Math 324: Multivariable Calculusmrrobins/324Winter18/notes.pdf · Math 324: Multivariable Calculus Matt Robinson February 26, 2018 1. CONTENTS CONTENTS Contents ... where the piece

Math 324: Multivariable Calculus

Matt Robinson

February 26, 2018

1

CONTENTS CONTENTS

Contents

1 Double Integrals over Regions 31.1 Double Integrals and Riemann Sums . . . . . . . . . . . . . . . . . . 31.2 Integration in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 101.3 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . 131.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Triple Integrals 242.1 Applications of Triple Integrals . . . . . . . . . . . . . . . . . . . . . 30

3 Coordinate Systems 313.1 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Derivatives 444.1 Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 464.2 The Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Vector fields 51

6 Integration over Curves 546.1 Integrating Vector Fields over Curves . . . . . . . . . . . . . . . . . . 586.2 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7 Operations on Vector Fields 707.1 The Del Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

8 Integration over Surfaces 768.1 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808.3 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858.4 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 878.5 The Fundamental Theorems of Vector Calculus . . . . . . . . . . . . 90

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1 DOUBLE INTEGRALS OVER REGIONS

1 Double Integrals over Regions

In this section we build off what was covered in Math 126, namely with a review ofdouble integrals, iterated integrals, integration in polar coordinates, and applicationsof double integration.

1.1 Double Integrals and Riemann Sums

Recall that if y = f(x), then

ˆ b

a

f(x)dx = limn→∞

n∑j=1

f(xj)∆xj

gives the signed area under the curve f(x) on the interval [a, b].

Now consider a function z = f(x, y) and a rectangle R in the xy-plane. Then

¨R

f(x, y)dA

represents the signed volume under f(x, y) on R.

3

1.1 Double Integrals and Riemann Sums 1 DOUBLE INTEGRALS OVER REGIONS

In terms of Riemannan sums, we then have

¨R

f(x, y)dA = limm,n→∞

m,n∑j,k=1

f(xj, yk)∆xj∆yk,

where as m,n→∞, the area ∆xj∆yk → 0 for each j ≤ m, k ≤ n.We will not do Riemann sums in class, but let’s recall how to compute double

integrals in practice.Recall that an intergral of the form

ˆ d

c

ˆ b

a

f(x, y)dxdy,

is called an iterated integral and is computed as follows

ˆ d

c

ˆ b

a

f(x, y)dxdy =

ˆ d

c

[ˆ b

a

f(x, y)dx

]dy,

where the piece in the inner bracket is single-variable function in y.If R = [a, b]× [c, d] is a rectangular region in the plane, then we need to a way to

convert the double integral ¨R

f(x, y)dA

into an iterated integral for which we can explicitly compute.

4


Theorem 1 (Fubini’s Theorem):Suppose f is continuous on the rectangle R = [a, b]× [c, d]. Then

¨R

f(x, y)dA =

ˆ d

c

ˆ b

a

f(x, y)dxdy =

ˆ b

a

ˆ d

c

f(x, y)dydx.

Example 2:Compute the volume of the solid whose base is the rectangle R = [1, 5] × [2, 7] andwhose top lies in the plane z = −x+ 2y + 10.

In this example, we have that z = f(x, y) = −x+ 2y + 10. So

V =

¨R

f(x, y)dA

=

ˆ 7

2

ˆ 5

1

(−x+ 2y + 10)dxdy

=

ˆ 7

2

[−1

2x2 + 2xy + 10x

]x=5

x=1

dy

=

ˆ 7

2

[(−25

2+ 10y + 50

)−(−1

2+ 2y + 10

)]dy

=

ˆ 7

2

(8y + 28)dy

= [4y2 + 28y]y=7y=2

= (196 + 196)− (16 + 56)

= 320.

Example 3:z = x sinxy over R = [0, 1]× [0, π].

5


¨R

x sinxydA =

ˆ π

0

ˆ 1

0

x sinxydxdy requires by parts

=

ˆ 1

0

ˆ π

0

x sinxydydx

=

ˆ 1

0

− cosxy|y=πy=0 dx

=

ˆ 1

0

(− cos πx+ 1)dx

=

[− 1

πsin πx+ x

]x=1

x=0

= 1.

Example 4:z = xex

y, R = [0, 1]× [1, 2].

ˆ 1

0

ˆ 2

1

xex

ydydx =

(ˆ 1

0

xexdx

)(ˆ 2

1

1

ydy

)= ln 2

ˆ 1

0

xexdx

= ln 2

[xex|10 −

ˆ 1

0

exdx

]= ln 2

[e− ex|10

]= ln 2 [e− e+ 1]

= ln 2.

