Handouts Div Curl Spherical Double in Teg

8/13/2019 Handouts Div Curl Spherical Double in Teg

http://slidepdf.com/reader/full/handouts-div-curl-spherical-double-in-teg 1/26

Calculus and Anal tic

Muhammad Ali Jinnah University,

Islamabad Campus

Geometry

MT 1043 – Fall 2012

.

Department of Mathematics

Mohammad Ali Jinnah University

1. Introduction (Grad) The vector differential operator ∇, called “del” is defined in

three dimensions to be:

Note that the components are partial derivatives

If a scalar function, f ( x, y, z ), is defined and differentiable at all

points in some region, then f is a differentiable scalar field. The

del vector operator , ∇, may be applied to scalar fields and the

result, ∇ f , is a vector field. It is called the gradient of f .

8/29/2013 11:06 AM R. Ali |Calculus & Analytic Geo 3



1. Introduction (Grad)


1. Introduction (Grad)

8/29/2013 11:06 AM 5R. Ali |Calculus & Analytic Geo



2. Divergence (Div)

If F( x, y) is a vector field, then its divergence is written asdivF( x, y) = ∇·F( r) which in 2-dimension is:

y x F y x F y x +⋅∂

+∂

=⋅∇ )),(),(()(),( iiF

It is obtained by taking the scalar product of the vector

operator and the vector field F( x, y). The divergence of avector field F is a scalar field.

= 2 +

y

F

x

F y x

∂

∂+

∂

∂=

∂∂21

,

8/29/2013 11:06 AM 6

26)2()3(

),(

2

21

+=∂

∂+

∂

∂=

∂

∂+

∂

∂=⋅∇

x y y

x x

y

F

x

F y xF

R. Ali |Calculus & Analytic Geo

2: Divergence (Div)

Exercise 1. Calculate the divergence of the vector fields F( x, y)

z

F

y

F

x

F z y x

∂

∂+

∂

∂+

∂

∂=⋅∇ 321),,(F

, , .




3. Curl The curl of a vector field, F( x, y, z ), in 3-dimensions

may be written curlF( x, y, z ) = ∇×F( x, y, z )121323),,(

F F F F F F z y x ⎟

⎞⎜⎜⎛ ∂

−∂

+⎟ ⎞

⎜⎛ ∂

−∂

−⎟ ⎞

⎜⎜⎛ ∂

−∂

=×∇ k jiF

321 F F F z y x ∂

∂

∂

∂

∂

∂=

k ji


It is obtained by taking the vector product of the

vector operator applied to the vector field F( x, y, z ).The curl of a vector field is a vector field.

∇×F is sometimes called the rotation of F and written as rot F

CurlExample: Given the vector field F( x, y, z ) = 3 x2 i + 2 z j – x k, the

curl of F is:

8/29/2013 11:06 AM 9

Quiz: Which of the following is the curl of F( x, y, z ) = x i + y j + z k

(a) 2i – 2j + 2k, (b) x i + y j + z k (c) 0 (d) i + j + k




Class Assignment1. Calculate the curl of the following vector fields F( x, y, z )

(a) F = x i – y j + z k, (b) F = y3 i + xy j – z k,

(c) , (d) F = i + 2 z – y ,

2. Let f be a scalar field and F( x, y, z ) and G( x, y, z ) be vector

fields. What, if anything, is wrong with each of the following

expressions:

222 z y x ++=F


(a) (b)

(c)

y x f 43 −=∇ k jiF z y x −−=⋅∇ 2

FG ⋅∇=×∇

Class Assignment3. Find the divergence of G = 2 x3i – 3 xy j + 3 x2 z k

4. Find the divergence of r/r 3 where r = |r| and

r = x i + y j + z k

5. Find the curl of F = x2i + xyz j – z k at the point

(2,1,– 2).

6. Show that the following vector field is irrotational,

i.e. its curl is zero


F = y i + ( x – z ) j – y k



Solutions to Exercises (1-a & 1-b)


Solution- C.A. 1-a




Solution- C.A. 1-b


Solution- C.A. 1-c




Solution- C.A. 1-d


Solution- C.A. 2-a & b




Solution- C.A. 2-c


Partial Derivatives





Multiple Integrals




d cb xa R ≤≤≤≤ ,:

Double IntegralsDouble Integrals over Rectangles:

The simplest type of planner region is a rectangle. Given a function f ( x, y) defined over a rectangular region R :

subdivide R into small rectangles using

a network of lines parallel to the x- & y-axes.

These rectangles form a partition of R .

A small rectangular piece of width ∆ x &

height ∆ y has area ∆ A = ∆ x∆ y

8/29/2013 11:06 AM 22

The Riemann Sum over R can be written as:

When the limit as n→∞

, (or || P ||→

0 or∆ A →

0), it is called thedouble integral of f over R written as


Double Integrals as VolumesWhen f ( x, y) is a +ve function over a rectangular region R in the

xy-plane, the double integral of ƒ over R may be interpret as the

volume of the 3-dimensional solid region over the xy-plane

oun e e ow y an a ove y e sur ace z = , y

Approximating solids with rectangular

boxes leads us to define the volumes of

more general solids as double integrals.

