more variable case is that you integrate one variable over an inter�val� and intervals have end�points at their boundary� whereas in higherdimensions� the boundary can be some fairly complicated thing� Youmight notice that the boundary of a one dimensional interval or curveif it has one is a zero�dimensional thing� a �nite set of points� Theboundary of a surface without holes if it has one is a one dimensionalcurve� and �nally� the boundary of a region in R

� is a surface� We havegot pretty crash hot at integrating functions of several variables overcurves� surfaces� and three dimensional regions� and the next stage isto integrate more complicated things than real valued functions�

The things we integrate over have a generic name� they are calledsmooth manifolds or smooth manifolds with boundary� This includeslines� line segments� circles� intervals� curves� spheres� balls� disks� tori�and almost all the things you have ever integrated a function over�These manifolds have been embedded in R

n for n � �� We also

��

� CHAPTER �� FORMS AND � FORMS ON R�

have objects which are piecewise smoothly embedded manifolds� sincedoing things piecewise is natural for integration� which is all aboutchopping things up into bits anyway� Thus squares and polygons andpolyhedra are piecewise smoothly embedded smooth manifolds�

The functions integrated have been de�ned on the space the embeddedmanifolds were sitting in� This jargon is going to be useful or I wouldn�ttrouble you with it� Note that the boundary of a manifold Mn ofdimension n� �M � is either empty or is a manifold of dimension n� ��

The things we are going to integrate next� more complicated than func�tions� are called forms� and they are very useful in describing more com�plicated objects than we have considered so far� Actually� real valuedfunctions are ��forms� and the next step is to look at ��forms� whichare so far as you are concerned the same things as vector �elds� Thenwe move on to ��forms� ��forms� and there we pretty much stop� It ispossible to go further� but we shan�t�

I shall start o� with some explanation of why we care about the simplestcase� ��forms�

Suppose you are driving due North in a car and the wind is blowingfrom the East� Then apart from the possible di�culty of staying onthe road� the wind will not actually slow you down� If the wind blowsfrom the South� it will assist you� and if it blows from the North it willslow you down� How much depends on the air resistance of your car� alarge truck will have more wind force to resist than a small car� but thelarge truck will probably have a more powerful engine to overcome thewind force� If the wind is blowing from the South�East� we can splitthe force into two components� one from the East which has no e�ect�and one from the South which assists us�

Suppose� more generally� that there is a windforce vector �eld all overthe plane� where the direction and amount of oomph of the wind onyour car is given by the map F � R � �� R

� which takes x� y� a possiblelocation of the car� to P�Q the windforce vector at that point� andhas P � x� y�� Q � x� y��

�� INTRODUCTION� A NEW IDEA ��

Exercise �� Sketch the vector �eld described above�

Now suppose I know that for every one unit of windforce against me�I spend an extra � cent in petrol for every kilometre travelled underthat windforce� And if the wind assists me� I pay a cent less� Howmuch money would I save due to the wind if I travel due north fromthe origin to the point �� ve kilometres North�

Well� at the point �� y the P component is irrelevant because this istransverse to my path� and the Q component is just y�� This beingpositive� I draw the arrow of the windforce pointing due North� theway I am going� so the force is with me� Eat your heart out� LukeSkywalker� As a result� I am saving money at the rate of y� cents perkilometre travelled� At the start this is zero� at the end it is �� and at� kilometre it is �� two kilometres it is �� three � and four � � I wantthe average of all the numbers not just the integer points multipliedby the distance travelled� Alternatively� I want

R �� y

� dy � �� This

is over forty cents� and worth knowing about� I can blow it on riotousliving when I get there��

That was easy� Now let�s make it a bit harder�

Suppose I travel along the straight line from �� to �� with thesame windforce vector �eld� What is my saving this time�

Well� the P component is due East and has value x � y�� but I amtravelling at forty��ve degrees� so the projection in my direction isthis number multiplied by cos��o which is �p

�� Similarly� there is

a contribution from the Northerly or Q component of x � y��p��

Now I have x � y consistently� so this last component doesn�t make adi�erence� and the calculation reduces to �p

�

R �� x

� dx� or a savings of

around ��

Before we work out wild things to spend it on� remember that we aregoing to need it in order to return the way we came� where the force of

�This enthusiasm may tell you something about academic salaries�

�� CHAPTER �� FORMS AND � FORMS ON R�

wind will be against us�

Now let us suppose that I have a constant force �eld due to wind or

anything else which is to say that whichever point

xy

�I am at� the

force vector at that point is the vector Fx� y � u �

u�u�

�where u�

and u� are just numbers� And suppose I am driving in a straight line in

direction v �

v�v�

�where again v� and v� are just numbers� In fact�

assume I go from the origin to the point v� The assist I get for thejourney from the force �eld is going to be the component of the force inthe direction v times the length of v� This is the length of the vector u�times the cosine of the angle between the two vectors� times the lengthof v� But this is just u � v� where � is the dot or inner product of thetwo vectors�

Now for the killer� Suppose I actually travel in a smooth curve� To beexplicit� let x � �a� b� �� R

� be a di�erentiable map from the interval�a� b� into the plane� How much are my savings after I�ve gone aroundthe bend described by x� I still assume some� constant in time� vector�eld F� which is now not necessarily constant in space� I also assumethat the drag or assist does not depend on how fast I am travelling�which looks a little unlikely� but what the hell�

If I travel along the smooth curve x�� then at time � time units I am

at location x� �

x��x��

�� where x�� x� are di�erentiable functions of

time �� At this time t� I am instantaneously travelling in the direction

given by the derivative of this� �x� �

dx�d�dx�d�

�� and if I travel for a brief

time �� I am pretty much going in a straight line vector �x�� whichis

dx�d�dx�d�

��

over that time� And the value of the force �eld is pretty much constantand is Fx�� That is� x� is where I am at time t and Fx� is


the force �eld at that point� For a short period �� I am in the samesituation described by the constant force �eld straight line drive casejust discussed� so the amount of oomph of the windforce vector overthat short time is

F � x� � �x� ��

The total gain is found by summing this and then taking limits� thatis� by integrating over the whole time I travel along the curve�

In other words� the integral of a vector �eld F on R� along a curve

x � �a� b� �� R�� is given by

Z b

aFx� � �x�d�

By the same argument as I used for integrating functions over curves�the result does not depend on the parametrisation as long as it is one�one and travels in the same direction along the curve� Instead of in�voking the time and a car travelling� I could have just set up littleline segments along the path� looked at the projection of the force �eldalong each of the line segments and added them all up� Repeating withsmaller line segments and taking limits makes it clear that the curvedoes not need a parametrisation for the idea of integrating a vector�eld over a path to make sense� On the other hand� parametrising itmakes it very easy to compute� as the next example shows�

Exercise ��

Let the curve be the circle of radius � units� and let the vector �eld

be given by Fx� y �

x� y�

x� y�

�� I travel around the circle in the

positive� counter�clockwise direction� starting and �nishing at the point�� What is the integral of the vector �eld over the curve�


Solution I shall parametrise the circle in the usual way by de�ning

x� �

� cos�� sin�

�

Then the required integral is

Z ��

�

� cos�� sin��

� cos� � � sin��

�� sin�� cos�

�d�

Working out the dot product we get

��Z ��

�c� � �cs� s��s � c� � �cs � s�c d�

where c is short for cos�� s for sin�� Simplifying using c��s� � ��we get

��R �� sin� d� � ��

R �� cos� d� �

��R �� sin�� cos� dt� ��

R �� cos�� sin� d�

which turns out to be zero� This is of course a great pity� If you made afew cents pro�t in driving around a circle� you could drive around lotsof times and get to be rich� And if you lost a few cents� you simply goin the other direction to get rich��

Your text book uses r for x� and so do some other old fashioned books�To my mind this vaguely suggests polar coordinates� so I avoid it� Yourtextbook also uses i� j� k to denote the standard basis vectors in R

��As a way of telling us what the components of a vector are� this has alittle to recommend it� but not much�

De�nition

If F is actually a force �eld on Rn� and � is a path through the �eld

parametrised by x � �a� b� �� Rn� the above path integral�Z b


�This may go some way to explaining why academic salaries are so low�


is de�ned to be the Work Done by the �eld over the path� This makessense for any positive integer n�

As with integrals of functions� if a path is a closed loop so the end pointand the starting point coincide� we may write

IcFx� � �x�d�

just to terrorise you�

�� Advice Learning This Stu

There are two ways you can go from this point� the �rst is to look atthe equation Z b


and try to memorise it while having only a tenuous notion of what theF and x mean� which is the worst way of proceeding� Try this and failmiserably�

The next best thing is to go and do a lot of examples� I shall give yousome� and there are others in the text book� This burns the thing intoyour memory� so that you both know the jargon and how to do thesums� If you do this� I guarantee that you will pass the exam�

The very best thing is to go through the argument about cars and windthat gives rise to the formula� see why it has to be that way� then do afew really nasty examples which you make up yourself�

It is up to you to decide which is the minimal e�ort way to pass theexam� The second way is for thickos� The third way is less time con�suming but requires that you think about what is going on� If thinkingis something you have trouble with� if it gives you a headache andyou just get lost and muddled� bash away at the sums by the secondmethod�


A variant of the �rst way is to scribble down lecture notes� copy some�body else�s assignment and start trying to memorise the formulae a fewdays before the exam� This pretty much guarantees that you will fail�If this is your preferred method� pull out now� and avoid the rush�

I tell you these things in a kindly way because� contrary to popularbelief� I am a nice guy� Also� I hate waste and would like you all topass�

�� Notation

Instead of writing the vector �eld as F � R� �� R

� and puttingFx� y � F�� F�� it is conventional to write F� � P x� y and F� �Qx� y� and instead of writing x for the parametrisation of the curve�we write x�� y�� whereupon we write the derivative� �x� in the formdxd�� dyd��

Then the formula for the integralZ b


can be written in an older notation as�Z b

a

P x� y

dx

d��Qx� y

dy

d�

�d�

This is mildly confusing� since x� y may be a point in R�� or a function

from R taking values in R�� and there are going to be points of R � which

are not on the curve� So it is a bit muddled� But that was the eighteenthand early nineteenth century for you� Actually� we are still muddled�but some of us are more muddled than others� I propose to keep enoughmuddle so you can make a reasonable guess at what the old fashionedbooks are on about�

It seems a pity to have to master the dot�product notation only tothrow it away at this stage� but this is what is done� Don�t blame me�

�� NOTATION ��

Now the fact is that the result of integrating the vector �eld over thecurve depends on the direction we go along the curve but does notdepend on the parametrisation of the curve� We could� as I suggested�take sequence of points on the curve� approximate the curve with lotsof line segments joining the points� work out the vector �eld at the mid�point of each of these line segments� take the dot product of the �eldat the midpoint with the vector de�ned by the line segment� add themall up� do it again for more points and shorter line segments� keep ongoing until the electricity fails� and in the limit we get the right answer�if the world makes any kind of sense� No parametrisation in sight�

This leads us to want to write the formula in a way that makes nomention of any parametrisation� We do this by writing�

ZP x� y dx�Qx� y dy

or� slightly shorter� ZP dx �Q dy

but the answer to the sum is the same� If you want to know how Ispecify the curve �� then I shall evade the issue� It doesn�t matter� Inprinciple I could give you a humungous list of all the points on it� orI could give you a graph� or a sequence of graphs� If I give it to youparametrically� the obvious way� then that rather suggests that thereisn�t much point in the latter notation� Because it suggests that if youdo it with di�erent parametrisations that are in the same direction� andbecause the suggestion is true� and also because everyone else uses it�we shall use the notation of the last formula for the integral of a vector�eld over a curve�

Such integrals are known in the literature as Line Integrals� if youwant to look up what you have been doing in other books� �

�This could be a terrible mistake� Most of the old fashioned books are prettymuddled about the kind of thing they are doing when they use the classical notation�Eighteenth and early nineteenth century Mathematicians were pretty hazy about


�� Exercises

Time for some exercises to build your mental muscles and make all thismore de�nite�

Exercise ��

In the problems below� sketch the vector �eld and make a crude esti�mate of the answer before doing the calculation� Some upper and lowerbounds on the value would be very professional� It could save a lot ofwork� too�

�� Let a force �eld on R� be given by saying that the force is always

towards the origin� and of magnitude proportional to �r� where

r is the distance from the origin� At the origin� the force hasmagnitude zero� The path consists of the unit circle� traversed inthe positive �anticlockwise� direction� Using a bit of gumption�write down the work done over the path�

�� The force �eld is as above� but the path is along the square withvertices at �� in the positive direction�Again� a bit of gumption will help�

� With the same vector �eld� along the straight line path from �� to ��

� With the paths of the last three cases� but with the magnitude ofthe force replaced by �

r��

�� With the last vector �eld and the path from �� to �� thatgoes around the circle of radius �� and goes in the top half of theplane�

lots of the �ne details� they just latched onto a couple of clear examples� and this iswhat you should do� too� Twentieth century Mathematicians tend to be very clearabout what kind of thing they are doing� not nearly as good at actually doing it�and totally in the dark about whether it�s worth doing in the �rst place�

�� SURPRISES� CLOSED AND EXACT FORMS ��

�� With the last vector �eld and a di�erent path with the same endpoints�

�� With the vector �eld given by P x� y � �y�Qx� y � x� and paththe unit circle in the positive direction�

�� With the same vector �eld and path the circle centred on the originand of radius R�

�� With the vector �eld of the last question� going around the squaredescribed earlier�

�� With the vector �eld given by P�x�y� � x�� Q�x�y� � y� and paththe straight line between the points �� and ��

�� With the last vector �eld and path any curve whatever going from�� to ��

�� As above� di�erent path� same end�points�

�� Same vector �eld� three di�erent paths from �� to ��

�� Same vector �eld� three di�erent paths from �� to ��

�� Surprises� Closed and Exact Forms

If you didn�t do all of the last section exercises� go back and bloodywell do them� You will have noticed� if you got them right� that forsome of the vector �elds� the integral between two points p and q seemsto depend on p and q but not on the path� whereas for other vector�elds the path makes a di�erence� This is right� there are indeed twodi�erent sorts of vector �elds� and it would be very nice to be able totell which is which� Obviously� if I give you some ghastly path from p toq and ask you to integrate some vector �eld or� more accurately� ��form�over the path� and if you spot that the ��form is one of the goodies�then you can choose a more convenient path and use that� Instead ofreferring to the goodies and baddies among ��forms� let us call a ��form


closed if whenever you integrate around a closed curve� a loop� you getthe result zero� Now it follows immediately that the integral along apath from p to q� for any points p� q does not depend on the path� forif it did we could go along one path� back along a di�erent path whichgave a di�erent answer� and have gone around a loop to emerge with anon�zero sum� So we can test a ��form to see if it is closed� and closed��forms are Good Things� at least from some points of view� Certainly�they are simpler to integrate over paths�

Now given a closed ��form P dx�Q dy on R�� take any point you fancy

and call it p� Now for any other point q� calculate the integral of the��form over any path from p to q� This gives a number which dependsonly on q� In other words� it de�nes a function fp�P�Q � R � �� R � And ifI do it for a di�erent choice of starting point p� then your function andmine will di�er only by a constant� which can be found by a methodwhich is evident to the meanest �intellect�

So closed ��forms over R� give rise to so called potential functions f �

And conversely� all di�erentiable functions over R� give rise to vector

�elds� for if you give me such a function f � I can di�erentiate it and geta row matrix of partial derivatives� which is a ��form� or alternatively�I can take the gradient of it� which is a vector �eld� It is plain thatthere is not a great deal of di�erence in which way I look at this� andif I want to be fussy useful in understanding more clearly what ishappening here� I can write

��f

�xdx�

�f

�ydy

�

for the ��form or

�f �

fxfy

�

for the vector �eld� There is a di�erence between column matrices androw matrices� but not one which it is pro�table to get one�s knickers ina knot about for this course�

�Are you awake�

�� SURPRISES� CLOSED AND EXACT FORMS ��

��forms which are obtained by the above process of di�erentiating func�tions or ��forms are called exact ��forms�

So we can go up from ��forms to ��forms by integrating� at least whenthe ��forms are closed� and we can go down from ��forms to ��forms�giving rise to a particularly nice kind of ��form� the exact ones� Iam assuming that all functions are smooth� which means that we candi�erentiate them as often as we feel the urge� It would be nice if allthe closed forms on R

� were exact and if the process of integration toget fp�P�Q from the form P dx� Q dy when followed by the process ofdi�erentiating to get the gradient �eld� got us back to where we started�After all� the fundamental theorem of calculus does precisely this in therather special case where the function is over R instead of R ��

Well� a little experimentation is called for� Let�s take fx� y � x� �y�� Then this gives rise to the ��form df � �x dx � �y dy� If weintegrate the ��form from �� to �� by the obvious parametrisation�xt � t� yt � �� t � �� we get

R �� x � �� This is indeed f��

Similarly� if we take the straight line path from �� to a� b by theparametrisation � � �� R

� de�ned by �� a�� b�� we get�x� � a� �y� � b� and the integral becomes�

Z �

��a� a� �b� b d�

which is just a� � b�� which is fa� b� So on the basis of a particularcase� it certainly looks as if it is working the way we might hope�

Exercise ��

�� Choose a slightly more complicated function fx� y� Derive the��form by putting P x� y � f

x� Qx� y � f

y� Choose any point

as your base point p �if I were you� I�d pick somewhere easy likethe origin� and then take some choices of q�

By choosing paths from p to q and integrating along those paths�con�rm that the result does not depend on the path but is alwaysfq� fp�


�� Do it again for a di�erent choice of function f �

You may now be prepared to believe�

Theorem �� Every exact form on R� is closed� and if f is a ��form�

the form df � fx

dx � fy

dy when integrated over any path from p to

q has the value fq� fp�

The exact form df is called the exterior derivative of f � And I remindyou that it is just the gradient of f if we disregard the di�erence betweenrows and columns� and think of ��forms as vector �elds�

There is an obvious picture of the e�ect of integrating an exact ��formover a path�

I used to run �kilometres every morning early� while it was still dark�so nobody would see me� I ran up Broadway� then up Princess Rd�� theup Dalkeith Rd� then another one and another until I came past thepolice station and back up Broadway� Uphill� every inch of the way�When I tried going the other way� it was still uphill every inch of theway� So I gave up running in the mornings�

Suppose that I am back in my car intending to head North for �vemiles� There is no wind� all is quiet and peaceful� But there is a vector�eld� it is obtained from noting that the journey is downhill in someparts and uphill in other parts� I get a boost from the terrain and theforce of gravity when I travel downhill� and this saves me some numberof cents on petrol per kilometre travelled� as before� Now to work outhow much I save� all I have to do is to work out the total di�erence inaltitude between my �nishing point and my starting point� There isa negative sign which goes in� because the gradient is uphill� when theforce is against me� Turn the terrain upside down in order to get thesign right�

It is immediately obvious that if the form is exact it is closed� that is�if I go in a closed loop in a vector �eld which derives from a potential�

�� FORMS ��

I must get back to the same height I started from� So exact forms� oralternatively vector �elds which are gradient �elds� do not allow us toget any net work out of a system by going around a closed cycle� InPhysics� we say the vector �eld is conservative� So to use a di�erentjargon� all gradient vector �elds are conservative�� This is merely sayingthat all exact forms are closed� You mights ask if all closed forms areexact� On R

� this is true�

Some more opportunity for a little thought�

Exercise ��

�� Consider the ��form �xy � y� dx � x� � �xy � ey dy� Is itexact� �Try integrating�

�� I claim that the ��form P dx� Q dy is exact provided Py

� Qx�

Explain why this test for exactness must be passed by any exactform�

�� forms

It is essential that you do the following exercises so as to get a feel forwhat we are about to do� Without the exercises� you will stay vagueand muddled� with them� it will be fairly simple�

�It is easy to get impatient with a proliferation of jargon� The reason for thejargon about �forms and �forms �which come next is that if you ever need to goto higher dimensions or work on spaces other than R

n� you �nd the baggage of theold notations gets in the way of understanding what is going on� The di�erencebetween vector �elds and �forms starts mattering in this case� and pretending theyare the same gets you into more trouble than its worth� So just in case you needto extend your education one day� you get to learn two lots of jargon rather thanone� If you think Mathematicians� Physicists and Engineers should get their acttogether� I agree� But the others insist on me changing to their systems� which isobviously stupid�


Exercise ��

�� Take the ��form df � x dx � y dy� and integrate it over the unitcircle in the positive direction� What do you expect to get� Whathappens when you actually do it� What ��form �function� hasthis as its derivative�

�� With the same ��form� take the circle of radius r and centre ��r� � which also passes through �� Integrate the form aroundthis loop� in the positive direction �i�e� anticlockwise�� and showthat the result does not depend on r

� With the similar looking ��form �y dx � x dy� integrate aroundthe paths of the last problem� Is this ��form exact�

� With the same ��form as the last problem� draw a square in theplane� and integrate around the square in the positive sense� Whatresult do you get if you divide the result by the area of the square�

�� Repeat with a di�erent square�

�� Repeat with some closed loop built up out of line segments that isnot square� Explain what is going on here�

The last exercises� not particularly di�cult to do� will have left youpuzzled� For some mysterious reason� the integral around a closedcurve in the positive direction for the particular form seems to workout as always twice the area of the loop you go around� It doesn�t seemto matter what shape the loop is�

Suppose the ��form has the property that integrating around the unitsquare in the positive direction always gives the answer �� and thatthe result does not depend on where the square actually is� Then it iseasy to see that the result of integrating the ��form around the squarewith vertices �� in that order� must be ��The diagram of �gure �� makes it clear why� We can decompose thelarge square into four smaller squares� and since the internal lines are

�� FORMS ��

Figure �� Integrating �y dx� x dy around a big square�

in matching pairs in opposite directions� the contributions must cancelout to give the integral along the outside big square� telling us thatthe integral around the big square is the sum of the integrals around thesmall squares� which is always �� If the integral around any square doesnot depend upon where it is� then we can run the argument backwards�and chop up the unit square into smaller squares� and so on� So theintegral around the square with vertices �� must be �� And for any square with area �A� the integral aroundthe boundary is ��A� And if now I take any region ! bounded by aclosed loop �! with an area !� then I can cover it with little squaresand have the same argument to compute the integral around �! to be� !� In algebra� I

��

Z��dA � � !

The picture in �gure �� illustrates a stage in the approximation of this�

The above leads the more imaginative thinker to imagine some amountof circulation in in�nitely tiny cells� which for this particular ��formis constant� and always twice the area of the cell� and that we areintegrating this constant circulation around little squares or for thatmatter litle circles�


Figure �� Integrating �y dx� x dy around a loop

Let�s investigate this particular ��form more closely� Since it doesn�tmatter where you are� let�s stand at the origin� Now the vector attachedto the origin is the zero vector� If we move a distance � due Eastright� we are at the point �� and the vector here is �� whichis a positive little arrow pointing straight up� If we had gone � to theWest left� the little arrow would have been pointing down�

There is therefore a tendency to turn any little paddle at the origin ina clockwise sense� arising from the Q component of the �eld� and itstorque is �� We must not forget the contribution for changes in y�If we move North a distance � from the origin� the �eld at �� is�� and at �� is �� So the P component of the �eld willalso tend to turn the little paddle at the origin in the same direction�the positive sense� with torque the same value� �� If we add thesevalues and consider the �amount of twist per unit area�� we get thenumber ��

What we have done here is to obtain a new geometric thing� the amountof circulation of the form about a point� For the ��form we were lookingat� this is constant i�e� does not depend on the point� In general wewould not be surprised if it varied from point to point�

�� FORMS ��

If F is the ��form� what we have done is to produce something called a��form� dF� I use the same d symbol as I used to get from ��forms to��forms� although the operations are on quite di�erent things� Lookingat what we have done in the particular case� we write�

De�nition �� If F � P x� y dx � Qx� y dy is a ��form on R��

and if P and Q are di�erentiable functions of both variables �so theyhave continuous partial derivatives� then the derived ��form is calleddF and is de�ned to be

dF �

��Q

�x� �P

�y

�dx � dy

The � sign is usually read as �wedge� or �veck��

If you take the ��form where we did our experimenting��y dx�x dy wehave Q � x� P � �y and the result of applying the exterior derivativeis to get the magic number �� Well� actually� we get � dx � dy�

The dx dy you might �gure is reasonable� since to use this� you willintegrate over the inside of the region covered by a loop so as to getthe integral of the original ��form around the boundary of the region�the loop itself� You want to know what the � sign is doing in there�Well� it is to tell you two things� one is that this is a ��form� not justany old function� since the next thing we are going to do is probably tointegrate it over a ��dimensional surface� And the second thing is totell you which way round we are going� because dx � dy � �dy � dx�as we really need to worry about orientations and directions in thisgame� If you turn the surface upside down or� equivalently� go aroundthe loop in the opposite direction� you get a sign change� We didn�thave to worry about that with the old double integrals� Since that isall that the � means� provided you bear it in mind� you can go aheadand integrate as if it wasn�t there� There is an implied orientationsense which is positive� If Gx� y dx� dy is a ��form we de�ne for anytwo�dimensional region of R �Z

UG dx � dy �

ZUG dA �

ZUG dx dy


the last part by Fubini�s Theorem�

We can now state an important and useful result�

Theorem �� Green�s Theorem

If F is a di�erentiable ��form on R� of the form

F � P dx�Q dy

and if the derived ��form is

dF �

��Q

�x� �P

�y

�dx � dy

Then for any region U bounded by a simple closed curve �U � we haveZUdF �

ZUF

where the orientation of �U is taken in the positive sense�

It is easy to see that what we are doing here is very similar to thechange of variable process� we take a local approximation to somethingwhich is pretty much constant in a su�ciently small region� a sort ofderivative in both cases� and we integrate this all over a region to getsomething instantly comprehensible in the simple case where the thingis globally constant�

Exercise ��

�� Use Green�s theorem to directly con�rm that the integral of theform F � �y dx � x dy around any closed loop is twice the areaenclosed by the loop�

�� Use Green�s Theorem to show that any exact di�erentiable ��formon R

� is closed�

�� FORMS ��

� Let U be the unit square� with vertices at �� and�� F � x�y dx�x�y� dy is a di�erentiable ��form on U � con��rm that Green�s Theorem holds by integrating the form aroundthe boundary in the positive sense� then by taking the exteriorderivative ��form and integrating that over U � and showing thatthese are the same number�

� Make up a U of your own and a non�exact ��form over U andverify Green�s Theorem in this case�

In order to show you how much work there is in these� not much I�lldo the last but one problem as an example�

Example �� U � I� is the unit square� The ��form to be inte�grated is

F � P dx�Q dy � x�y dx� x�y� dy

Integrating F over �I� is most easily done in four steps

base xt � t� yt � � � t � �� Then �xt � �� yt � � and thecontribution to the whole integral is

Z �

�t�� t��dt � �

right xt � �� yt � t � t � �� xt � �� yt � � and we get

Z �

��t� � ��t�� dt �

�

�

top xt � �t� yt � � � t � �� xt � �� yt � � and thecontribution is

Z �

�t�� t�� dt � ��

�


left xt � �� yt � �t � t � �� xt � �� yt � �� and thecontribution is

ZP �x�Q �y dt �

Z �

�� dt � �

Total � ��

Now we take the derivative of F to get the ��form

F � P dx� Q dy� P � x�y� Q � x�y�

Then�Q

�x� �x�y��

�P

�y� x�

So

dF � �x�y� � x� dx � dyIntegrating this over I� we get

Z �

�

Z �

��x�y� � x� dx dy

�Z �

�

hx�y� � x��

i��dy

�Z �

�y� � �� dy

�

y�

�� y

�

��

� � �

��

as required�

�� HOLES ��

Figure �� Integrating a ��form around two loops

�� Holes

The cases I have described above had regions that were simply con�nected� which is posh for saying that they didn�t have holes in them�The boundary was just a deformed circle� a simple closed curve� If wewere to take a region consisting of a disk with a hole in the middle�for example fx � R

� � � � kxk � �g� then we need to know if Green�sTheorem will still hold� The answer is yes� but there is a small problem�If we chop up the region into little squares as in �gure �� we need togo around the inside boundary as well as the outside� as in �gure ��

The point to note is that the boundary� �U � consists of two circles�and the orientation of the outside one is opposite to the orientation ofthe inside one� The whole business of de�ning the orientation of theboundary in general is fairly complicated� and I shall leave you to workit out by commonsense rather than trying to give you complete rules forit� It does mean that we have a stronger version of Green�s Theorem�

Theorem �� Green�s Theorem

If U is any subset of R � having a boundary which is a union of a �nitenumber of disjoint sets� each an image by a one�one piecewise di�eren�


tiable map of S�� then for any smooth ��form f de�ned on U �ZUdf �

ZU

f

provided the orientation of the boundary loops is positive for externalbounding loops� negative for internal bounding loops�

If you feel particularly courageous or drunk with power� try taking areally bizarre region U with holes in it and several bits� some inside theholes and some having holes inside holes� Then try to work out whatthe right de�nition of �external� and �internal� is for the above theoremto be true� If you do not feel that brave� just check it out for somemoderately complicated region U �

�� Connections

For a path� the boundary is the two end points� and the one at the startis reckoned negative and the one you arrive at is reckoned as positive�

To integrate a function over two points� you add up the values� takingaccount of sign� Now if the function f is a ��form on R � it is an ordinaryfunction� It has the exterior derivative df which is just the function df

dx

followed by a dx� Integrating this function over an interval �a� b� isthe one dimensional analogue of integrating a ��form on R

� over a twodimensional region U � Green�s Theorem in this case is just�

ZUdf �

ZU

f

again� and this tell us that the integral of the derivative of f over theinterval is the integral over the end points of the function f � This isjust the sum of the values� taking account of the orientation of the endpoints� i�e� fb � fa� Thus the good old Fundamental Theorem ofCalculus� which we have been using more or less non�stop so far� is

�� THE EXTERIOR DERIVATIVE AND K FORMS ��

just Green�s Theorem in one dimension� Green�s Theorem is Green�sTheorem in two dimensions� This leads to speculation about whatGreen�s Theorem is called in Three Dimensions� We will consider thisvital matter shortly�

�� The exterior derivative and k forms

The process of taking the exterior derivative of a ��form to get a ��form� does not on the face of things look much like taking the exteriorderivative of a ��form to get a ��form� This is a mistake� The processis really the same� if looked at from the right perspective�

When I have f � R � �� R and take the exterior derivative� all I do isdi�erentiate partially with respect to x and then stick a dx after it� dothe same with respect to y and stick a dy after it� and then add themup�

When I have P dx � Q dy and I take the exterior derivative� I �rstpartially di�erentiate the P partially with respect to x and stick a dxafter it� There is another dx already there� so I put a � in betweenthem to get

�P

�xdx � dx

Then I do the same again with respect to y and add them up�

�P

�xdx � dx�

�P

�ydy � dx

Now I have to do the same for the Q� again I add everything up to get�

�P

�xdx � dx�

�P

�ydy � dx�

�Q

�xdx � dy �

�P

�ydy � dy


This is the answer� If it doesn�t look like it� it is because we need todo a little cleaning up� Since dx � dy � � dy � dx� it follows thatdx � dx � �� and likewise dy � dy � �� Now putting everything ina nice tidy� fashion� with the dx� dys in alphabetical order� we get thecorrect expression for the derived two form�

A little thought suggests that we could keep on taking exterior deriva�tives until the cows come home� getting higher and higher k�forms� andthis is true� However on R

�� the only possible three�form will havedx � dx � dy with two dxs� or another version with two dys� or threeof one or the other� And these can only give zero� So there is only onethree form on R

�� and it is zero everywhere� which makes for no fun atall�

The situation on R� is� however� more interesting�

Exercise ��

Write down a ��form on R�� Now take the exterior derivative to get a

��form on R�� which will look like P dx�Q dy �R dz� Now calculate

the derived ��form�

Not very exciting� Write down a ��form on R� which is not exact�

and take the exterior derivative to get a ��form� Just do it as for R��

taking partial derivatives with respect to variable thingy and insertingdthingy � in front of all the other dwotsits� and adding up the result andsimplifying� You will wind up with A dx � dy�B dx � dz �C dy � dzfor some functions A�B�C which depend� naturally enough� on P�Q�R�

Now take the exterior derivative again by the same procedure to getW dx � dy � dz� for some function W �

Not a very stimulating W� Try taking a di�erent ��form and taking itsexterior derivative�

OK� now you have discovered a few important things about forms onR�� Do you recognise any of those operations you have just been doing�

Seen them before� The business of getting from a ��form to a ��form


s

t

x

y

g

Figure �� Change of Variables for forms

is just the gradient vector �eld of a scalar �eld� in Physics jargon� Howabout the other two cases�

�� Change of Variables

It is going to be necessary to do for ��forms and ��forms what we didearlier for functions ��forms� namely to work out how they behavewhen we do transformations and stretchings and such things to themanifold we integrate over� This is actually easier than for functions�because the orientation business makes it simpler�

First� suppose we have a curve in R�� say c � �a� b� �� R

� and that theinterior of the curve is called D� I shall call this the X�Y space� Nowsuppose that D is actually the image of the unit disk B� in R

� by amap g � R � �� R

� which is smooth and one�one� The general situationis shown in �gure �� I shall call the domain of g the S�T space�


�� forms

Given a ��form F on the X�Y space and a desire to make our lifesimpler� instead of integrating F around the perimeter of D� we couldtry to integrate the composite form F � g around the unit circle in theS�T space� This is� after all� a neat way of �nding areas and suchlike�so we ought to be able to make it work for forms�

Suppose the ��form is given by

F � P x� y dx�Qx� y dy

and the map g is given by

gs� t � xs� t� ys� t

Then it is easy to change to get

P xs� t� ys� t dx�Qxs� t� ys� t dy

and the question is� what do we do about turning the dx and the dyinto suitable functions of ds and dt�

We study what is going on here by the usual method which mathemati�cians use to �gure out what is going on in complicated cases� namely�forget about the hard problems� and study a really easy one �

We examine the simplest case I can think of that isn�t too trivial� andthat is when gs� t � as� bt for some positive numbers a� b� This map�as we well know� merely stretches the X�axis by a factor of a and theY�axis by a factor of b� turning the circle into an ellipse�

Then it is easy to see that the dx ought to be a ds and the dy oughtto go over to b dt� because that is the amount of alteration we do tothe length of little �ss and �ts� and so we ought to have that� for say

�Having solved the easy problem� they are then expected to go back and usethe approach developed with the easy case to solve the hard case� Sometimes theyforget�


the ��form F � x� � y� dx � x� � y� dy and E being the ellipsenx� y � R

� � x�

a�� y�

b��

o� the integral of this form over E ought� if

there is any justice in life� to beZS�as� � bt� a ds� as� � bt� b dt

Well� there isn�t much justice in life� but if you put s � cos �� t � sin �and integrate

Z ��

�a� cos� � � b� sin� �a� sin � � a� cos� � � b� sin� �b cos � d�

then you are doing the integral in the domain space� Alternatively� youcan parametrise the ellipse in the X�Y space by putting x � a cos �� y �b sin �� and dx

d�� a sin � d�� dy

d�� b cos � d� and you come out with

exactly the same integral� So there is some justice in life��

In our endless quest for clarity and understanding� we now take thenext and last but one step� We try it for a linear map g given by

gs� t �

xy

��

a ub v

� st

�

and look to see what it does to our ��form Pdx�Qdy� Thinking aboutthe idea of little �xs and �ys and what the map will do to them� wecome out with the rule�

dxdy

��

a ub v

� dsdt

�

So we deduce that given the above map g� and given c a curve in theS�T space� and any ��form P x� y dx�Qx� y dy� we haveZ

g�c�P x� y dx�Qx� y dy �

ZcP as�ut� bs�vta ds�u dt�Qas�ut� bs�vtb ds�v dt

So you�d better be careful�


In particular� if u � b � � this agrees with our last result� and if weinterchanged the S and T axes� we also get the right result�

Exercise �� Check That last claim carefully�

Finally� suppose g is just some smooth map which deforms the S�Tspace in some more interesting manner� What should we do� Wellthe P and Q behave just as you�d expect� they just get composed withg� The ds and dt get treated in precisely the above manner� theyget multiplied by a matrix which depends on where you are in the S�T space� the matrix being� you guessed it� the derivative of g� Thisimportant result is not surprising if you think it through carefully� andit merely generalises simple rules you have known for a while now� Itis worth stating carefully however�

Theorem �� Forms and Maps

If c is a curve in the S�T space and g � R � �� R� is a smooth one�one

map of the S�T space into the X�Y space� written


and ifP x� y dx�Qx� y dy

is a ��form on the X�Y space� thenZg�c�

P dx�Q dy �

ZcP gs� t

�x

�sds�

�x

�tdt �Qgs� t

�y

�sds�

�y

�tdt

�Zc

P gs� t

�x

�s�Qgs� t

�y

�s

�ds�

P gs� t

�x

�t�Qgs� t

�y

�t

�dt

�Zc

� P gs� t

�x

�s�Qgs� t

�y

�s

�ds

d��

P gs� t

�x

�t�Qgs� t

�y

�t

�dt

d�

�d�


The important thing to note is that the ds and dt transform quitesimply and in an obvious way according to the derivative of g� whenthe functions P and Q are transformed according to g� If you�d had toguess� this is probably what you�d have guessed anyway� Certainly� themore you think about it� the more reasonable it seems�

Example ��

Problem

Let c� � s�� t� � cos �� sin � for � � � � �� be a curve in theS�T space and let gs� t � xs� t� ys� t be de�ned by�

st

��

�s�

t�

�

Let F � x� � y�dx � x� � y�dy be a ��form on the X�Y space�Integrate this form over gc

Solution

The problem is to integrate the form

x� � y� dx � x� � y� dy

over the path �� x�� y� � cos� �� sin� �� We can either usethe given parametrisation or we can think of it as

xs� tys� t

��

s�

t�

��

dxdy

��

�s� �� t�

� dsdt

�

In this case we write the solution asZ ��

�cos � � sin ��cos� �� cos � � sin ��sin� � d�

which is the same thing�

Big Deal�


Exercise �� Repeat the above argument for the case where

gs� t � xs� t� ys� t � s� � t�� s� � t�

�� forms

There is no extra brain�strain in handling ��forms� We jsut use thederivative to �x up dx and dy and use the fact that

dx � dy � �dy � dx

and we are done� This obviously ought to be right from our intuitivenotions of what the two form is� an in�nitesimal amount of twist as�sociated with a point� frequently although not necessarily obtained byintegrating around a little loop in a vector �eld and dividing by the areaof the loop� There is a little tidying up to be done using the propertiesof � � The following theorem is� however� easy to believe�

Theorem �� forms and maps

If U is a region in the S�T space and g � R � �� R� is a smooth one�one

map of the S�T space into the X�Yspace� written


and ifP x� y dx � dy

is a ��form on the X�Y space� thenZg�U�

P dx � dy �

ZUP xs� t� ys� t

�x

�sds�

�x

�tdt �

�y

�sds�

�y

�tdt

�ZUP xs� t� ys� t

�x

�s

�y

�t �

�x

�t

�y

�s

�ds � dt


This gives the result of multiplying by the determinant of the matrixwhen g is a linear transformation� Note that we do not take the absolutevalue� since if the map reverses the orientation� we want to know aboutit" It would mean we are going in the opposite direction� Note thatit has� in general� the Jacobian occurring as a multiplying factor� justas for the integration of functions or ��forms� but without the absolutevalue operation� again� to preserve information about the orientation�

�� Some simple rules

The above sums were not particularly hard� and we used only a fewsimple principles in doing them� It is worth making a list of what weassumed here�

��

dx � dy � �dy � dx

�� It follows thatdx � dx � �

��

a dx � b dy � ab dx � dy

for any numbers a�b�

��

df ��f

�xdx�

�f

�ydy

��

dP dz �

��P

�xdx�

�P

�ydy

�� dz

for any z�


The associativity� distributivity and linearity of the various operationsare also generally taken for granted too� things like df � g � df � dgare probably worth your while extracting carefully from the text bookor these notes�

�� What is a dierential form

Students are sometimes puzzled to know exactly what a di�erentialform is� There are a number of di�erent answers to this question� Thequestion is a good one and demonstrates some maturity of thought�

One answer is�

De�nition �� A di�erential form on a di�erential manifold is asmooth section of the alternating tensor bundle of the manifold�

This is true but not very helpful� It raises more questions than itanswers� It suggests that if you de�ne �di�erential manifold� and �alter�nating tensor bundle� which could involve more terms you have neverheard of you will eventually nail the idea down� On the other hand�this could take quite a while�

Another answer is� what do you care what a thing is� With mathe�matics� what you do is learn how to recognise certain kinds of thingand you learn what you can do to them� Do you know what a numberis� If you were to ask for a de�nition of a number� I could give you onewhich would be about as helpful as the de�nition of a di�erential form�

De�nition �� A �real� number is an equivalence class of cauchysequences of rational numbers�

Now you have been using real� that is decimal� numbers for yonks� andat no time did you need to know what one was� you needed to be able


to recognise them� and to know what you can do to them� and how todo it� What you can do is to add them� subtract them� multiply them�and divide any one by another as long as the other isn�t zero� you canalso tell when one is positive or negative� And this is all you need tobe able to do� It is all it is possible to do with real numbers� know thisand you know� in a certain sense� all there is to know about them� Ithas certainly been enough to keep you busy and usefully employed overthe years�

Pure Mathematicians are people who want to know what things reallyare� Applied Mathematicians only want to know how to use them� Sowe are taking an Applied point of view here� To recognise a di�erentialform you look to see if it has dx or dy or dx � dy or similar sorts ofthings at the right hand edge of it� And what you do is to integratethem over manifolds or calculate the exterior derivative of them� Eachform is a k�form for some k� and you can work out what the degreeof the form is� k�forms get integrated over k dimensional manifolds�This isn�t very precise� but it is good enough for engineers and appliedmathematicians� It was good enough for the guys who invented thethings� the Pure Mathematicians came along and tidied up a long timeafterwards� We are all quite grateful to them� because they showed ushow to make it easy to learn the stu�� but and this is typical PureMathematicians only swept the streets after someone else had builtthem� They are useful people to have around� they are good at clearthinking� But the things they think about are not exciting to mostpeople�

There are therefore two ways of understanding di�erential forms� theabstract or Pure Mathematical one� and the Instrumentalist or AppliedMathematical one� But there is a second question which you ought toask about di�erential forms� which is�

Why does anyone care about di�erential forms� learning how to recog�nise them� integrate them over manifolds and compute the exteriorderivative is a fair amount of work� Why bother�

The answer to this can be worked out by thinking about the same


question for numbers� or any other mathematical object� They areuseful for handling reality� If they weren�t� nobody would bother withthem� Pure Mathematics is a fun game� but nobody would pay PureMathematicians� even the small sums they do get� if the stu� wasn�tuseful�

A deeper understanding of di�erential forms comes from using themto describe physical systems� This is what numbers are for too� youunderstand real numbers a lot better if you measure lengths and areasthan if you just learn to do arithmetic�

��forms you already know about� they are just vector �elds in all thecases you are going to meet� How about ��forms� How can you use a��form for describing a physical system�

Well� probably the easiest case to think about is the pressure in a liquid�There are many other cases� but this is a practical consideration inmany areas� so let�s look at it� It requires that we work in R

�� or a bitof it� but this will not alarm you if you are smart enough to be worriedabout these matters�

If you go down under the ocean in a bathysphere� the water is pressingdown on the outside of the bathysphere� and the amount of press� thepressure� increases as you go deeper� Let�s simplify by assuming thatthe water is at rest� which is not quite true� but it will make thinkingabout the issues easier�

Now what exactly do we mean by the �pressure�� How would we mea�sure it� As you know� we measure it in pounds per square inch if weare American� or HectoPascals if we are European or Australian� Sowhat do you do� Well you take a square inch of bathysphere and lookat the force on it� In pounds weight� say� If it seems that the pressureis changing across the square inch� we take a tenth of a square inch�measure the force in pounds� and multiply by ten to get the pressure�If you want to do it in kilograms per square centimetre� go ahead�

If you do this under water in your bathysphere� you might ask if itmatters where you put your square inch� A square inch� ought to be


a little thing about as big as a postage stamp� and the centre of it isthe place where the pressure is being measured� but what about thequestion of what direction the postage stamp is in� The postage stamphas� let�s suppose� a sticky side and a printed side� Let�s imagine aneedle sticking out of the printed side� The direction in which theneedle points gives us information about the orientation of the squareinch� Does pressure vary depending on what the orientation is� Wellactually� it doesn�t� but there are other things besides pressure� whichvery well might� Let�s call such a thing a Meta�Pressure� The fact thatpressure is a particular sort of Meta�Pressure where the orientation isirrelevant� is a fact about water that is at rest� It doesn�t have to holdfor all things that we might want to measure by sticking a little squareinch into the world and looking at the force on it� For some of thosethings� the orientation could make all the di�erence� It certainly wouldif the water were moving� for instance�

In a sense� the pressure acts on the square inch and is always normal toit� so the numerical value we get out doesn�t depend on the orientation�Even this is not quite true� If the square inch is under water on the wallof our bathysphere� if it is horizontal� the pressure is not changing overthe square inch� Chop the square inch up into hundredths� measure thepressure on each hundredth of a square inch� and it will be the same�If the square inch is vertical� this won�t be true� the pressure on thetop of it will be lesss than the pressure on the bottom� Not much inwater� but a bit� So the total pressure on our square inch will be anaverage over smaller bits of it� and so on� We clearly need the idea ofthe pressure at a point� but at the same time� pressure has something todo with area� and Meta�Pressure has everything to do with orientationof the area�

Well� if we take a teeny�weeny� itsy�bitsy� in�nitesimal postage stamp�with sides ds and dt� and if the system is the usual right hand framewe can write this as ds � dt� and if we have for each point x�y�z in R

�

a number associated with a particular orientation of the in�nitesimalpostage stamp� we can specify the orientation by having some amountof dx �dy� some amount of dx � dz� and some amount of dy � dz�Then sticking the ds � dt stamp in is given by the parametrisation of


a surface in general� say the bathysphere� And the ��form

P x� y� zdx � dy �Qx� y� zdx � dz �Rx� y� zdy � dzis exactly what we need to talk about the Meta�Pressure�

It is dimensionally right� because it has the little area built into it�so it is a measurable force which is associated with an in�nitesimalarea� the total force on a surface will be obtained by integrating theMeta�Pressure ��form over the surface�

So if you think of a ��form as being pretty much a vector �eld� a ��formis rather like a generalised pressure� This is not too di�erent from avector �eld� of course� but it is nice to have the in�nitesimal area inthere telling us that it acts over areas� whereas our vector �elds justact at points and have nothing to do with areas� but lots to do withdirections� This is why we integrate ��forms along curves� but ��formsover surfaces� They are easily muddled in R

�� because each surface hasa unique normal direction at each point� but they correspond to phys�ically di�erent kinds of things� and the physical di�erence is broughtout quite nicely by the notation� Good notations do this� they makeit harder for you to stu� up� Also� if you start looking at these thingsin R

�� the ��forms are speci�ed by four functions giving projections onthe axes� while ��forms need six functions because they are associatedwith pairs of axes�

I shall come back to water and pressure later� it should be noted thatbecause the pressure is always normal to the surface� we can considerit to be a simple scalar �eld� or ��form� on the region occupied by thewater� This requires us to distinguish between the pressure as a ��formand the pressure thrust as a ��form�

If you want to know about ��forms� you now know the right kind ofquestion to ask� it is not �What is a ��form�� The right question is��What is an example of a physical system where it would be naturaland useful to use a ��form to describe it precisely�� I leave you to �ndone� It must be something associated with in�nitesimal volumes�

Please note that it is very easy to ask the wrong questions� A huge


amount of thought goes into asking the right ones� Usually one asksthe wrong ones and keeps stu�ng up until one �nds a better question�To ask �What is a di�erential form�� is a good question� but to ask�What does one use di�erential forms for�� is a better one� Most philo�sophical sounding questions are the wrong ones� most engineers havegood instincts which tell them these are not the right questions� youcan�t make a buck out of the answers� But only the best engineers canask the right questions� and asking the wrong ones �rst is a necessarystep�

That�s enough Philosophy for the rest of the course�


Chapter �

Forms on R

�

�� Some examples of Exterior Derivatives

Just to drive some points home which are at the heart of the course�we are now entering the lower ventricle� kindly ensure your seat beltsare fastened� some examples�

Example ��

�� Let f � R � �� R be de�ned by fx� y� z � x�y� y�z� z�x� Whatis df�

Solution

df ��x�y � y�z � z�x

�xdx�

�x�y � y�z � z�x

�ydy�

�x�y � y�z � z�x

�zdz

� �xy � z� dx� x� � �yz dy � y� � �zx dz

�� For the same f � what is d�f�

Solution

��

�� CHAPTER �� FORMS ON R�

ddf � d�f

��xy � z�

�xdx � dx�

��xy � z�

�ydy � dx�

��xy � z�

�zdz � dx

��x� � �yz

�xdx � dy �

�x� � �yz

�ydy � dy �

�x� � �yz

�zdz � dy

��y� � �zx

�xdx � dz �

�y� � �zx

�ydy � dz �

�y� � �zx

�zdz � dz

� � � ��xdx � dy � ��zdx � dz

� �xdx � dy � � � ��ydy � dz

� �zdx � dz � �ydy � dz � �

� �

The change of sign as we swap dz � dy with dy � dz and soon� and the writing out of the dx � dx terms despite their beingzero is ridiculously ine�cient and the thing can be speeded upconsiderably with a bit of practice� I�ll leave out the interveningsteps next time