So far, we’ve been dealing with the relatively straight-forward case of regions thatare rectangles. Let’s turn our attention to integrating over a more general region.

Suppose we wanted to integrate over a region D that looked like

6


then we have the integral

¨D

f(x, y)dA =

ˆ b

a

ˆ g2(x)

g1(x)

f(x, y)dydx,

which is the signed volume under f(x, y) on D.

Example 5:Evaluate ¨

D

(2− 3x+ xy)dA

if D is the region bounded by the x-axis, the line x = 1 and the line y = 3x.Always draw a picture of D!We have that 0 ≤ x ≤ 1 and 0 ≤ y ≤ 3x. Then

¨D

(2− 3x+ xy)dA =

ˆ 1

0

ˆ 3x

0

(2− 3x+ xy)dydx

=

ˆ 1

0

[2y − 3xy +

1

2xy2]y=3x

y=0

dx

=

ˆ 1

0

(6x− 9x2 +

9

2x3)dx

=

[3x2 − 3x3 +

9

8x4]x=1

x=0

= 3− 3 +9

8

=9

8.

7


Note that since we’re not integrating over a rectangle, Fubini’s Theorem does notapply. We can’t say that the above integral is equal to

ˆ 3x

0

ˆ 1

0

(2− 3x+ xy)dxdy.

This integral doesn’t even make sense.Now, suppose we wanted to integrate over a region D that looked like

then we have the integral

¨D

f(x, y)dA =

ˆ d

c

ˆ h2(y)

h1(y)

f(x, y)dxdy.

Let’s look at the previous example again.


D

(2− 3x+ xy)dA

if D is the region bounded by the x-axis, the line x = 1 and the line y = 3x.

8


From the same picture, we have that 0 ≤ y ≤ 3 and y3≤ x ≤ 1. Then

¨D

(2− 3x+ xy)dA =

ˆ 3

0

ˆ 1

y/3

(2− 3x+ xy)dA

=

ˆ 3

0

[2x− 3

2x2 +

1

2x2y

]x=1

x= y3

dy

=

ˆ 3

0

[2− 3

2+

1

2y − 2

3y +

1

6y2 − 1

6y3]dy

=

ˆ 3

0

[1

2− 1

6y +

1

6y2 − 1

6y3]dy

=

[1

2y − 1

12y2 +

1

18y3 − 1

24y4]y=3

y=0

=

...

=9

8.

Note that in these previous two examples, if f(x, y) is any continuous function,then ˆ 1

0

ˆ 2x

0

f(x, y)dydx =

ˆ 3

0

ˆ 1

y/3

f(x, y)dxdy,

where the only way to make sense of the bounds and equality is with a picture of D.


D

y2exydA, D = {(x, y) : 0 ≤ y ≤ 3, 0 ≤ x ≤ y}.

First, let’s draw the picture of D. Then

9

1.2 Integration in Polar Coordinates 1 DOUBLE INTEGRALS OVER REGIONS

¨D

y2exydA =

ˆ 3

0

ˆ y

0

y2exydxdy

=

ˆ 3

0

yexy|x=yx=0 dy

=

ˆ 3

0

(yey2 − y)dy

=

[1

2ey

2 − 1

2y2]y=3

y=0

=1

2(e9 − 9− 1 + 0)

=1

2e9 − 5.

1.2 Integration in Polar Coordinates

Let’s first recall what polar coordinates are. Instead of using the ordered pair (x, y)to describe a point in the plane, we use the ordered pair (r, θ), where r is the radius(or distance from the origin) and θ denotes the angle of rotation from the positivex-axis. Furthermore, recall our basic polar identities

x2 + y2 = r2, x = r cos θ, y = r sin θ.

We know Cartesian rectangles look like a normal rectangle

R = [a, b]× [c, d] = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d},

but a polar rectangle looks like

R = [a, b]× [α, β] = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β},

a piece of a pie-shaped wedge!So to integrate a function z = f(x, y) on a polar rectangle (perhaps to find a

volume), say of the form

R = [a, b]× [α, β] = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β},

we make the conversion x = r cos θ, y = r sin θ, dA = rdrdθ and define

¨R

f(x, y)dA =

ˆ β

α

ˆ b

a

f(r cos θ, r sin θ)rdrdθ.

10


Don’t forget the extra r term! Even though we know the extra r term comes fromthe area of a piece of a wedge, it also has geometric significance that we’ll revisit laterin the course.