The volume of the solid shown here is the

8/29/2013 11:06 AM 23

ou e integra o ƒ(x, y over t e aseregion R.




Double Integrals as Volumes This is because: As n increases, the Riemann sum

approximations approach the total volume of the solid shown


Fubini’s Theorem for Calculating Double Integrals

Where A( x) is the cross-sectional

area at x. For each value of x, we may

calculate A( x) as the integral


Which is the area under the curve z = 4 – x – y

in the plane of cross-section at x. In calculating

A( x), x is held fixed and the integration takes place w.r.t

y.





Fubini’s Theorem (First Form)

Example:

8/29/2013 11:06 AM 28

C.A. 1- Repeat calculations by reversing the order of integration (Sol.)


Double IntegralsDouble Integrals over Bounded Non-

rectangular Regions: A rectangular grid

partitioning a bounded nonrectangular

region into rectangular cells (see fig.).

As before we have the Riemann sum:

Its limiting value as n →∞

is the double integral of f ( x, y) over the

region R.

8/29/2013 11:06 AM 29

The Additivity Property for rectangular

regions holds for regions bounded by

continuous curves. (see fig.)




Double Integrals as Volume

If f ( x, y) is positive andcontinuous over R we

define the volume of the

solid region between R

and the surface z =f ( x,y)

to be ∫∫ R f ( x,y) dA, as

before. (see fig)


Let R be the region bounded above and below by the curves

y = g 1( x) & y = g 2( x) and on the sides by the lines x = a & x = b ,

we can again calculate the volume by the method of slicing :st -

Double Integrals

. .

area A( x) and then integrate this area

from x=a to x=b. The area of the

vertical slice shown is:

an get t e vo ume as t e terateintegrals




Double IntegralsSimilarly if R is the region bounded curves x = h1( y) & x = h2( y)

and the lines y = c & y = d , then the volume calculated by themethod of slicing is given by the iterated integrals


Fubini’s Theorem (Stronger Form)




Example 2Find the volume of the prism whose base is the triangle in the xy-

plane bounded by the x-axis and the lines y = x & x = 1 andwhose top lies in the plane:

= = – – ,

Sol.


Evaluate

Where R is the triangle in the xy-plane bounded by the x-axis the

line y = x and the line x = 1.

Example 3

.

(Solution)

If we reverse the order of integration,

we’ve to calculate:

It cannot be expressed in terms of elementary functions (there is no

simple antiderivative). There is no general rule for predicting which order of integration will be

the good one.

If the order you first choose doesn’t work, try the other.

Sometimes neither order will work, & then use numerical approximations.




Finding Limits of Integration

2. Find the y-limits of integration. Imagine a

8/29/2013 11:06 AM 36

ver ca ne cu ng roug n e

direction of increasing y. Mark the y-values

where L enters & leaves. These are the y-limitsof integration and are usually functions of x

(instead of constants).


Finding Limits of Integration3. Find the x-limits of integration. Choose x-

limits that include all the vertical lines

through R. The integral shown here is




Finding Limits of Integration

To evaluate the same double integral as an

iterated integral with the order of integrationreversed, use horizontal lines instead of vertical

lines in Steps 2 and 3. The integral is


Example 4:Sketch the region of integration for the integral

& wr te an equ va ent ntegra w t t e or er o ntegrat on

reversed.

Sol. The region of integration is given by the inequalities x2 ≤ y ≤ 2 x and

0 ≤ x ≤ 2. It is therefore the region bounded by the curves y = x2 and y = x

between x = 0 & x = 2.




Example 4 (cont.):

C.A. 2. Solve these integrals.

Solution (a)


o u on . .

Properties of Double Integrals




Home Assignment

Exercise 15.1: Q.1 to Q.10

Q.13, Q.16, Q.17, Q.19

Q.22, Q.24, Q.26, Q.30


Sol. C.A. 1




Sol. Example 2:For any x between 0 and 1 , y may vary from y=0 to y = x (fig. b)

Hence,

When the order of integration is reversed (Fig. c) the integral for the volume

is:


Sol. Example 3:




Sol. C.A. 2 (a):

∫∫ ∫ +=+

2

2

0

22

0

2 ]24[)24( 22dx y xydydx x x

x

x

x

+−+=0

22))](2)(4())2(2)2(4[( dx x x x x x x

)464(

2

0

23 ++−= ∫ dx x x x


8)000()81616(2

4

3

6

4

4

0

234

=++−−++−=

⎥⎦

⎢⎣

++−= x x x

Cylindrical Coordinates




Example

Describe3

. ,

2. θ = θ o,

3. z = z o

Constant-coordinate equations

in cylindrical coordinates

yield cylinders and planes.


Example

Express x2 + ( y – 1)2 = 1 in cylindrical coordinates.




Spherical Coordinates


Spherical Coordinates

Constant-coordinate

e uations in s herical

coordinates yield spheres,

single cones, and half-

planes.




Converting Cartesian to Spherical


Converting Cartesian to Spherical


Documents

Handouts Div Curl Spherical Double in Teg