� The ��form F on R� is de�ned by

F � �x�yz dx� xy� � z� dy � x�yz� dz

What is dF�

Solution

dF � ��x�z dx � dy ��x�y dx � dz

� y� dx � dy ��z dy � dz

� �xyz� dx � dz � x�z� dy � dz

� y� � �x�z dx � dy

� �xyz� � �x�y dx � dz

� x�z� � �z dy � dz

�� SOME EXAMPLES OF EXTERIOR DERIVATIVES ��

� The ��form F on R� is de�ned by

F � �x�yz dx � dy � xy� � z� dx � dz � x�yz� dy � dz

What is dF�

Solution

dF � �x�y dz � dx � dy��xy dy � dx � dz��xyz� dx � dy � dz

� �x�y � �xy � �xyz� dx � dy � dz

�� The ��form G on R� is de�ned by

G � y��x�z dx � dy��xyz��x�y dx � dz�x�z��z dy � dz

What is dG�

Solution

dG � ��x� dx � dy � dz

� �xz� � �x� dx � dy � dz

� �xz� dx � dy � dz

� �

The above examples should be thoroughly understood by the tradi�tional method of imitating them on k�forms of your own choosing� forvarious k� on R

n for various n� The following things should emerge�

� d�f � �� for any k�form f �

� The number of functions needed on R� to describe a ��form is ��

to describe a ��form is �� and to describe a ��form is � again�

� The number of functions needed on R� to describe a ��form is ��

to describe a ��form is three� to describe a ��form is three� andto describe a ��form is � again� You may notice something aboutthese numbers�


� An implication of the above is that ��forms on R� are pretty much

the same as ordinary functions or ��forms� likewise ��forms on R�

are pretty much just functions� ��forms and ��forms on R� are

pretty much both the same thing as vector �elds� Just what�pretty much� means is open to some debate however�

If they do not emerge� it is because you have been doing somethingWRONG and should reform immediately�

�� Closed and Exact k forms

You can probably see how to write down a general k�form on Rn for

huge k� n and also how to take exterior derivatives� This is a harmlessexercise� if a little pointless as far as you are concerned� It should beobvious that we can de�ne what we mean by exact forms for any k�a k�form G is exact i� there is a k � ��form F such that G � dF�We can also de�ne closed k�forms� A k�form G is closed i� dG � ��This means that the derived form is actually the zero form� Last timewe de�ned a closed ��form on R

� as one which had integral around anyloop equal to zero� Well� if F is closed in the new sense� it has a derived��form which integrates over any area to give zero which is the sameas the integral around the loop bounding the area� Conversely� if theintegral of the derived form is zero no matter what region we integrateit over� then it must be the zero ��form� if life is to play fair with us�So the two de�nitions of �closed� are the same in the case of ��forms�

The discovery that d� � �� taking ddf and �nding out that it is alwayszero� tells us that every exact form is closed� The question as to whetherevery closed form is exact is open to your informed speculation�

�� STOKES� THEOREM ��

�� Stokes� Theorem

�� History

This theorem appears to have been �rst noted by William Thompson�otherwise Lord Kelvin in �� Proving it was set as a question onthe Smiths Prize Examination paper for �� at Cambridge� an exam�ination taken by the best mathematicians of the day� Stokes set theexam� It has become known as Stokes� Theorem just as Alzheimer�sdisease is called Alzheimer�s disease� not because Alzheimer was the�rst person to get it� but because he was the �rst person to publishit� Stokes didn�t even publish a proof� although Kelvin did and so didMaxwell� My information here is taken from Michael Spivak�s bookCalculus on Manifolds� Spivak being a contemporary Mathematician ofsome stature�

Stokes� Theorem is just what Green�s Theorem is called in dimensionthree� Green�s Theorem� you will recall� is about going around theboundary of a disk embedded in R

� and integrating a ��form over theboundary� This� says Green� is the same as what you get if you integratethe derived ��form over the disk itself� A sexier version let you havelots of disks� possibly with holes in them� but the extension is not toohard� We shall get there shortly�

�� Integrating ��forms and ��forms in R�

It is not hard to visualise a curve in R�� and integrating a ��form on

R� over the curve should not be essentially di�erent from doing it for a

��form over a curve in R��

You can take a disk or a rectangular region and embed it in R�� as

when you drop a piece of kleenex on your head� This also embeds theboundary of the disk or rectangular region� a circle or rectangle� in R

��


You will have noted that ��forms get integrated over ��dimensionalthings� namely curves� ��forms get integrated over ��dimensional thingssuch as surfaces� and we can reasonably guess that ��forms will getintegrated over volumes� This is true� and easy to remember�

We know how to integrate ��forms over regions like disks and squaresin R

�� it�s just good old double integration� Kid�s Play�

You may also throw your mind back to integrating ordinary functionsde�ned on R

� over surfaces� Recall �nding the average temperature ofa ramp in R

��

Recall that we had a quite successful strategy in doing such integralsof ordinary functions� It goes like this�

To integrate a function over some curved and curly surface M in R��

� First express your surface as the image by a smooth embedding ofa nice� easy surface� preferably a square or a disk i�e� a subset ofR�� embedded in R

�� In other words� parametrise the surface� Youmay have to do it with several disks having overlap of measurezero� but that�s OK� Suppose M � gD and g is the embedding�

� Pull back your function f on the surface M to a function f � gon D� there is a stretching factor involved here which must beconsidered�

� We already know how to integrate functions on bits of R �� Do it"

This gives the integral of the function in R� over M �

The question now is� can we do the same for ��forms as for ��forms onsurfaces in R

�� The answer is yes� but there are di�erences� because��forms aren�t functions� and we have to worry about more than just

�If you skipped that one as too hard for your poor little brain� you made a bigmistake� Go back to square one� do not pass GO� stick your nose in the electricitysocket and switch your brain to the ON position�


the area stretch� we also have to worry about the orientation� Thismakes things easier rather than harder� and you will be pleased to hearthat we have done almost all the work in the last chapter�

Let�s look at the case of a two form in R� and let�s think about in�

tegrating it over the unit square� I�� embedded in R� at height zero�

i�e� over the surface fx� y� z � R� � � � x � �� y � �� z � �g� For

de�niteness� let�s try out the ��form

F � y��x�z dx � dy��xyz��x�y dx � dz�x�z��z dy � dz

and try to integrate it over I�� I shall write P � y� � �x�z� Q ��xyz� � �x�y and R � x�z� � �z for purposes of referring to thedi�erent parts of the ��form�

Now it is pretty clear that the last two components� Q�R� are not goingto have any e�ect on the integral� and we need worry only about the�rst� part� P � In fact the answer ought� on any reasonable basis� to beZ

Dy�

since z � �� This is just

Z y��

y��

Z x��

x��y� dx dy

Which is just �� The answer to integrating a k form over a k�dimensional manifold jolly well ought to be a number� and it is� More�over� it�s a reasonable number� because for x� y in the unit interval andz � �� the amount of twist over the unit square should be between �and � from a glance at the function P �

Now we are ready to consider what happens if we stick the unit squareinto R

� at an angle� Obviously there will now be a contribution fromthe Q and R components of the ��form in general� First we stick it into

�If you doubt this� and why not�� just think of drawing little squares� integratingthe function over the squares� and adding up the values� When the square is in thedx� dy plane� We integrate the P �x� y� zdx � dy part of it and ignore the othercomponents� And similarly for the other bits�


the space by a linear map� Using s� t for the parameters� suppose wesend it to R

� by the matrix ��a ub vc w

��

Now there is a stretching to be considered and also the projection ofeach component of the twist on the embedded parallelogram� Both willinvolve a dot product� We can take care of everything in one go bywriting out the following equations�

x � as� ut � dx � a ds � u dt

y � bs� vt � dy � b ds � v dt

z � cs� wt � dy � c ds� w dt

Now we work out the form


composed with the embedding given by the matrix by putting

dx � dy � a ds� u dt � b ds� v dt

� ab ds � ds � av ds � dt� ub dt � ds� uv dt � dt� av � ub ds � dt

We also need to do this with the terms dx � dz and dy � dz since theywill not be othogonal to the parallelogram now� And we add everythingup to get the right answer� This leads to �

dx � dz � a ds� u dt � c ds� w dt

� aw � uc ds � dt�dy � dz � bds� vdt � c ds� w dt

� bw � vc ds � dt

Now we can pull back the form via the linear map speci�ed by thematrix to get the right ��form over R

�� This gives usZI��P Ls� tav � ub �QLs� taw� uc �RLs� tbw � vc� ds�dt


where L is the map speci�ed by the matrix� This works out to�

ZI�

bs� vt� � �as� ut�cs� wtav� ub

� �as� utbs� vtcs� wt� � �as� ut�bs� vtaw� uc

� as� ut�cs� wt� � �cs � wtbw� vc

ds � dt

This is a polynomial in two variables� s and t� which is now directlyintegrable as a straight function of two variables�

Exercise �� Do it�

Notice that the �rst case we investigated had a � �� v � � and all theother numbers zero� which just gives

ZI�t� ds dt � ��

which is correct� The second case had a and v positive numbers� all therest zero� which gives Z

I�vt�av ds dt

which is also correct�

Note also that this is just a minor extension of the change of variablesformula in the last chapter� We are doing exactly the same but we havea slightly higher dimensional space we are going to�

The only step remaining is to deal with the case where the embeddingis not linear but is smooth and is locally replaceable by the derivative ofthe map which embeds the disk or square to get the surface� We needonly replace the numbers a� b� c� u� v� w by the corresponding partialderivatives� This leads to the formula�


Theorem �� If D is a disk or rectangle in R� and g � R

� �� R�

embeds D in R�� and if

P dx � dy �Q dx � dz �R dy � dzis a ��form on R

�� thenZg�D�

P dx � dy �Q dx � dz �R dy � dz

is ZDP gs� t

�x

�sds �

�x

�tdt �

�y

�sds�

�y

�tdt

�Qgs� t �x

�sds �

�x

�tdt �

�z

�sds �

�z

�tdt

�Rgs� t �y

�sds�

�y

�tdt �

�z

�sds �

�z

�tdt

The last expression can be obtained by saying that what we do is tocalculate the di�erential terms from��

�� dxdydz

��

��

xs

xt

ys

yt

zs

zt

�� dsdt

�

And then using the properties of the � operation� it is straightforwardto reduce the expression to a straight oriented double integral with afunction followed by ds � dt�

Example �� Let the unit disk in R�� B� � fx � R

� � kxk � �g beset to the northern hemisphere of the unit sphere in R

� by the map

g � B� �� R�

s� t �� x � s� y � t� z �q�� s� � t�

This can also be thought of as the graph of the function z �q�� x� � y�

if you �nd that simpler�


Let the ��form F be de�ned on R� by


Integrate F over the image of g� i�e� over the northern hemisphere ofthe unit sphere in R

��

Solution We �rst work out the terms one at a time

�� xyz

��

��

st

q� � s� � t�

��

�� dxdydz

��

��

xs

xt

ys

yt

zs

zt

�� dsdt

��

��

� ��

�sp��s��t��

�tp��s��t��

�� dsdt

�

F � y� � �x�z dx � dy � �xyz� � �x�y dx � dz � x�z� � �z dy � dz

� P dx � dy �Q dx � dz �R dy � dz

P �� t� � �s�q� � s� � t�

Q �� st�� s� � t�� s�t

R �� s�� s� � t�� q� � s� � t�

dx � dy �� x

�sds�

�x

�tdt �

�y

�sds �

�y

�tdt

� ds � dt

dx � dz �� x

�sds�

�x

�tdt �

�z

�sds�

�z

�tdt

� � tq�� s� � t�

ds � dt

dy � dz �� y

�sds�

�y

�tdt �

�z

�sds �

�z

�tdt

�sq

� � s� � t�ds � dt


Figure �� Stoke�s Theorem� Don�t Shoot the Artist��

So the whole integral becomes

ZB�

�t� � �s�

q� � s� � t�

� �st�� s� � t�� s�t�t�q�� s� � t�

� s�� s� � t�� q� � s� � t�s�

q� � s� � t�

�ds � dt

Exercise �� By using the Polar map P from an r�� space to theS�T space above� convert the above integral into polar form� Con�rmthat

ds � dt � r dr � d�

�� Stokes� Theorem Modern Dress

Stoke�s Theorem in R� is exactly the same as Green�s Theorem in R

�

and the proof is exactly the same too� after all� the de�nition of a ��formis arranged to make it come true� the idea of in�nitesimal squares orcircles and taking the ratio of the integral around the perimeter dividedby the area as being the amount of twist at a point still makes sense for


a surface in R�� Now if you add up the contribution from lots of squares

side by side� as in �gure �� where the artist had enough trouble withonly two squares and they are not exactly in�nitesimal� you can seethat the argument that says that the amount of circulation integratedover the area ought to equal the line integral of the vector �eld of whichthe ��form is the circulation around the boundary� Or�

Theorem �� Stokes� Theorem� Version �

If U is the image of a disk embedded in R� by a piecewise smooth map�

and if f is a smooth ��form on R�� thenZ

Udf �

ZU

f

Exercise ��

�� Let f � �y dx � x dy � �dz be the smooth ��form on R� which

it is� and let U be the northern hemisphere of the unit spherein R

�� Integrate f around the unit circle� Integrate df over thehemisphere U � Hence con�rm that Stoke�s Theorem works in thisrather easy case�

�� If f � �y dx�x� z dy�x� y dz� verify Stoke�s theorem onthe same manifold U �

� Make up an embedding of the unit square in R� into R

� which is�a� smooth and �b� not altogether trivial� Make up a ��form onR� which is also not too simple� Verify Stoke�s Theorem for your

choice of form and manifold�

Example �� I shall do the �rst one of the above exercises asan example� the trick is in the lay�out and after that� they are all thesame until it comes to doing the integrals�

First we integrate the ��form over the boundary� The boundary is justS� the unit circle�


STEP ONE Choose a parametrisation of the boundary

I choose the unadventurous

x � cos �� y � sin �� z � � � � � � � ��

STEP TWO Work out dx and dy and dz

dx � � sin � d�� dy � cos � d�� dz � �

STEP THREE Substitute for x� y� z and dx� dy� dz in the expressionfor f � and integrate over the range of the parameter

Z ��

�� sin �� sin � d� � cos �cos � d�

Simplify Z ��

�� d�

STEP FOUR Evaluate the integral to get a number

��

Next we �nd the ��form df by using the rules for the exterior derivative

f � �y dx� x dy � � dz

STEP ZERO apply exterior derivative

df � � dx � dy � � dx � dy � � dx � dz � � dy � dz

Simplify � dx � dy � � dx � dz � � dy � dz


STEP ONE Choose a parametrisation of U

I choose the unadventurous application of the graph

x � s� y � t z �q� � s� � t� � s� � t� � �

STEP TWO Work out dx and dy and dz

dx � ds� dy � dt� dz ��sq

� � s� � t�ds�

�tq�� s� � t�

dt

STEP TWO�A Work out dx � dy� dx � dz and dy � dz�

�If you have to deal with �forms� you have to work out more of thesethings� likewise for higher order forms� should that be necessary oneday��

dx � dy � ds � dt�

dx � dz ��tq

� � s� � t�ds � dt�

dy � dz �sq

�� s� � t�ds � dt

In this case there wasn�t much point in working out the last two terms�because they get multiplied by zero anyway� but I am being mindlesslymechanical at the moment�

STEP THREE Substitute for x� y� z and for dx � dy� dx � dz anddy � dz in the expression for the ��form df � and integrate over therange of the parameters

Z t��

t��

Z s�p��t�

s��p��t�

�ds dt


STEP FOUR Evaluate the Integral to get a number

��

The observation that the two numbers are the same �nishes the veri��cation that Stoke�s Theorem works in this particular case�

Reducing everything to mindless computation of partial di�erentiationsand iterated integrals manages to conceal the ideas very e�ectively�There are people who think this is a good idea� it evens things up a bitif you are thick� Notice that the computational process I have givenwould just as easily allow you to take f a ��form on R

�� integrate itover the boundary of a six�dimensional ball� then calculate the exteriorderivative� df � and integrate it over the whole six dimensional ball� thencheck to see if the two numbers were equal� It would probably take abit longer� but it still comes down to partial di�erentiation and iteratedintegrals� with rather a lot of both of them� You will not be asked todo this in an examination� if only because the exams are not allowedto take longer than the degree�

If this doesn�t make you appreciate computer algebra packages nothingwill�

The question rather springs to mind� is it true in dimension thirty sevenfor ��forms and six�dimensional manifolds� The answer is yes�

Theorem �� Stoke�s Theorem Version �

If U is a smooth manifold of dimension k embedded in Rn by a piecewise

smooth embedding� and if f is a k � ��form on Rn� thenZ

Udf �

ZU

f

Stoke�s Theorem in this form was never known to poor old Stokes andreally ought to be named after Cartan or deRham or one of the guyswho sorted this out in the twenties�


Exercise �� Go and �nd the Dieudonne History of Modern Math�ematics and �nd out who did �rst sort out Stoke�s Theorem in the aboveform�

�� Stoke�s Theorem Classical Dress

The version of Stoke�s theorem I have given� and the setting out of thecalculations� make it fairly straightforward to both do the sums andto see why it is going to work� at least in the three dimensional casefor ��forms and surfaces� In this particular case� the classical formswere worked out by appeal to paticular physical interpretations� Thisis often how theorems in Mathematics get started� before the starksimplicities are arrived at� a physicist or engineer devises a hack thatworks� The job of the mathematician is to throw away the irrelevanciesand confusion and make it simpler� For this subsection� we work inR� and I put the irrelevancies back in� You can decide which you like

better�

The �rst observation is that it takes three functions to specify a ��form�usually thought of as a vector �eld� on R

�� It also takes three functionsto specify a ��form on R

�� This suggests that ��forms could also bethought of as vector �elds� and there is an obvious way of doing this� Ilook at the expression dx�dy and I remember the standard orientationof the axes in R

� so I identify dx � dy with the dz component of anew vector �eld� Similarly� I identify dx � dz with �dy and dy � dzwith dx� If we do this� we can represent the exterior derivative ratheringeniously�

F � Pdx�Qdy �Rdz �� PQR

��

� ��

xyz

��


dF �� F

�

��

xyz

��

�� PQR

��

If you �nd this easier to remember than the rules I have given you� thenthis will allow you to do the calculation of ��forms from ��forms fairlyrapidly� You note that it gives the right answer�

You can easily see that if g � D� �� R� is a parametrisation of the

surface in R�� then at a point s� t in the disk� the vectors

��

xsyszs

�� and

��

xtytzt

��

are tangent to the surface in two independent directions� We can de�nean outward normal to the surface to be the vector

v �

��

xsyszs

��

��

xtytzt

��

and then we de�ne the unit outward normal to be�

n �v

kvkWhen we calculate the ��form transformed to the parameters� s� t� itis an easy calculation to see that this comes out as

�� F � vand that the integral of the ��form over gD� is�

ZD�

�� F � vds dt

Since the area stretch of the map g at the point gs� t is preciselykvk� we can take an area measure on the image of g� gD� and call it


dA� I just de�ne dA at the point gs� t to be the factor kvk� wherev � g

s� g

t� a common shorthand for the above de�nition of the normal

vector v�

If F is a vector �eld� F �

�� PQR

�� and if x �

�� xyz

�� and dx �

�� dxdydz

��

then we can write the ��form corresponding to F as F � dx�

I can now write out Stoke�s Theorem in dimension three as�

Theorem �� Stoke�s Theorem� Version Three

If F is a vector �eld on R� and if U is a surface �two dimensional

manifold with boundary� embedded in R�� thenZ

UF � dx �

ZU��F � n dA

This does not change in any essential way how you actually calculatethe integrals of ��forms or ��forms� but you might try both ways andsee which is quicker�

Exercise ��

�� Suppose a surface in R� is given as the graph of a function f �

R� �� R over some region in the plane� This gives an obvious

parametrisation from that region� Calculate the area stretchingfactor in terms of f � i�e� calculate dA in terms of ds� dt and f �

�� Con�rm that if gs� t � xs� t� ys� t� zs� y is a map embed�ding a disk U in R

� and if F is a vector �eld on R�� then the

integral of the ��form corresponding to dF� dx over the imagegU can be written

ZU�� F � g � �g

�s� �g

�t ds dt


Figure �� Some water� lurking

�� Gauss

If you have been keeping a close watch on what is going on in thecourse� you will have noticed that we went from ��forms to ��formsand got a Stoke�s Theorem out which was rather a minor variation onthe fundamental theorem of calculus� whether done in R

� or R�� We

went from ��forms to ��forms on R� and R

�� and again had a Stoke�sThorem� which was about twist and curl� Well� there isn�t anywheremuch to go after that in R

�� since we run out of dimensions� But inR� we ought to be able to go from ��forms over surfaces bounding solid

objects to integrating ��forms through the volume� And indeed we can�and Gauss got there �rst�

To grasp the idea� a little excursion�

First� suppose you take a large plastic bag and put it in a swimmingpool� Close the bag after it has �lled with water� and leave it at thebottom of the pool� lurking� Now take away the plastic bag but shadethe water that used to be inside it� See �gure �� for a picture�

Now suppose the pool is stagnant� no water �ow� and no �sh swim�The water just sits there� Now there is a net force of gravity on the


water� tending to pull it down and �atten it out at the bottom of thepool� There is also a pressure of the surrounding water on the surfacebetween the shaded water and the rest� orthogonal to the surface� Thisis caused by the force of gravity on the rest of the water� trying to pullthat down to the bottom and squash it �at� These two forces balanceout so that the shaded water doesn�t actually go anywhere� The �rstforce is gravity acting on all the water inside the bag or inside whereit was� and you would have to integrate over all the water in the bagto decide how much that amounted to� The other force acts over thesurface of the bag or where the bag was and is just the pressure ofthe water� as would be exerted on you if you were the same shape asthe bag of water and in the same location�

Now this is either one of the incredible mysteries of life that these twoforces should exactly cancel out� or it�s dead obvious that they have to�depending on your point of view� Anyway� this is Gauss� Theorem� Itsays something about integrating one thing over a volume and �ndingit equal to another thing integrated over the boundary of the volume�The relationship between the ��form over the surface� and the ��formthroughout the volume� is �xed up to make it all come true�

Example ��

Problem

Suppose the volume is the cylinder with �at ends standing verticallycentred on the Z axis in R

�� The radius and height of the cylinder areboth one unit� and the bottom of the cylinder is at z � �� The pressurethrust on the surface is normal to the surface and of magnitude ��z�The diagram of �gure � shows the tank made of water�

Calculate the total upthrust on the cylinder� and show that this is equalto the weight of a cylinder full of water of the same size�

Solution

The downthrust on the top is the area of a circle of radius � times the


X

Y

Z

Figure �� Some water� in a tank

pressure which is zero� The upthrust on the circular base is the areatimes the pressure which is � � � � ��

The upthrust on the cylindrical part is normal to the surface and ob�viously cancels out in the X and Y directions by symmetry� and thethrust is directed vertically anyway� So the total upthrust is just ��The volume of water is the height times the area of the base� which isalso ��

Example ��

Problem

The tank is now placed on its side� its axis of symmetry along the Xaxis� and its base in the x � � plane� See �gure ��

Compute the upthrust due to the pressure on the tank and show it isequal to the weight in this con�guration� Show that the same resultshold�

solution � Bare Hands�

The volume hasn�t changed and is still � units� We have to integrate


X

Y

Z

Figure �� Fallen tank

the pressure over the curved portion� and show we still get the answer�� Since the axis of the cylinder is the X axis� we only need to worryabout the circular part� where the situation is as shown in �gure ��The lengths of the vectors have been scaled down to make it easier tosee the idea�

If we split up the circular region into little bits of arc�length �� andplace a vector at location �� we see that the length of the vector is� � � � z where z � sin�� and its projection in the upward verticaldirection is �� sin theta� so the amount of upthrust due to the segmentof length �� is sin� � � sin �� We also have to turn this into an areaby multiplying by a little bit of x� �x� Taking sums and then limits ofsums� we integrate

Z x��

x��

Z ��

��sin� � � sin � d� dx

to get the total upthrust� This is

Z x��

x��

Z ��

��

�� cos��

�� sin �d�dx

which comes to � again�


Figure �� section of vector �eld

Solution � Higher Powered� We observe now that the upthrust ofthe water on the cylinder is the resultant of the pressure thrust over thewhole surface� We represent this pressure thrust by a ��form�

We �rst parametrise the curved part of the cylinder by

g � �� R�

s� t �� x � s� y � cost� z � sint

We compute the normal vector with

�g

�s�

��

�� g

�t�

��

�� sintcost

��

to get

�g

�s� �g

�t�

�� cost� sint

��

We notice that this has the right sign for an inward pointing normal�

Next we write our ��form giving the pressure as if it were a vector �eld�


also in terms of the parameters s� t to get

� � sint

��

�� cost� sint

��

And �nally we integrate the dot product with the vertical to get theupthrust summed over the surface

Z �

�

Z ��

�� sint

�� cost� sint

��

��

�� dt ds

The �at ends contribute nothing to the upthrust since the normal tothe surfaces are orthogonal to the vertical� Alternatively� the pressurethrust one one face is equal and opposite to the pressure thrust on theopposite face� so the sum of the contribution from the �at faces is zero�

It should be fairly clear how to compute the upthrust on a more generalshape now�

The last approach was fairly general� but it did still project in thevertical direction at the last stage to get the upthrust� This neglectsthe remote possibility that the horizontal components don�t sum tozero� so we �nish up with the complete solution according to Stokes�

Solution �� Last Version

Example ��

We parametrise the curved surface �rst with

g� � �� R�

s� t ��B� x � s

y � sin tz � cos t

�CA


The acute reader will have noticed that this is slightly di�erent fromlast time� because last time we wanted an inward normal� and now weare going to directly verify the Gauss Theorem� so we want the correctorientation of the boundary of a solid object� which is� by convention�outward�

The next problem is to get the general ��form for the pressure thrustof water� I have pointed out that the pressure is the scalar �eld whichat location x� y� z in the water has value c � z� where c is the heightof the surface with respect to the origin� In this case c � �� Now thepressure thrust is orthogonal to any surface placed in the �uid� If thein�nitesimal postage stamp were placed at x� y� z and were horizontal�the direction would be given by �dx � dy� The negative sign becausethe natural orientation of such a little element is vertically up� and theforce is in the opposite direction if this is the outside of some object�If the in�nitesimal object were pointing in the positive y direction� thedirection of the opposing thrust would be dx � dz� and if it were in thepositive x direction it would be �dy � dz� In general� all we have todo is to add these components up� So the unit normal thrust pointinginwards over any surface can be written as

�dx � dy � dx � dz � dy � dz

It follows that the ��form describing the pressure thrust in stationarywater is given by

c� z�dx � dy � dx � dz � dy � dz

orz � cdx � dy � dx � dz � dy � dz

where c is the height of the surface of the water� Naturally� this onlymakes sense under the water� Over the water� the form is zero� unlessyou want to count atmospheric pressure� where the rules are slightlymore complicated� In our case� c � ��

The �rst job then is to integrate the form given over the curved surface�this is going to require us to get expressions for dx� dy� dz and then tocalculate the wedges of them� Rolling up our sleeves� we begin


x � s � dx � ds

y � sin t � dy � cos t dt

z � cos t � dz � � sin t dt�

dx � dy � cos t ds � dtdx � dz � � sin t ds � dt

dy � dz � ��

F � z � �dx � dy � dx � dz � dy � dz� cos t� �cos t� sin t ds � dt

The integral over the surface is then given by

Z t��

t��

Z s��

s��cos� t� cos t sin t� cos t� sin t ds dt

�Z ��

�cos� t� cos t sin t� cos t� sin t dt

�Z ��

�

� � cos �t

��

sin �t

�� cos t� sin t dt

��

�

Z ��

�dt

� �

To do the rest� we need to parametrise the end disks

g� � D� �� R

�

s� t ��B� x � �

y � sz � t

�CA

does the one side� and


g� � D� �� R

�

s� t ��B� x � �y � �sz � t

�CA

does the other side� To verify that we have the right orientation�

�g��s

�

�B� ��

�CA

�g��t

�

�B� �

��

�CA

�g��s

� �g��t

�

�B� ��

��

�CA

which is indeed pointing out from the cylinder� Check the other sideyourself�

We can do exactly the same thing with each of the �at surfaces as wedid with the curved surface� It would show that you are a bit thick toactually do this however� because if you do them both at once you willrapidly discover what is obvious to the thoughtful student� which is thatthe integrals are of opposite sign and equal absolute value so must sumto zero� Working out each of them separately takes a fair amount oftime� and then adding a number to its negative gets a rather simpleresult� It is better to notice they are going to cancel out before wastingyour valuable time doing a nasty integral� This is only one example ofthe general principle that a little thought can save an awful lot of timeand energy�

This gives the answer � for the total integral� This is� of course� anumber� not a vector� By Gauss� Theorem� it is equal to the exteriorderivative of the form giving the pressure thrust� Let�s verify that


F � z � �dx � dy � dx � dz � dy � dzdF � � dz � dx � dy

� � dx � dy � dz

And integrating this over the volume of the cylinder gives its volume�Which is � and hence we have the right answer�

Exercise ��

�� Do the above calculations for a hemisphere with a �at base im�mersed in water with the �at base horizontal and at height zero�and with the same assumptions about the pressure� Show againthat the upthrust is equal to the weight of the water displaced onsuch a sphere�

�� Take the same hemisphere but turn it sideways so that the formerbase is now in the Y�Z plane� Show again that the upthrust isequal to the weight of the water displaced�

� Using the above result� show it also holds for a sphere�

Returning to the general issue of Stoke�s Theorem� Suppose

F � P dx � dy �Q dx � dz �R dy � dzis a ��form on R

�� We take the exterior derivative of F�

dF ��P

�zdz � dx � dy � �Q

�ydy � dx � dz � �R

�zdx � dy � dz

which after a bit of swapping and sign changing gives�

dF � �P

�z� �Q

�y��R

�z dx � dy � dz

Stoke�s Theorem in this case tells us as usual that�


Theorem �� The Gauss Divergence Theorem

If U is a region in R� which is a smooth embedding of a ��ball� B�� and

if F is a ��form on R�� thenZ

UF �

ZUdF

If we do the same translation procedure as for Stoke�s theorem for ��

forms� we get that the ��form F comes out as a vector�

�� R�QP

�� and

then dF �� F� or�

dF �

��

xyz

��

��

R�QP

��

When we integrate a ��form over a surface we have to put the unitnormal vector n into the expression when we translate� so this givesthe classical form of the Gauss Theorem�

Theorem �� The Gauss Divergence Theorem Classical Dress

If U is a region in R� which is a smooth embedding of a ��ball� B�� and

if F is a vector �eld on R�� thenZ

UF � n dA �

ZU�� F dV

If F is a vector �eld on R��

F

��xyz

��

��F�x� y� zF�x� y� zF�x� y� z

��

We de�ne div�F�� the divergence of F� to be the function f � R � �� R

given by�

fx� y� z ��F�

�x��F�

�y��F�

�z


and write it as �� F�

If we write the Upthrust vector as

F

��xyz

��

��F�x� y� zF�x� y� zF�x� y� z

��

��

��

� � z

��

in the above examples� equivalent to the ��form

�� z dx � dy � � dx � dz � � dy � dz

we have that the upthrust or integral over the surface is

ZU

� � z dx � dy � � dx � dz � � dy � dz

which with the obvious s�t parametrisation used in the example becomes

x � s dx � ds

y � cost dy � sint dt

z � sint dy � cost dt

We therefore get the integral of the ��form over the surface as�

Z �

�

Z ��

�� sint ds � � sint dt

Since Pz

� �� integrating the derived ��form is rather easy�

ZU�� dx � dx � dz

which is just the volume of the region U multiplied by �� And theseare equal according to Stoke�s Theorem� which must be true after allthis trouble�


Where did that bloody minus sign come from� Well� when I wrotedown the ��form� I was trying to model the pressure of the outside onthe inside� which is the opposite way round to the obvious orientationof the boundary of a region in R

�� The pressure of the inside on the out�side is directed towards the interior of the region� whereas the naturalorientation of a surface bounding a region is towards the outside fromthe inside� So if I want to model the thing sensibly� I need to put in aminus sign when I describe the pressure �eld� If you were awake� you�dhave caught it� because I did it right when I dealt with the cylinderunder water�

Exercise �� Do the last exercise about hemispheres using thegeneral rule about ��forms instead of messing about with vector �elds�

�� Tying Up Stokes

There are often� in Mathematics� many ways to think about something�In this case there are a lot of ways of looking at divergence� If youimagine that the vector �eld corresponding to the ��form F is some�uid �owing through space� with the components