Similarly to Cartesian coordinates, we could have more general regions. Supposewe had a region D given by

D = {(r, θ) : α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ)}.Then ¨

D

f(x, y)dA =

ˆ β

α

ˆ g2(θ)

g1(θ)

f(r cos θ, r sin θ)rdrdθ.

Example 8:Compute

˜DxydA if D is the region in the first quadrant that lies between the circles

x2 + y2 = 4 and x2 + y2 = 2x ((x− 1)2 + y2 = 1).Converting both circles to polar coordinates we have r = 2 for the first and

r = 2 cos θ for the second. Then our bounds are 0 ≤ θ ≤ π2

and 2 cos θ ≤ r ≤ 2.Hence ¨

D

xydA =

ˆ π/2

0

ˆ 2

2 cos θ

(r cos θ)(r sin θ)rdrdθ

=

ˆ π/2

0

cos θ sin θ

ˆ 2

2 cos θ

r3drdθ

=

ˆ π/2

0

cos θ sin θ1

4r4∣∣∣∣r=2

r=2 cos θ

dθ

=

ˆ π/2

0

cos θ sin θ(4− 4 cos4 θ)dθ

= 4

ˆ π/2

0

(cos θ − cos5 θ) sin θdθ

= −4

ˆ 0

1

(u− u5)du

= 4

ˆ 1

0

(u− u5)du

= 4

[1

2u2 − 1

6u6]u=1

u=0

=4

3.

Example 9:Find the volume of the solid bounded by the plane z = 0 and the parabaloid z =1− x2 − y2. Set up the intergral in both Cartesian and polar coordinates.

11


‘After drawing the picture, we see the Cartesian domain is given by

D = {(x, y) : −1 ≤ x ≤ 1,−√

1− x2 ≤ y ≤√

1− x2}

and so

V =

¨D

(1− x2 − y2)dA =

ˆ 1

−1

ˆ √1−x2−√1−x2

(1− x2 − y2)dydx.

Note how difficult this would be to evaluate. Where as, in polar coordinates, we havethat

D = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π},

and

V =

¨D

(1− x2 − y2)dA

=

ˆ 2π

0

ˆ 1

0

(1− r2 cos2 θ − r2 sin2 θ)rdrdθ

=

ˆ 2π

0

ˆ 1

0

(1− r2)rdrdθ

=

ˆ 2π

0

[−1

4(1− r2)2

]r=1

r=0

dθ

=1

4

ˆ 2π

0

dθ

=π

2.

Example 10:Find the volume of the solid that lies under the parabaloid z = x2 + y2 above theregion D in the plane bounded by the circle (x− 1)2 + y2 = 1.

First noting that (x− 1)2 + y2 = 1 is equivalent to x2 + y2 = 2x, after convertingto polar coordinates, we see that

r2 = 2r cos θ,

or ratherr = 2 cos θ,

and the circle makes a complete revolution as θ ranges from −π2

to π2. Therefore, in

polar coordinates our region D is given by

D = {(r, θ) : 0 ≤ r ≤ 2 cos θ, 0− π

2≤ θ ≤ π

2}.

12

1.3 Applications of Double Integrals 1 DOUBLE INTEGRALS OVER REGIONS

Thus, the volume is given by

V =

ˆ ˆD

(x2 + y2)dA

=

ˆ π2

−π2

ˆ 2 cos θ

0

r2rdrdθ

=1

4

ˆ−

π

2

π224 cos4 θdθ

=

ˆ−

π

2

π2(2 cos2 θ)2dθ

=

ˆ−

π

2

π2(1 + cos 2θ)2dθ

=

ˆ−

π

2

π2(1 + 2 cos 2θ + cos2 2θ)dθ

=

ˆ−

π

2

π2

(1 + 2 cos 2θ +

1

2+

cos 4θ

2

)dθ

=3π

2.

1.3 Applications of Double Integrals

The volume of a uniform object (same height) is given by

Volume = (area of D)h

so if the height is identically one, then

¨D

1dA = (area of D)1 = area of D.

Example 11:Compute the area of D.

13


Area(D) =

¨D

1dA

=

ˆ 1

0

ˆ ex

x2dydx

=

ˆ 1

0

(ex − x2)dx

=

[ex − 1

3x3]x=1

x=0

= e− 1

3− 1 + 0

= e− 4

3.

Can we switch the order of integration here? Yes, but it will be the sum of two

14


integrals.

Then we have that for D1 that 0 ≤ y ≤ 1 and 0 ≤ x ≤ √y and for D2 that1 ≤ y ≤ e and ln x ≤ x ≤ 1. So

¨D

dA =

¨D1

dA+

¨D2

dA

=

ˆ 1

0

ˆ √y0

dxdy +

ˆ e

1

ˆ 1

lnx

dxdy

...

= e− 4

3.

We now turn our attention to an application of double integrals in the form offinding the mass of flat plates (negligible height) which are called lamina.

Suppose D is a region in the xy-plane and we have a lamina in the shape of D.The density of the plate varies, in the particular, at the point (x, y), the density isgiven by the function ρ(x, y), measured in units of mass per unit area.

15


Example 12:Find the mass of the lamina in the shape of a triangle with vertices (2, 3), (8, 3) and(6, 5), if the density is given by ρ(x, y) = x+ y.

We have that 3 ≤ y ≤ 5, 2y − 4 ≤ x ≤ 11− y. So the mass is given by

m =

ˆ 5

3

ˆ 11−y

2y−4(x+ y)dxdy

=

ˆ 5

3

[x2

2+ yx

]x=11−y

x=2y−4dy

...

= 54

Note that this can analagously be used to find the total charge of an object byreplacing density with “charge density”.

Recall that x is proportional to y if there is a constant k such that y = kx. x isinversely proportional to y if there is a constant k such that y = k

x.

Example 13:A lamina D occupies the portion of the first quadrant enclosed by the polar curver = cos 3θ. The density of the lamina at each point is inversely proportional to itsdistance from the origin. Let k be the constant of proportionality and find the massof the lamina.

Sincedist((x, y), (0, 0)) =

√x2 + y2,

we have that

ρ(x, y) =k√

x2 + y2.

Our region D is described by

16


and so the mass m is given by

m =

¨D

k√x2 + y2

dA

=

ˆ π/6

0

ˆ cos 3θ

0

k√r2rdrdθ

=

ˆ π/6

0

ˆ cos 3θ

0

kdrdθ

= k

ˆ π/6

0

cos 3θdθ

=k

3sin 3θ

∣∣∣∣θ=π/6θ=0

=k

3.

The previous application was dealing with finding masses of lamina, so now we’llturn our attention to the closely related topic of finding the moments and the centerof mass of a lamina.

Definition 14:The moment about the x-axis of a lamina (which is a measure of the tendency

17


of the lamina to spin about the x-axis) is given by the equation

Mx =

¨D

yρ(x, y)dA

and the moment about the y-axis is given by

My =

¨D

xρ(x, y)dA.

The center of mass of the lamina is the point (x, y), where

x =My

m, y =

Mx

m.

Note that it is possible for the center of mass to lie outside of the lamina:

Example 15:A lamina occupies the region in the first quadrant bounded by the circle x2 +y2 = a2.Its density function is ρ(x, y) = xy2. Find the center of mass.

18


Our region D is the polar rectangle [0, a]× [0, π/2]. Let’s first compute the mass.

m =

¨D

xy2dA

=

ˆ π/2

0

ˆ a

0

r cos θr2 sin2 θrdrdθ

=

ˆ π/2

0

cos θ sin2 θdθ

ˆ a

0

r4dr

=a5

5

ˆ π/2

0

sin2 θ cos θdθ

=a5

15sin3 θ

∣∣∣∣θ=π/2θ=0

=a5

15.

Then the moment about the x-axis is

Mx =

¨D

yρ(x, y)dA

=

ˆ π/2

0

ˆ a

0

r sin θr cos θr2 sin2 θrdrdθ

=

ˆ π/2

0

sin3 θ cos θdθ

ˆ a

0

r5dr

=

(1

4

)(a6

6

)=a6

24

19


and the moment about the y-axis is

My =

¨D

xρ(x, y)dA

=

ˆ π/2

0

ˆ a

0

r cos θr cos θr2 sin2 θrdrdθ

=

ˆ π/2

0

cos2 θ sin2 θdθ

ˆ a

0

r5dr

=a6

6

ˆ π/2

0

cos2 θ sin2 θdθ

=a6

6

ˆ π/2

0

(1 + cos 2θ

2

)(1− cos 2θ

2

)dθ

=a6

24

ˆ π/2

0

(1− cos2 2θ)dθ

=a6

24

ˆ π/2

0

(1− 1

2(1 + cos 4θ)

)dθ

=a6

48

ˆ π/2

0

(1− cos 4θ)dθ

=a6π

96.

Then the center of mass is given by

x =My

m

=a6π

96

15

a5

=5πa

32,

y =Mx

m

=a6

24

15

a5

=5a

6.