F

�� xyz

��

�� F�

F�

F�

��

and if you draw a little box around the point

�� xyz

�� then integrating

the �eld over the surface of the box tells you how much �uid� net� isemerging from the box� Dividing by the volume of the box tells yousomething about the rate of production of �uid inside the box� Makethe box smaller and smaller� and we are talking about the amount of

�uid generated at the point

�� xyz

�� If there are no souces or sinks for

the �uid� then this will be zero�


We often think of electric charge as concentrated on a point� and ifwe investigate the �eld between some collection of charges� we have avector �eld telling us the strength and direction of the force on a unitcharge at each point of the space� So long as we integrate over a surfacewithout boundary which does not contain a charge inside it� we oughtto get the result zero� since the divergence inside the region boundedby the surface will be zero� So Physicists and Engineers dealing withelectricity use this sort of idea a lot� There is a lot more than that tothe applications of these ideas to electricity� including electromagnetictheory� but I shall stop because you are mechanical and civil engineers�

An important family of cases of applications of the Gauss DivergenceTheorem comes up when the vector �eld which gets integrated over theboundary is actually a gradient �eld� Recalling the Gauss Theorem�Z

UF � n dA �

ZU�� F dV

suppose we have that F � �f � Writing the theorem out again ratherclumsily by expanding the notation we get�

ZU

�B�

fxfyfz

�CA � n dA �

ZU

�B�

xyz

�CA �

�B�

fxfyfz

�CA dV

Evaluating the dot product on the right hand side� we get

� ��f ��f

�x��f

�y��f

�z�

This is usually written�

��f ��f

�x��f

�y��f

�z�

by Physicists and Engineers who� we are all agreed� are a scru�y lotwho make the language up as they go along� Since lots of vector �eldsthat we meet in real life are gradient �lds� it is not surprising that


the expression �� keeps cropping up� There has been a move amongPhysicists to write it as �� but you don�t need to know that except toavoid panic if someone springs such a notation on you one day�

There are also many applications of the various forms of Stoke�s The�orem� all you need to do is to �nd the right forms� and parametrisethe right regions� The theorem works for regions which are morecomplicated than just disks or balls of various dimensions� doing theparametrisations can be messy but the principles are obvious� or shouldbe�

�� Integration of forms over manifolds

�� A Summary

Given a manifold M and a k�form F� we need the manifold to have di�mension k in order to integrate F overM � M can then be parametrisedby expressing it as the image of a one�one map from a ball or cube inRk� We have been doing this for k � �� for some time now� If it is a

really evil shape� we may have to break the manifold up into bits andparametrise each bit separately� An example is the cylindrical tank ofthe last section�

To integrate the form is really very simple and involves reducing theintegral� via the parametrising map� g say� to an integral over the k�ball� U � We simply compose the form F with g to get the function partof it�

Now we have to replace the dxs and dys with dss and dts in the expres�sion for the integral� The �rst thing to do is to get each of the dxs anddys and possibly dzs separately� these di�erentials are transformed

�� INTEGRATION OF FORMS OVER MANIFOLDS ��

by the derivative of g in the same way as F is transformed by g�

�� dxdydz

��

��

xs

xt

xs

xt

xs

xt

�� dsdt

�

It would be quicker and neater to write

x � gs� dx � Dgds

These deal with the case of a surface� i�e a ��manifold� Other dimensionsare done in the same way�

Having obtained expressions for the dxs and dys� we substitute for thesein the expression for the form� We may very well then have to simplify�using the properties of the � relation�

This turns the original into an integration of a transformed k�form overa ball or cube in R

k� I integrate this by simply forgetting about all thelittle � signs�

All the various forms of Stoke�s Theoremwork this way� and I personally�nd this a lot easier than the way this is explained in your text�book�but whatever turns you on� as they used to say in the sixties�

�� A note on i� j� k

It is often a good idea to write vectors as columns� if nothing else itallows you to be ready to act on them by matrices� This used to cause

acute pain for typesetters� Putting

��xyz

�� smack in the middle of a

line caused the poor devils no end of trouble� so they refused to doit� There are three ways around this� one way is to write your vectors


horizontally as rows� The fact that it should be a column is indicatedby putting a little �T� for �Transpose� at the end� as in

x� y� zT �

�� xyz

��

This gets you o� the hook but is a bit ugly� Still� it kept the typesettershappy�

Another approach works so long as you stick to R� and R

� and is henceeven more useless� but was very popular about half a century ago�What you do is to declare once and for all that� in R

�� the standardbasis vectors have special names��

��

�� i�

��

��

�� j�

��

��

�� k�

And in R� we use i and j for the obvious two basis vectors� This has

the advantage that we can write xi� yj� zk instead of

�� xyz

��

There are a few other minor advantages of this notation� but they aremostly obsolete� So I haven�t used it�

The third and by far the best way is to get rid of the typesetters� andthis is what has been done� TEXand LATEX do it far better than humanbeings and their complaints can be dealt with much more cheaply�

One of the problems of teaching Mathematics to Engineering studentsis that they are also taught by engineers who got their own educationfrom mathematicians about thirty years before� And those mathemati�cians got their education about thirty years before that� The use of theExterior Calculus as I have described it was invented by mathemati�cians in the twenties and thirties� made it to the Physics textbooksGravitation� by Misner� Wheeler and Thorne does a reasonable com�promise in the sixties and seventies� and should get to be standard

�� INTEGRATION OF FORMS OVER MANIFOLDS ��

in Engineering courses around the �rst or second decade of the nextmillennium� And you will probably have to explain what you are do�ing to your lecturer in Engineering� so you had better understand it"Making mysterious things simple is quite hard� but you don�t usuallyget much credit for it� It is� on the whole better to go the other way asthe mathematician Piet Hein explained�

To make a name for learningWhen other roads are barred�Take something very easyAnd make it very hard�


Chapter �

Partial Di�erential Equations

�� Introduction

A Vector Field on R is a way of assigning to each point of the space R �a little arrow telling you how fast you are travelling along a curve ateach point� If I express this as a function� I can write

dx

dt� vx

where vx tells you the size of the arrow at the point x� and xt isthe curve along which you are travelling so as to be travelling at speedvx when you are at x� This does the usual muddle of not being clearabout whether x is a point on the line or a function telling you howyou move along the line� Ah well� you get used to it� Anyway� we getsuch expressions as

dy

dx� �y

or

�x � sinx

out of this by changing some variable names�

��

�� CHAPTER � PARTIAL DIFFERENTIAL EQUATIONS

We may also want to deal with the case where the vector �eld changesin time�

dx

dt� vx� t

which gives expressions such as

dy

dx� �xy

or�x � �x� t

with a few changes of names to confuse you�

These are Ordinary Di�erential Equations� and you meet them in allsorts of places� ODEs don�t have to be on the real line� they generallyare not� and what are sometimes called Systems of ODEs are just vector�elds on R

n for various n� They are extremely important things tounderstand� The reason is that we often know something about whatthe amount of shove is at every point in the space� the rules tellingyou how the state of a system is changing or trying to change in timecan be expressed as a vector �eld� And we may know where we arestarting from� Now it is very reasonable indeed to want to know whatis going to happen in the future� We can hope to �nd out what thetime development of the system is going to be by solving the ODE�

ODEs are Science�s answer to the Crystal Ball or the Tarot pack� orTea Leaves� or consulting wise men or economists� or any of the manyand varied and usually dotty ways people have of foretelling the future�Since Engineers send space craft to Saturn using these methods and weget some nice pictures back to prove it� whereas most fortune tellersare scru�y twits operating out of caravans� we may safely conclude thatODEs have something going for them which Tea Leaves and CrystalBalls can�t match��

The solution to an ODE is a one�dimensional curve in a space of statesof some system� The space gets decomposed into a whole lot of possible

�And just take a look at the economy if you believe in economists� or the e�ciencyof business if you believe in management consultants�

�� INTRODUCTION ��

curves� each of them the future time development of some system witha particular starting point� I am simplifying a bit here so as to dealonly with initial value problems� as they are called� There are otherapplications of ODEs which I am skipping over�

There is an obvious generalisation of ODEs to the situation where in�stead of something varying in just one dimension� time in our presentdiscussion� it can vary in two or more dimensions� A solution to sucha problem would be some surface or higher dimensional manifold sit�ting in a space� You will be glad to hear that although the higherdimensional cases are of more practical importance� in this course westick to lower dimensional cases where it is easier to get a feel for whatis happening�

Example ��

I take a loop of wire and twist it about a bit� Then I dip it in soapsolution and get a nice soap �lm in the wire� If I hold the wire up inR�� there is a function de�ned over the �shadow� of the loop and the �lm�

which tells us the height of the soap �lm everywhere� More generally� Ihave a function from S� into R

� which embeds the circle in three�space�and this extends to a function from the unit disk� D� into R

��

The illustration of �gure �� shows you the possibilities�

The soap �lm extension is only one among an in�nite number of pos�sible extensions �blow on the �lm to distort it to get some others�� thequestion is� what made the �lm choose the particular shape it did� Theanswer is that surface tension was busy trying to minimise the area�given the boundary� Now this is a purely local thing� like a vector �eld�while the surface that you actually get is a global solution� So thereought to be a way of setting up something a bit like an ODE and �ndinga way to solve it which would give a solution to the soap �lm problem�

There is indeed a whole body of Mathematics dedicated to precisely thissort of problem and its higher dimensional analogues� and it is called


Figure �� Blowing Bubbles

the study of Partial Di�erential Equations� Just as ordinary di�erentialequations do di�erentiation of the time or some other single variablebecause the solution is a curve� so the PDEs have partial derivativesoccurring in them because the solution will be a surface or some higherdimensional manifold� It is more complicated than ODE theory forseveral reasons� one of them being that the boundary of a curve isjust a pair of points unless the curve is closed� when it doesn�t havea boundary� whereas the boundary of a two dimensional thing like adisk is a circle� which is a lot more complicated than a couple of points�

Actually� most PDEs are so hard we don�t have the foggiest ideas abouthow to solve them�� we can only do a few easy ones� But those we cansolve are very� very� useful�

Example �� I take a solid ball of iron and sit it on a table� Ev�erything is at room temperature� Now I heat up the table just under theball by applying a blow�torch� the temperature of which is rather a lot

�Well� closed form or analytic solutions in terms of standard functions are veryrare� and even solutions in terms of explicit in�nite series are often impracticable� But numerical methods can give us a solution to high accuracy in manycases� Determining whether the numerical solution is a stable� safe one is stillunder investigation�

�� INTRODUCTION ��

higher than room temperature� say �� How does the temperature ofthe interior point x� y� z of the solid ball change in time� It obviouslystarts o� at room temperature at time zero� and then goes up fairlyfast� and the closer x� y� z is to the blow torch� the faster it goes up�It would be nice to have some details�

If I have a function f � R � �� R � and if I di�erentiate it� I get a rowmatrix of partial derivatives�

�f

�x��f

�y

�

It makes sense therefore to guess that if there is a Partial di�erentialequation the solution to which is a disk or ball mapped into R

n� the theequation itself will have partial derivatives in it� Hence the name�

Example �� It can be shown that if f � D� R� �� R is a

function which describes the height of a soap �lm above the z � � plane�then provided there are no other forces but surface tension operating�and providing the function f on the boundary is not too di�erent froma constant function� then f approximately satis�es the condition

��f

�x��f

�y��

orfxx � fyy � �

if this notation is more to your taste�

If you are told that f � S� �� R is a particular function� and that�fx�

� �fy�

� �� then the instruction �solve for f � means to �nd the

unique �you hope� function f � D� �� R which has this property andis as speci�ed on S� � �D��

PDE problems where we know relationships such as this locally� andwe are given all values of a function on a boundary and want to �nd


the function on the interior� are called Dirichlet Problems after the manwho made a speciality of tackling them in the nineteenth century� This�incidentally� was the �rst bloke to work out what a function is� Theghastly function which is � on the irrationals and � on the rationalswas his idea� He was German� not French as the name suggests� Hewas born in �� so this is all recent stu�� only a century or so old��

Another kind of problem� not a Dirichlet problem but related is�

Example �� If I heat up one half of a copper rod to ��o and keepthe other half at �o while doing so �pity about the midway point� andthen take away the freezer and the �ame� the function of length givingthe temperature will start o� as a step function and gradually even outuntil the bar is a nice ��o everywhere� assuming no heat is lost to theoutside world� Given information about how heat is conducted throughthe material� we ought to be able to compute the function at any timeafter t�� We think of time as the positive reals� so we know the value ofthe function �x� t completely at t � �� the step function� where � is thetemperature� and it is known that the Heat Equation must be satis�ed

��

�t� c�

��

�x�

So again we have a partial di�erential equation�

Partial Di�erential Equations then occur quite naturally as ways ofdescribing Physical systems� We have two jobs to do�

� From a physical situation� set up the equation which describesthe system

�You may reasonably suspect that this is a joke� On the other hand� most ofwhat you did in �rst year was known to Newton in �� when he had more orless given up on Science and Mathematics as less important than Theology� The�rst arti�cial satellite had been invented by Newton many years before� It took theEngineers about three hundred years to catch up� Seen from that point of view�you are doing quite well�

�� THE DIFFUSION EQUATION ��

� Solve the equation

We next consider some simple cases of the �rst part� setting up theequation�

�� The Di�usion Equation

�� Intuitive

In this subsection I am going to give you a loose� intuitive� sloppyapproach� as done by all the best engineers and mathematicians� In thenext subsection I shall do it in a more respectable algebraic manner�so as to guarantee intellectual respectability� Some people worry aboutthses things�

Imagine� then� a long tube closed at both ends and containing a largenumber of bees which were put in at one end before it was closed� If xis used to measure the distance down the tube� t is the time� let fx� tmeasure the density of the bees at location x and time t� To get thedensity of the bees at a point� we take a little bit of tube of length�x centred on x� count the number of bees� and divide by �x� Thenwe take the limit as �x gets closer to zero� Anyone who objects toanything as silly as this on the grounds that the answer will almostalways be zero� and that bees take up some space and aren�t points� issimply refusing to enter into the spirit of things and will lose out onsome innocent fun�

If we put the origin� �� at the left hand end of the tube� let Nx� t bethe number of bees between location � and location x at time t� Thenwe have that �

fx� t ��Nx� t

�x

Now look at the bees in some such small slab� as shown in �gure ��

� � CHAPTER � PARTIAL DIFFERENTIAL EQUATIONS

x x x + Δ

+ 2 Δx

x

Escaped Bee

Figure �� Dynamics of a swarm of bees

We suppose that the bees move about at random� quite independentlyexcept that possibly they may bounce o� each other if they collide�They are just as likely to be going one way as another at any time� andthey buzz around in the way that bees� atoms and small children atparties are prone to do�

It is fairly plausible that the number of bees going from the slab betweenx and x��x into the slab to the right of it� from x��x to x� ��x�over any time interval from t to t��t� is proportional to the di�erencebetween the number of bees in the two slabs� The actual number ofbees will depend on such things as the mean bee velocity� but if half thebees are going one way and half the bees are going another� then therewill be approximately �x � fx�� bees going right across the barrierand �x�fx��x�� going to the left from the second slab� if the beesare going fast enough�

The rate of �ow of bees then past a point x will be simply proportionalto the rate of change of density at x� f

x� If the density is increasing�

the bees will tend to go backwards� so N will tend to increase and wecan write�

�N

�t� c�

�f

�x

�� THE DIFFUSION EQUATION � �

where N is the number of bees between � and x� and c� is a positiveconstant telling us something about the mobility of the bees�

Now we have that N is of course related to f � in fact f is the spacederivative of N � f � N

x� We therefore try to use these facts to say

something about the change of density in time�

Di�erentiating the last equation partially with respect to x� we get�

�

�x

�N

�t� c�

��f

�x�

and reversing the order of partial di�erentiation� which is OK if thefunction f is continuously di�erentiable� we get

�

�t

�N

�x� c�

��f

�x�

and given that we recognise the de�nition of f lurking in the equationwe can �nish up with�

�f

�t� c�

��f

�x�

This equation is known as the Di�usion Equation in one dimension�We can con�rm the reasonableness of it as a description of heat� atomsand even� to a crude approximation� bees� by experiment and argu�ment� Experiment is more convincing to everybody except theologiansand philosophers� and gives the expected answers� If you are a whizprogrammer� you can set up a program where there are a number ofslabs next to each other� say A�B�C� � � � and there are some number ofbees at time zero in each slab� say NA� NB� NC� � � �� The rules are thatthat there is a jump to time � during which each bee makes a randomchoice between moving into the preceding slab or vanishing if thereisn�t a preceding slab� moving into the following slab or vanishingif there isn�t one� or simply staying where it is� The probabilities ofgoing left or right are equal� Now iterate the process for some initialdistribution of the bees in the slabs and watch what happens� Eventu�ally all the bees leave� If you want you can make it circular so thereis a conservation of bees� or you can treat the end points di�erently�


You are doing a discrete simulation of the di�usion equation� which ispretty reasonable for bees� Note that the continuous approximation forbees is fundamentally daft but still works� and it works even better foratoms�

The chain of reasoning I have given is pretty much what the eighteenthcentury mathematicians did to justify the di�usion of heat along a rod�the main di�erence being they said it in French and left out the bees ��

Bees are reasonably well described by the Di�usion Equation� but soare a lot of other things� including heat conduction which is largelya matter of vibrating atoms� and hence the di�usion equation is alsoknown as the Heat Equation� The di�usion of gases through pipes andatoms of one substance in another� from dyes in water to doping agentsin silicon� are also described by the same equation� Bees are easier tovisualise� but perhaps not so important in the grand scheme of thingsas heat conduction or atoms� Much depends on your point of view�

The next stage of development of the argument is to consider a thinplanar slab of bees� which can now move in two dimensions insteadof being compelled to go either backwards or forwards� And the �nalstage for most books is to go to the full three dimensional case� wherethe bees can �oat free�

In order to treat the two and three dimensional cases it is necessary toconsider the space� R

� or R�� to be decomposed into little squares or

boxes in a manner which is by this time rather familiar to you�

�It is remarkable that the French did such a lot of the mathematics of thissubject� but you don�t know the half of it� Most of them weren�t mathematicians�they were lawyers� medics� engineers and blokes who generally speaking did it forfun in the evenings after a hard days work� �Gauss� who was not French� was aprivy councillor� If you know what a privy is� you are doubtless wondering howyou counsel them� but this is your problem� You have to have a fairly high IQto think that this sort of thing is entertaining� but it was thought to be the sortof activity which re�ective gentlemen should do� In England there weren�t anyre�ective gentlemen� the gentlemen were horsing around killing foxes� dressing up insilly clothes and fancy hairdos� and gambling� Much like modern Australia� exceptthat Australians chase balls instead of foxes and aren�t gentlemen� Of course� theydidn�t have television in those days�


�� Saying it in Algebra

Watch me like a hawk here� This is tricky but cool�

Suppose we have a three dimensional space and that there are bees�ying around in it� Let T x� t denote the density of the bees at locationx at time t� Let U denote a region of the space think of a solid ballshaped region if you want to visualise this� then by de�nition thenumber of bees inside the region u at time t is just�Z

UT x� t dV

The �ow of bees �ying into the region U at time t is� by de�nition�

�

�t

ZUT x� t dV

Each bee has to �y through the boundary of U to get into U � Thegradient �eld of T gives us the direction in which the density of bees isincreasing� bees will �y down the gradient just as in the one dimensionalcase� So the rate of �ow of bees into U is justZ

Uc��T � n dA

for some positive constant c�� By the Gauss Divergence Theorem� thiscan be written as�

c�ZU��T dV

Equating the two expressions for the �ow of bees into U we get�

�

�t

ZUT x� t dV � c�

ZU��T dV

and interchanging the partial derivative orders again�

ZU

�T

�t� c��T dV � �


If the integral of a continuous function f over every region U is zero�then f must be zero� Suppose it weren�t zero at some point x� Thenit must be non�zero in some little region around x� and if fx � ��take U to be the region around x where f is positive� Then

RU f � ��

contradiction� Likewise if fx � �� It follows therefore that

�T

�t� c��T

which is the di�usion equation in three dimensions� Note that theargument actually works for any dimension�

Now we do it for heat� Let T x� t denote the temperature of a point xat time t in some region of R �� This is a function T � R � � R �� R � Itgives rise to a gradient vector �eld on R

n which will change in time� Wewrite this as �T � It matters� because heat rolls down the temperaturehill�

If we �x� again� some de�nite region U � the amount of heat in theregion U in R

n is given by a simple rule� the speci�c heat of a solidis the amount of heat it takes to raise a unit mass of the solid by atemperature of �o� so in a region U if we assume the speci�c heat andthe density are constants� � and �� we conclude that the heat in theregion U at time t is given by

HU �ZU��T x� t dx

and we can regard the heat �ow into U as

dHU

dt�

ZU��

�T x� t

�tdx

HU is� for a given box U � just a function of time ��

Heat �ows into the box U down the temperature gradient at a rate pro�portional to the conductivity of the material� K say� and the gradient

�It might be a good idea to think of the amount of heat as the number of beesand the temperature as the bee density� with some constants thrown in�


of the temperature� �T � at some point x� is in the opposite directionto the heat �ow� If we want to get the vector telling us the rate of �owof heat at the point x at time t� we can call it v and write

v � �K�T

Now the rate of �ow of heat out of the box U is going to be

ZUv � n

where n is the outward normal� which is equal� by the Divergence the�orem to Z

Udivv � �K

ZU��T

This succession of �s is written� with a rather shaky excuse� as �� asbefore� It is clear that ��f is shorthand for� in the case of two variablesx and y� �f

x�� f

y��

We may therefore equate the heat �ow into the box� dHU

dtto the tem�

perature T in two di�erent ways�

dHU

dt� K

ZU��T �

ZU��

�T

�t

Or to put it another way�

ZU

K��T � ��

�T

�t

��

Since this holds for all U � the function inside the brackets must be thezero function� and so we get the general heat equation�

K��T � ��T

�t

� CHAPTER � PARTIAL DIFFERENTIAL EQUATIONS

where � is the speci�c heat� � is the density of the material and K isthe conductivity of the material� This is just the di�usion equation�but you have some information about the positive constants and theproperties of materials� Whether you prefer to think of temperature orbee density is entirely optional�

�� Laplace�s Equation

In the case where there is no heat supplied to or leaving the object� andthe system is in a steady state� we get the famous equation of Laplace�

��T � �

which for a general function f � R� �� R can be written at greater

length as

��f ��f

�x��f

�y��f

�z��

This equation also applies to a large number of other situations� in twodimensions it applies� to a good approximation� for nearly �at surfaces�to soap �lms� it also applies to the electric �eld produced by a set ofpoint charges except at the charged points themselves� to gravitational�elds similarly� and hence has importance in dynamics� Wherever thereis some sort of minimum energy con�guration there is often a func�tion satisfying Laplace�s Equation� or some equation approximated byLaplace�s Equation� describing the state�

Example ��

The function f � R � �� R given by fx� y � x�� y� satis�es Laplace�sEquation everywhere� The function with fx� y � x��y� does not� Anyconstant function does� and if g � R � �� R

� is linear and invertible� andf satis�es Laplace�s Equation� so does f � g�

Exercise ��

�� LAPLACE�S EQUATION � �

�� Show that fx� y � x�x��y� satis�es Laplace�s Equation whereit is de�ned� Sketch a portion of the graph�

�� Show that fx� y � sinx coshy satis�es Laplace�s Equation�Sketch a portion of the graph�

� Find two more functions which satisfy Laplace�s Equation on someregion in R

� and two which do not�

It is useful to think of �� as an operator which takes a function f �Rn �� R to a new function� ��f � R n �� R � Then we want to know�

what functions get killed by the Laplacian Operator�� That is� whichfunctions f get sent to the zero function by ��

Functions which satisfy Laplace�s Equation are known as HarmonicFunctions and the study of Harmonic Functions is called Potential The�ory� Many mathematicians have spent the best years of their lives �nd�ing out things about harmonic functions� mostly just from curiosity� butthe results are often very handy to engineers� so I shall mention a fewof them here�

First� we can see immediately that

ZU��f �

ZU�f � n

and that �f � n is just the derivative of f in the direction n� So fora harmonic function� these are always zero� We use the notation f

nfor

�f � n

If you think about what this means in dimension �� you can see thatif you draw any simple closed curve in the plane� and integrate theslope of any harmonic function f along the outward normal around thecurve� you have to get zero� It follows immediately that a function suchas x� � y� is not harmonic� since the unit circle centred at the originhas got f

na positive constant� It suggests that you might be luckier

with something like xy or x� � y� which at least has the right sort ofbehaviour at the origin� We can get a little more mileage out of this


by making it a little more complicated� if we look at two functions�f� g � R n �� R we can de�ne F � f�g� This multiplies the vector �eld�g by the value of the function f at each point� Now by straightforwardmanipulations�

divF �� f�g � f��g ��f�g

and

F � n � n � f�g � f�g

�n

for any vector n� If we take some region U and apply the divergencetheorem to F� we get�

ZUf��g ��f�g �

ZU

f�g

�n

for n the normal to the boundary� which equation is called Green�sFirst Identity�

Repeating this with the functions in the reverse order� to the vector�eld g�f we get

ZUg��f ��f�g �

ZU

g�f

�n

Subtracting this from Green�s First Identity we get

ZUf��g � g��f �

ZU

f�g

�n� g

�f

�n

This is called Green�s Second Identity� These are useful in Fluid Me�chanics for those with good memories�

Now putting f � g� a harmonic function� in the �rst identity we get

ZU�f� �

ZU

f�f

�n

Now suppose f is zero on �U � Then the right hand side is zero� andso the left hand side is zero too� Hence �f � � and so f is constant�Since it is zero on the boundary� f � ��

�� LAPLACE�S EQUATION � �

This tells us that if f is harmonic and is zero on the boundary of aregion� it is zero throughout the region�

If f� g are two harmonic functions on U which agree on �U � then theirdi�erence is zero on the boundary� so their di�erence is zero everywhere�so the functions are equal� In other words�

Theorem ��

The solution to a Dirichlet problem for Laplace�s Equation is unique ifit exists�

A third nice fact about Harmonic functions is that if you take a pointx and look at the value at fx and then take a sphere centred on xand integrate the value of f over the sphere� then the result is alwaysequal to the value of fx multiplied by the area of the sphere� In twodimensions� it is a circle instead of a sphere� and the perimeter of thecircle is what we must multiply by� A simpler way to put this is that theaverage value on the boundary of a ball is the same as the value at thecentre of the ball� This result follows from integrating the directionalderivative normal to the sphere around the sphere to get zero for anin�nite set of spheres� and integrating� This result gives a nice parallelalgorithm for �nding a solution to the Dirichlet Problem for Laplace�sEquation in the plane�

Take a circle and suppose f is known on the circle and we want toextend it to the disk� Make a grid of processing elements� each joinedto its nearest neighbours as in �g� �� The elements which intersect thecircle are shaded� put in each element a number giving the value of thefunction fx� y at that point� Put random numbers in the processingelements inside the circle� and ignore any elements outside the circle�

An iteration of the system takes any element in the interior of thedisk and replaces the value inside it by the average of the values of itsfour neighbours� Elements on the circle itself have the numbers leftunchanged� We simply iterate this process� Eventually the numbers onthe inside stop changing� and when they do the grid gives a discrete


Figure �� Numerical Computation of Laplace�s Equation on a SystolicArray

approximation to a solution to Laplace�s Equation� Anyone who likesprogramming can fake the parallel processing on a PC� If you don�tlike the precision� do it with more processing elements� If you wantto generalise to something more complicated than a circle� the generalprinciple is clear� and if you want to increase the dimension� it is easyto see how to modify the algorithm�

Real hardware parallel machines �Systolic Arrays� have been builtat Stanford and Carnegie�Mellon Universities� Students of Robotics atthis University have used the method in software for �nding trajectorieswhich avoid obstacles� You can try this for the problems in the nextsection to get out numerical solutions� They are not as neat as analyticsolutions in terms of functions of course�

Exercise ��

Show that the only harmonic functions on the line are a�ne �linearplus a shift� and con�rm the third nice fact for this case�

Con�rm the third nice fact by integrating around the circle of radius rcentred on �a�b�� for the function fx� y � xy�

�� THE DIRICHLETPROBLEM FOR LAPLACES EQUATION��

a

b

Figure �� Soap Film on a rectangular wire

�� The Dirichlet Problem for Laplaces

Equation

We cannot hope to solve the general Dirichlet Problem for Laplace�sEquation� but we shall treat a few simple cases�

Suppose we take a rectangle in the plane� and lift up one of the sides ofthe rectangle by a function� See �gure �� It will be useful to think ofit as a wire frame� We are going to �nd the equation of the soap �lmwhich will be formed when the whole thing is dipped in soap �

Formally� we have � � x � a� � � y � b as the region U � We have thatthere is some unique function f � U �� R which is unknown� but thatwe have�

�I am simplifying here� the Partial Di�erential Equation for Soap �lms or areaminimisation is nonlinear� the general problem for solving it for given boundary conditions is known as Plateau�s Problem� The PDE is approximated wellby Laplace�s Equation provided the nonlinear e�ects are small� which will happenif the function f is not too di�erent from an a�ne function� and solving Laplace�sEquation for a given boundary condition is often a good start on the Plateau Problem� From now on� I shall cheerfully talk of soap �lms as if they were exactly solvedby Laplace�s Equation


�� fx� � � �� x � �� a�

�� f�� y � �� y � �� b�

�� fx� b � �� x � �� a�

�� fa� y � hy� �y � �� b�

�� fx�

� �fy�

� �

for some given function h� I have illustrated h in �gure �� with anice parabolic function� but let us keep h general at the moment� Theproblem is to �nd f � the soap �lm function� I remind you that thisis not being done because we care about soap� but because very muchmore signi�cant problems can be done using the same methods� and itis useful to have clear pictures of a simple sort in your mind�

The �rst thing we do is make an assumption which is not immediatelyjusti�able or even reasonable� but which actually works�

Separation of Variables

Suppose that the function fx� y can be written as a productof functions of x and y separately

Write

fx� y � pxqy

Then di�erentiating partially with respect to x gives

�f

�x� qy

dp

dx�

��f

�x�� q

d�p

dx�

and similarly

�f

�y� px

dq

dy�

��f

�y�� p

d�q

dy�


Now Laplace�s Equation gives us

q#p� p#q � �

Or#p

p� � #q

q� c

for some constant c�

Thus we have reduced the partial di�erential equation to two simulta�neous Ordinary Di�erential Equations� #p � c p and #q � �c q� which weknow how to solve��

The boundary conditions for these can be worked out from the bound�ary values for the original PDE� we have that f�� y � p�qy � ��for all the y � �� b�� and since q � � will not be a solution at x � a�we must have p� � �� Similarly q� � �� qb � �� We also deducethat paqy � hy� You may be coming to feel that what is goingto happen is that we are going to have that nice parabola at fa� ysimply scaled down progressively to zero as we reduce x to zero� Thisseems physically reasonable�

First we solve#q � �c q

a familiar old face� We recall that the general solution is

qy � A sinpcy �B cos

pcy

if we suppose c is positive� and we just swap p� q otherwise�

Now

q� � �� B � �� and qb � �� c � n�

b�� n � ��

I have used the notation �p and �q rather casually� we are di�erentiating withrespect to di�erent variables here� I interpret the dot as �Di�erentiate with respect tothe �single variable�� Other� sterner� folk insist that we use Newton�s dot notationonly when the variable is time� I have tried it in other notations� and it is longerand harder to read�


So we get solutions for q of the form

qy � An sinn�y

b

Going back to the equation for p� we have that

px � C sinhpcx �D cosh

pcx

is the general solution and we know that p� � � and hence thatD � ��

But we now know something about c� from the boundary conditionson q� so we can conclude that for every positive integer n� there is asolution in waiting�

fnx� y � cn sinhn�x

b sin

n�y

b

These �solutions in waiting� as I have called them� exist for all positiveintegers n� and for all constants cn� and the sum of any set of themis also a solution�in�waiting� In order to be a real solution� we haveto �nd a sum of them which also satisfy the �nal remaining boundarycondition� they have to agree with the function h� the parabola in the�gure� We cannot reasonably expect the sum of a �nite collection ofsine functions to be a parabola� but we can get closer and closer� weare just doing Fourier Theory�

This means that

fa� y ��Xn��

cn sinhn�x

b sin

n�y

b � hy

Each Fourier coe�cient is cn sinhn�xb� given by

cn sinhn�a

b �

�

b

Z b

�hy sin

n�y

b dy


1

π

Figure �� Function on the boundary

If we can do the integrals� we can calculate the coe�cients as far as welike� and in some happy cases we can get explicit solutions�

We can get a reasonable agreement with �gure �� if we put b � � andhy � siny� We therefore work through the following example�

Example ��

Problem

Let the harmonic function f � �� satisfy the following boundaryconditions

�� fx� � � ��x � ��

�� f�� y � ��y � ��

� f�� y � ��y � ��

� f�� y � siny��y � ��

Sketch the graph of f on the boundary of the rectangle� and calculatethe function f on the interior�


Solution The graph is illustrated in �g ��

We suppose that the solution is separable� fx� y � pxqy�

This tells us that#p

p� � #q

q� c

and we have no way of knowing whether c is positive or negative until weinvestigate the boundary conditions� since we could always interchangex and y in the problem�

Since we know that at x � � we have that f�� y � siny� we havep�qy � siny� which tells us that p� � � and

qy � siny

is a possibility� with c � �� In which case�

#p

p� �

and we havepx � A coshx �B sinhx

is a solution in waiting� It� or some �possibly in�nite� sum of suchsolutions� must satisfy the boundary conditions not so far used� Theseare straightforward we must have

p� � �� p� � �

and this immediately tells us that

A � �� B ��

sinh�

is a solution� So the �nal solution is

fx� y �sinhx

sinh�siny

It is straightforward to verify �� that the boundary conditions are allsatis�ed and �� that the function is harmonic �everywhere�� Since wehave a uniqueness theorem� we have produced the only possible solution�


Exercise ��

Verify that the given solution satis�es Laplace�s Equation�

If you have any soul in you at all� you will now stand up and clap forhalf an hour at something so wonderful�

Example ��

Problem

Let a square of side � units be made of metal� and let three of the foursides be kept at a temperature of �o� Let the last side have temperaturesinx at distance x along from one end� Find the temperature at thecentre of the plate when the system is in equilibrium�

Solution

e�� e��

e� � e��

This is essentially the same as the last worked example of course� Onlythe names have been changed to protect the guilty�� There is a scaledversion of the last solution because the plate is bigger� I have expandedthe sinh function out for those of you who like to see everything in termsof exponentials�

Go on� check it out� then clap" It isn�t quite as wonderful as Euler�sFormula� ei� � �� but it is pretty damned smart and can be veri�edby experiment� which is more than can be said for the Euler Formula��

The innocent are not in need of protection� their strength is as the strength often� because their hearts are pure�

�What is marvellous isn�t the answer� it is that somebody of the same speciesas you was smart enough to �gure it out� and you are smart enough to follow theargument� If this doesn�t strike you as astonishingly wonderful� you are probablydead but haven�t noticed yet�


2

π-

2

π-

X

Y

Z

Figure �� Function on the boundary

Exercise ��

�� Suppose the function f of the preceding worked examples had beenmodi�ed so that it is de�ned on the interval

��

and is zero along two opposite sides and is a cosine function alongthe other two opposite sides� as in �g �� Find an explicit formfor the harmonic function from �rst principles�

�� You have calculated the Fourier Series for a certain number offunctions by now choose some function where you have the FourierSeries already worked out and use it as an alternative to my func�tion hy � siny to obtain a Fourier Series solution to your veryown soap �lm problem�

� The constraints on the boundary look rather strong� and you mightwonder what you could do with a case where one edge was �xed tohave height h�y and the opposite edge was �xed to have heighth�y� Deal with the case where one end of a square of side � ismade to have height siny� the opposite is made to have height� siny and the other two are kept at height zero� Do this by

�� LAPLACE ON DISKS ��

�nding solutions to �a� the case where three sides are zero andone side is at height siny� which has been done� �b� the casewhere the opposite side is kept at height � siny and all othersides kept at zero and then �c� adding up the answers� After all�if two functions satisfy Laplace�s Equation� so does their sum��Of course� if the two functions h�� h� are the same� there might bea quicker method� as an earlier h�� h� are the same� there mightbe a smarter method� as in our earlier worked example�

� Laplace on Disks

Take the map P � R� �� R

� which takes r� � to x � r cos �� y �r sin �� otherwise the polar coordinates transformation� Suppose werestrict the map to R

� � �� as usual so as to make it one�one ontothe plane R

� except for a small problem at the origin� If a functionf � R

� �� R satisi�es Laplace�s Equation� what equation does f � Psatisfy� The answer is Laplace�s Equation in Polar coordinates� andit is worth knowing� because it gives us a chance to do for disks andsectors what we have just done for rectangles�

The map P has derivative�

cos � �r sin �sin � r cos �

�

We can write��f

�r��f

��

��

��f

�x��f

�y

� cos � �r sin �sin � r cos �

�

Now inverting the matrix by appealing to the Inverse Function Theo�rem we get�

��f

�x��f

�y

��

��f

�r��f

��

� cos � sin �

��r sin � ��r cos �

�


Or more fully��

�x� cos �

�

�r� �

rsin �

�

��

�

�y� sin �

�

�r�

�

rcos �

�

��

In particular

�f

�x� cos �

�f

�r� �

rsin �

�f

��

��f

�x�� cos �

�

�rcos �

�f

�r� �

rsin �

�f

��

� sin �

r

�

��cos �

�f

�r� �

rsin �

�f

��

��f

�y�� sin �

�

�rsin �

�f

�r�

cos �

r

�f

��

�cos �

r

�

��sin �

�f

�r�

cos �

r

�f

��

When these are evaluated and added many terms cancel out and weget�

��f

�x��f

�y��f

�r��

�

r

�f

�r�

�

r��f

��

Now if the left hand side is zero we get the Polar Form of Laplace�sEquation�

��f

�r��

�

r

�f

�r�

�

r��f

��

Which you should memorise�

Exercise ��

Complete the above calculation to derive for yourself the Polar form ofLaplace�s Equation�


Figure �� Soap Film on a circular wire

Now suppose we have a piece of circular wire bent so that its projcctionis a circle� as in �gure �� The shape indicated can be represented asthe graph of a function h � S� �� R � We assume again that a functionf � D� �� R exists with the following properties�

�� frr ��rfr �

�r�f��

�� f�� h�

�� fr� � � prq�

The �rst is just the Polar form of Laplace�s Equation� the second saysthat we know the value of a function satisfying the equation on theboundary of the disk� and the last says that the variables are separa�ble� This has the status of pious hope at this stage� I have usedthe shorthand notation for the partial derivatives partly because I amcrapped o� with the TEX expressions for the other form� and partlybecause it will be good for you to have to practise with it�

Putting the �rst and the last together� we get

r�q#p� rq �p� p#q � �


� r�q#p� rq �p � �p#q� r�

#p

p� r

�p

p� k

and � #q

q� k

For some constant k� We therefore have again reduced the originalPDE down to two ODEs�

r�#p � r �p� kp � �� #q � kq � �

both of which look fairly straightforward�

We consider the possibilities for k� it can be negative� positive or zero� Ifit is zero� we rapidly deduce that q� � m��c and this can only meanthat q is constant� otherwise the function could not be continuous onthe boundary� It has to have q� � q�� But if q does not depend on�� the height function around the circle would also have to be constant�If the constant were zero� there is a unique solution� the zero function�similarly� if the function is constant on the boundary it has to have thesame constant value throughout the interior� This is possible but notexciting enough to contemplate further�

If the constant k is negative� we get #q � ��q � for positive �� withexponential solution

q� � c�e � � c�e

� �

which is impossible to have continuous on the boundary for positive �except in the thoroughly uninteresting case when c� � c� � ��

Thus we may conclude that k � �� This forces solutions to the equationin q to be periodic�

q� � c� sinpk� � c� cos

pk�

sopk must be a positive integer n�


Going now to the equation for p�

r�#p� r �p � ��p � �

This is easily seen to have solutions of the form

pr � Ar �Br�

The r� terms go o� to in�nity as r � �� so we are left with termswhich have to be of the form rn for positive integers n�

Thus we conclude that any solution must be a sum of such solutions�so we get�

fr� � � A� ��Xn��

rnAn sinn� �Bn cosn�

And in order to get the given function h on the boundary� we have

f�� h� � A� ��Xn��

An sinn� �Bn cosn�

Which means that we have its ordinary Fourier Series� hence

An ��

�

Z ��

�h� sinn� d�� n � ��

and

Bn ��

�

Z ��

�h� cosn� d�� n � ��

The problem is solved� you may now cheer wildly and scream yourselveshoarse in support of something pretty smart�

Exercise ��


�� Suppose the function de�ned on S� in �gure �� is smoother thanit looks and is actually just sin �� Find the unique extension tothe disk which satis�es Laplace�s Equation�

�� Suppose we are given a semicircle� r � �� r � �� Suppose the temperature is maintained at zero onthe diameter� and is given by h� on the arc� Show how to solveLaplace�s Equation for this case�

� If in the above problem� h� � sin�� sketch the solution andcalculate it exactly�

� We are given a unit disk made out of metal� Suppose that the tophalf of the unit circle on the disk is kept at a temperature of ��o

and the bottom half at �o� Find the steady state temperature inthe inside of the disk� Be suitably �u�y about what happens atthe points where the two temperatures are adjacent�

� Solving the Heat Equation

The heat equation is more general than Laplace�s Equation� so a littlemore complication is to be expected� Recall that we had�

c��f � ft

In the case of a one dimensional bar� this comes to

c�fxx � ft

and we investigate this case �rst�

Let us suppose that we take a bar of metal of length �� and heat it up insome way so that the temperature at location x is hx for � � x � ��Let us suppose that the ends are kept at zero temperature from thestarting time to the end of time� If you have a bar of some di�erentlength� just change your units� If you want to know what temperatureunits I am using� it doesn�t much matter� although perhaps it would be

�� SOLVING THE HEAT EQUATION ��

a good idea to avoid the Kelvin scale since negative temperatures mightbe convenient and we would prefer some realism� I shall also assume�again in a spirit of optimism� that the function f can have its variablesseparated� i�e is a product of two functions� one of space and the otherof time� Then we have the boundary value problem of a function fsatisfying the following conditions�

�� c�fxx � ft

�� f�� t � �

�� f�� t � �

�� fx� � � hx

�� fx� t � pxqt

It is useful to draw the diagram of �gure �� in order to see the twodimensions of the problem�

Then we have�

fx� t � pxqt� fxx � q#p and ft � p �q

This givesc�q#p � p �q

So

c�#p

p�

�q

q� k

for some constant k�

This gives us two ODEs� just as for the case of Laplace�s Equation�

c�#p� kp � �

and�q � kq � �


Space

Time

Figure �� The Heat Equation for a rod

The second has solution qt � Aekt� which tells us that k � �� since arunaway temperature� for example� is hard to credit� If we put k � ��the �rst equation is

#p � ��

c�p

which has solution

px � A cos�

cx �B sin

�

cx

It is time now to apply the boundary conditions so as to get someinformation about �� Since f�� t � � for all t� we get p� � � andhence A � �� Since f�� t � �� we get

�

c� n�� n � ��

This gives a family of solutions to the Heat Equation�

fnx� t � cne�n��c�t sinn�x

The general solution is some in�nite sum of these for a choice of cnwhich makes the initial state at t � � equal to the given functionhx� So we choose to expand the given hx on the interval �� in


terms of sin functions� which gives us the required cn� and we are done�This will require some rescaling� since mostly you will have done yourFourier Theory on the interval �� rather than on �� but this isnot particularly di�cult�

Example � �� Suppose the initial value at time t � � is hx �sin�x� Then c� � �� and otherwise cn � �� So the complete solutionin this case is

fx� t � e��c�t sin�x

So at time t � � the temperature at the middle of the bar is

e��c�

I hope you agree that this is pretty clever stu��

Instead of keeping the ends at any temperature� we can insulate them sothat no heat leaves the bar� We can express this in terms of conditionsthat the derivative of the function px must be zero at the ends of thebar� This leads to a similar equation but with cosine terms instead ofsine terms as the solution� We give an easy example of this kind ofproblem� worked through from �rst principles�

Example � ��

Problem

Suppose an insulated rod of length �� units is heated to a temperatureof � � jxj for a distance x along the rod from the centre� We do notkeep both ends of the rod to be at zero for all time� but instead ensurethat no heat leaves the bar anywhere� but we remove the heater at timezero� Given that the density� speci�c heat and conductivity of the barare �� K respectively� calculate from �rst principles the temperatureof the point one quarter of the way along the bar after time units haveelapsed� What happens to the temperature at the ends of the bar�

Solution


Note that I have changed the length of the rod and measured from itscentre to make the sums easier�

K��T

�x��

�T

�t

is the heat equation in one dimension� where x is the distance along thebar� t is the time� and T x� t is the temperature at the point x at timet�

Writing T x� t � pxqt on the assumption of separability of vari�ables� This becomes

Kq#p � ��p �q

After rearranging we obtain

#p

p��

K

�q

q� k

for some constant k� The second equation we may solve immediately togive

qt � eKk�� t

and since �� K are positive� k must be negative for a physically intel�ligible solution� Putting k � �� the �rst equation then has solutionsof the form

p x � A cos�x �B sin�x

and any sum of such solutions for any number of di�erent � will alsobe a solution� We wish next to use the boundary conditions to work outwhat restrictions are implied on the possible � and the constants�

We have from the fact that the function T x� � � pxq� is symmet�ric� that p is symmetric� and hence that B � �� for any value of ��We also have the condition �p�� p� � �� since no heat leaves therod at the ends� This requires that

A � sin ��


which requires that � be an integer� which may be zero� There is noparticular reason to have negative integers� so we shan�t�

We seek next a Fourier expansion of the function ��jxj on the interval�� This is because we want to know what sum of functions

A� ��Xi��

Ai cos ix

can be the function px which is T x� � � � � jxj�

Thus the general solution so far is a sum

Xi��

Ai cosixe� Ki�

��

!t

To �nd the right sequence of coe�cients to satisfy the initial conditionfor the temperature distribution� that is� to calculate the Fourier Seriesfor p� we need to calulateZ �

��cosnx� � jxj dx

which is most easily accomplished by observing that the function is sym�metric about the origin and so is simply

�Z �

�cosnx� � x dx

� ��Z �

�cos nx dx� �

Z �

�x cos nx dx

which for n � � is

��Z �

�x cos nx dx

and integrating by parts we get

��x

nsin nx�

�

n�cosnx

��


��

n�� cos n��

which is zero when n is even and ��n� when n is odd� When n � �we have simply the area under the graph of � � jxj between �� and �which is �� We therefore have that the Fourier Series for � � jxj on�� is given by

� � jxj � �

��

�

�cos x�

�

��cosx�

�

��cosx� � � �

or if you prefer it more formally

p � �

��

�

�

�Xn��

�

�n � ��cos�n � �x

Thus we conclude �almost� by writing down the solution to the heatequation as

T x� t � �

��

�

�

�Xn��

�

�n� ��cos�n � �xe�

K��n��

��t

It is easier to verify that the space function p and the time function qboth satisfy� termwise� the required ODEs than it is to produce them�If you believe that I got the Fourier expansion of � � jxj right� then itfollows that

�

�

�Xn��

�

�n� ��

�

by evaluating at x � �� Alternatively� the sum of the reciprocals ofthe squares of the odd numbers is �� This doesn�t look very likelyo�hand� and you can check it out by adding up a few thousand termsto see if it is probably right�

The original problem �which you have probably now forgotten� it wasso far back in space and time� was to say what the temperature is ata point quarter of the way along the rod at time t � �� This meansthat x � �� is the point we care about� and we notice that cos�n �

�� AND IN CONCLUSION��

�� This tells us that the temperature at this point does notactually change in time� and that it started at �� and will remain soinde�nitely� This is reasonable� since what one would expect to happenis that� with no heat leaving or entering the system� the temperaturewill tend to uniformity� �Imagine all those trapped bees�� And theaverage temperature is �� to start out with� so the proposition thatthe bar will wind up at that temperature everywhere is believable� as isthe proposition that the point of average temperature will stay that way�This works because my original function p is linear over each half of thebar and symmetrical� To have noticed this at the beginning and simplywritten down the answer would have shown some genuine talent� Ifyou noticed this� congratulations� you are either very practised or verysmart�

The temperature at the end points is

T �� t � T �� t � �

��

�

�Xn��

�

�n� ��e�

K��n��

��t

which tends to �� from � as time goes by� Note that if K were � andno heat �owed� nothing would happen� If it is small� whatever happens�happens slowly� This seems reasonable�

�� And in Conclusion��

I have just got you to stick your toes in the water as far as PDEs areconcerned� There are people who make a lifetime�s work of solvingLaplace�s Equation for progressively more complicated boundary con�ditions� and they feel their time is well spent� There are applicationsof PDEs in mining� working out from gravity surveys where the bodythe ore body is buried� in Electromagnetism� in Quantum Physics�in studying waves in every medium you can imagine and a good fewyou can�t� in the twisting and bending of solids� and the list goes on�All I have had time to do is to lead you into the shallow water so you


can have a paddle with the kiddies� All the same� you have seen somepretty amazing things if you stop to think about it�

I hope you enjoyed the ride� even if it was a bit hectic in places"

Documents

Alder - Lectures on Integration of Several Variables (Cálculo)