So the center of mass is given by the coordinates(5πa

32,5a

8

).

20

1.4 Surface Area 1 DOUBLE INTEGRALS OVER REGIONS

Draw pic of center of mass

One final application deals with finding the moment of inertia which gives in-formation on the resistance to change of rotation. That is, if we increase mass orincrease how far the mass is from the axis, we increase moment of inertia (and makeit harder to rotate).

Definition 16:The moment of inertia for a single particle of mass m a distance of r units fromthe axis is defined to be mr2. For a lamina D with density ρ(x, y), the moments ofinertia with respect to the axes is defined as follows:

Iy =

¨D

x2ρ(x, y)dA = moment of inertia from the y-axis,

Ix =

¨D

y2ρ(x, y)dA = moment of inertia from the x-axis,

Io =

¨D

(x2 + y2)ρ(x, y)dA = Ix + Iy = moment of inertia about the origin.

1.4 Surface Area

Recall that the area of of the parallelogram determined by two vectors u and v isgiven by

|u× v|.

Now, suppose we have some surface in R3 given by the function z = f(x, y) overthe region R, and we wish to compute the surface area of z = f(x, y) over R. Now,let’s assume a very fine partition of R into subrectangles Rij. Then as usual, thearea of this subrectangle is given by ∆Aij = ∆xi∆yj. At a point (xi, yj) in Rij, wehave the point Pij = (xi, yj, f(xi, yj)) on the surface z = f(x, y). The surface areaof z = f(x, y) over the subrectangle Rij is approximately equal to the area of thetangent plane at the point Pij. That is,

∆Sij ≈ ∆Tij,

where ∆Sij is the actual surface area over Rij and ∆Tij the area of the parallelogramof the tangent plane at Pij above Rij.

Hence the total surface area if given by the sum

m,n∑i,j=1

∆Sij ≈m,n∑i,j=1

∆Tij.

21


Thus, we need a way to nicely characterize the area of the piece of the tangent plane∆Tij. Recall from Math 1261 that the two vectors 〈0, 1, ∂yf(xi, yj)〉 and 〈1, 0, ∂xf(xi, yj)〉lie in the tangent plane at Pij and so scalling by the subdivision, we have thatu = 〈∆xi, 0, ∂xf(xi, yj)∆xi〉 and v = 〈0,∆yj, ∂yf(xi, yj)∆yj〉 lie entirely in the tan-gent plane above Rij. Thus the area ∆Tij is given by

∆Tij = |u× v|= | 〈−∂xf(xi, yj)∆xi∆yj,−∂yf(xi, yj)∆xi∆yj,∆xi∆yj〉 |= | 〈−∂xf(xi, yj),−∂yf(xi, yj), 1〉 |∆Aij

=√

(∂xf(xi, yj))2 + (∂yf(xi, yj))2 + 1∆Aij.

Summing these up, and taking the partition finer and finer as in Riemann sums,we get the following definition

Definition 17:If f and its first partial derivatives are continuous on the region D in the xy-plane,the surface area of S described by z = f(x, y) over D is given by

Surface Area =

¨D

dS

=

¨D

√1 + [∂xf(x, y)]2 + [∂yf(x, y)]2dA.

Note that dS is called our surface area differential and is given by

dS =√

1 + [∂xf(x, y)]2 + [∂yf(x, y)]2dA.

Example 18:Find the area of the part of the paraboloid z = x2 + y2 that lies under the planez = 9.

Let’s first compute some partial derivatives of f(x, y) = x2 + y2. We have that

∂f

∂x= 2x,

∂f

∂y= 2y.

Thus

Surface Area =

¨D

dS

=

¨D

√1 + (2x2) + (2y)2dA

=

¨D

√1 + 4(x2 + y2)dA.

1See my 126 notes, Section 5.2 on Tangent Planes

22

http://sites.math.washington.edu/~mrrobins/324Autumn17/126notes.pdf


Projecting this surface down to the xy-plane, we see that our region is the diskof radius 3, witch is the polar rectangle [0, 3] × [0, 2π]. Turning the above into aniterated integral, we see that

Surface Area =

¨D

√1 + 4(x2 + y2)dA

=

ˆ 2π

0

ˆ 3

0

√1 + 4r2rdrdθ

=2π

8

2

3(1 + 4r2)3/2

∣∣r=3

r=0

=π

6(37√

37− 1).

23

2 TRIPLE INTEGRALS

2 Triple Integrals

Thinking about the applications we’ve just covered (finding mass and center of mass),it makes sense to now extend our notion of integration to higher dimensions, namelytriple integrals.

Consider the case where R is a rectangular prism in R3 (3-dimensional Euclideanspace). That is,

R = [a, b]× [c, d]× [r, s]

= {(x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}.

And we wish to know the mass of the object R, where the denisty is then a functionof three variables, ρ(x, y, z). Then the mass is given by the triple integral

˚R

ρ(x, y, z)dV,

where dV represents our infinitesimal volume

dV ≈ ∆V = ∆x∆y∆z,

taken in the form of a Riemann sum. This means our triple intgral is defined in theusual way via Riemann sums, i.e.,

˚R

ρ(x, y, z, )dV = liml,m,n→∞

l,m,n∑i,j,k=1

ρ(xi, yj, zk)∆xi∆yj∆zk.

To actually compute these triple integrals, we use similar methods as in the two-dimensional case.

Theorem 19 (Fubini’s Theorem):If f is continuous on the rectangular box R = [a, b]× [c, d]× [r, s], then

˚R

f(x, y, z)dV =

ˆ s

r

ˆ c

b

ˆ b

a

f(x, y, z)dxdydz.

Note: that just in the two-dimensional case, you can switch the order of integrationhere, however, there will be six different permutations all yielding the same result.

Example 20:Evaluate the triple integral

˝Rxyz2dV , where R = [0, 1]× [−1, 2]× [0, 3].

24

2 TRIPLE INTEGRALS

Let’s integrate as follows:

˚R

xyz2dV =

ˆ 1

0

ˆ 2

−1

ˆ 3

0

xyz2dzdydx

=

ˆ 1

0

ˆ 2

−19xydydx

=

ˆ 1

0

27

2xdx

=27

4.

We can also do triple integrals over more genreal regions besides boxes, however,since we’re in three-dimensions now, we’ll have three general tpyes.

Definition 21:A solid E is of type I if it lies above a region D in the xy-plane between two continuousfunctions of only x and y, that is,

E = {(x, y, z) : (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)}.

If f(x, y, z) is any continuous function on E, then

˚E

f(x, y, z)dV =

¨D

[ˆ u2(x,y)

u1(x,y)

f(x, y, z)dz

]dA.

25

2 TRIPLE INTEGRALS

This is interpreted as we first integrate with respect to z (holding x and y con-stant), leaving just to compute a double integral over D in the variables of x and y(as we’ve covered in the previous section).

Definition 22:A solid E is of type II if it lies above a regionD in the yz-plane between two continuousfunctions of only y and z, that is,

E = {(x, y, z) : (y, z) ∈ D, u1(y, z) ≤ x ≤ u2(y, z)}.

If f(x, y, z) is any continuous function on E, then

˚E

f(x, y, z)dV =

¨D

[ˆ u2(y,z)

u1(y,z)

f(x, y, z)dx

]dA.

Definition 23:A solid E is of type III if it lies above a region D in the xz-plane between twocontinuous functions of only x and z, that is,

E = {(x, y, z) : (x, z) ∈ D, u1(x, z) ≤ y ≤ u2(x, z)}.

26

2 TRIPLE INTEGRALS

If f(x, y, z) is any continuous function E, then

˚E

f(x, y, z)dV =

¨D

[ˆ u2(x,z)

u1(x,z)

f(x, y, z)dy

]dA.

Example 24:Evaluate

˝EzdV , where E is the tetrahedron bounded by the four planes x = 0,

y = 0, z = 0, and x+ y + z = 1.Note that this tetrahedron has vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1).

Thus if we project D on the yz-plane, we see that 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1− y and

27

2 TRIPLE INTEGRALS

we’re of type II between the two surfaces x = 0 and x = 1− y − z. Then

˚E

zdV =

¨D

[ˆ 1−y−z

0

zdx

]dA

=

ˆ 1

0

ˆ 1−y

0

ˆ 1−y−z

0

zdxdzdy

=

ˆ 1

0

ˆ 1−y

0

((1− y)z − z2)dzdy

=

ˆ 1

0

(1

2(1− y)3 − 1

3(1− y)3

)dy

=1

6

ˆ 1

0

(1− y)3dy

= − 1

24(1− y)4

∣∣y=1

y=0

=1

24.

Example 25:Evaluate

˝E

√x2 + z2dV where E is the region bounded by the surfaces y = 4 and

y = x2 + z2.Let’s first set this up as a type I integral, that is projecting E onto the xy-plane,

we see that −2 ≤ x ≤ 2 and x2 ≤ y ≤ 4, and we get the triple integral

˚E

√x2 + z2dV =

ˆ 2

−2

ˆ 4

x2

ˆ −√y−x2

−√y−x2

√x2 + z2dzdydx.

This integral is very difficult to evaluate, so let’s try setting this up as a type IIIintegral instead. That is,

˚E

√x2 + z2dV =

¨D

ˆ 4

x2+z2

√x2 + z2dydA =

¨D

(4− x2 − z2)√x2 + z2dA,

where D is the disk of radius 2 in the xz-plane. Thus, let’s switch to polar coordinates

28

2 TRIPLE INTEGRALS

in the xz-plane, where r2 = x2 + z2, 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Hence˚E

√x2 + z2dV =

¨D

(4− x2 − z2)√x2 + z2dA

=

ˆ 2π

0

ˆ 2

0

(4r2 − r4)drdθ

= 2π

(4

323 − 1

525

)=

27π

15

=128π

15.

Example 26:Set up but do not solve the integral

˝Ef(x, y, z)dV as an iterated integral in six

different ways, where E is the solid bounded by y2 + z2 = 9, x = −2, and x = 2.Note that this is a cylinder of height 4 in along the x-axis of radius 3.In the type I case, projecting to the xy-plane, we have a projection to the rectangle

[−2, 2]× [−3, 3] with −√

9− y2 ≤ z ≤√

9− y2 and so

˚E

f(x, y, z)dV =

ˆ 3

−3

ˆ 2

−2

ˆ √9−y2

−√

9−y2f(x, y, z)dzdxdy

=

ˆ 2

−2

ˆ 3

−3

ˆ √9−y2

−√

9−y2f(x, y, z)dzdydx.

In the type II case, projecting to the yz-plane, we have a projection of a circle ofradius 3 with −2 ≤ x ≤ 2. Thus˚

E

f(x, y, z)dV =

ˆ 3

−3

ˆ √9−z2−√9−z2

ˆ 2

−2f(x, y, z)dxdydz

=

ˆ 3

−3

ˆ √9−y2

−√

9−y2

ˆ 2

−2f(x, y, z)dxdzdy.

In the type III case, projecting to the xz-plane, we again have a prokection to arectangle of the form [−2, 2]× [−3, 3] with −

√9− z2 ≤ y ≤

√9− z2 and hence

˚E

f(x, y, z)dV =

ˆ 3

−3

ˆ 2

−2

ˆ √9−z2−√9−z2

f(x, y, z)dydzdx

=

ˆ 2

−2

ˆ 3

−3

ˆ √9−z2−√9−z2

dydxdz

.

29

2.1 Applications of Triple Integrals 2 TRIPLE INTEGRALS

2.1 Applications of Triple Integrals

Similar to case of calculating area via double integrals with

A(D) =

¨D

dA,

we can also define the volume of solids V (E) to be

V (E) =

˚E

dV.

We also have the various definitions for computing mass, center of mass, andmoments:

Definition 27:If the solid E has density function ρ(x, y, z), then the mass of E is given by

m =

˚E

ρ(x, y, z)dV.

The moments of E about the three coordinates planes are given by

Mxy =

˚E

zρ(x, y, z)dV, Mxz =

˚E

yρ(x, y, z)dV,

Myz =

˚E

xρ(x, y, z)dV.

The center of mass of the solid E is the point (x, y, z), where

x =Myz

m, y =

Mxz

m, z =

Mxy

m.

The moments of inertia about the three corrdinates axes are given by

Ix =

˚E

(y2 + z2)ρ(x, y, z)dV

Iy =

˚E

(x2 + z2)ρ(x, y, z)dV

Iz =

˚E

(x2 + y2)ρ(x, y, z)dV.

30

3 COORDINATE SYSTEMS

3 Coordinate Systems

When computing double integrals, we’ve seen many examples of when the domain Dis circular in nature, it makes sense to convert to polar coordinates to carryout theintegration.

What is the natural generalization of this to triple integrals?

3.1 Cylindrical Coordinates

The first natural notion of extending polar coordinates to R3, is keeping the samedefinition for (r, θ) as in polar coordinates and then let z stay the same. That is,

Definition 28:The cylindrical coordinate system on R3 represents a point P as an orderedtriple,

P = (r, θ, z),

where r is the distance from P to the z-axis, θ is the angle of roation from thepositive x-axis, and z the signed distance from the xy-plane.

31

3.1 Cylindrical Coordinates 3 COORDINATE SYSTEMS

The conversion between Cartesian coordinates and cylindrical coordinates workssimillar as in polar coordinates, that is,

(x, y, z)↔ (r, θ, z),

is given byx = r cos θ, y = r sin θ, z = z,

and conversely,

r2 = x2 + y2, tan θ =y

x, z = z.

Note, that you can rotate cylindrical coordinates to be along any axis, dependingon the desired situation. E.g., for cylindrical coordinates along the y-axis, we havethat

x = r cos θ, z = r sin θ, y = y.

Now, let’s see how to integrate using cylindrical coordinates: Suppose we have atype I solid E. That is,

E = {(x, y, z) : (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)},

and if f(x, y, z) is any continuous function on E, we have that

˚E

f(x, y, z)dV =

¨D

[ˆ u2(x,y)

u1(x,y)

f(x, y, z)dz

]dA.

Now, suppose we can write D in polar coordinates, that is,

D = {(r, θ) : α ≤ θ ≤ β, g1(θ) ≤ r ≤ g2(θ)},

then from the above, we get

˚E

f(x, y, z)dV

ˆ β

α

ˆ g2(θ)

g1(θ)

ˆ u2(r cos θ,r sin θ)

u1(r cos θ,r sin θ)

f(r cos θ, r sin θ, z)rdzdrdθ.

Example 29:Find the volume of the solid E that lies inside the cylinder x2 + y2 = 1, outside theparabaloid z = 1− x2 − y2, and below the plane z = 2.

32

3.1 Cylindrical Coordinates 3 COORDINATE SYSTEMS

In cylindrical coordinates, we see that E is the region of 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1,and 1− r2 ≤ z ≤ 2.

Computing the volume,

V =

˚E

dV

=

ˆ 2π

0

ˆ 1

0

ˆ 2

1−r2rdzdrdθ

=

ˆ 2π

0

ˆ 1

0

(1 + r2)rdrdθ

=2π

2(1 + r2)

∣∣r=1

r=0

= π.

Example 30:Evaluate ˆ 2

−2

ˆ √4−y2

−√

4−y2

ˆ 2

√x2+y2

xzdzdxdy.

We see that E is the cylinderical region of 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, r ≤ z ≤ 2.

33

3.2 Spherical Coordinates 3 COORDINATE SYSTEMS

Hence

ˆ 2

−2

ˆ √4−y2

−√

4−y2

ˆ 2

√x2+y2

xzdzdxdy =

ˆ 2π

0

ˆ 2

0

ˆ 2

r

r2z cos θdzdrdθ

=

ˆ 2π

0

cos θdθ

ˆ 2

0

ˆ 2

r

r2zdzdr

= 0.

Cylindrical coordinates are a good first notion of generalizing polar coordinates,however, there are other geometric generalization for which cylindrical coordinatesaren’t helpful. This leads us to our next natural generalization.

3.2 Spherical Coordinates

In contrast to cylindrical coordinates, where we just kept polar coordinates and addedthe z-component, we’re redefining a couple of the components here.

Definition 31:The spherical coordinate system on R3 represents a point P as the ordered triple,

P = (ρ, θ, φ),

where ρ is the distance from P to the origin, θ is the angle of rotation from thepositive x-axis (the same as in polar coordinates), and φ is the angle of rotation downfrom the positive z-axis. Note that by convention ρ ≥ 0 and 0 ≤ φ ≤ π; θ is calledthe polar angle and φ is called the azimuthal angle.

34


Let’s use some simple geometry to derive the conversion betrween Cartesian andsphereical coordinares.

From the above picture, we can see that

ρ cosφ = z, ρ sinφ = r,

and as usualx = r cos θ, y = r sin θ.

Putting it all together,(x, y, z)↔ (ρ, θ, φ),

is given byx = ρ cos θ sinφ, y = ρ sin θ sinφ, z = ρ cosφ,

and conversely,

ρ2 = x2 + y2 + z2, tan θ =y

x, tanφ =

r

z,

where r is given as the usual radial coordinate.

35


Now, let’s see how to compute a triple integral using spherical coordinates. Sup-pose we wish to evaluate ˚

E

f(x, y, z)dV,

in spherical coordinates. We know how to sitch from Cartesian coordiantes to spher-ical coordinates using the formulas

x = rρ cos θ sinφ, y = r sin θ sinφ, z = ρ cosφ,

and we can describe the region E in terms of spherical inequalities. Thus we so farhave ˚

E

f(x, y, z)dV =

˚E

f(ρ cos θ sinφ, ρ sin θ sinφ, ρ cosφ)dV.

However, unlike the cylindrical case, where our cylindrical-volume differential waseasily obtained from our polar-area differential to be

dV = rdrdθdz,

we don’t yet have formula to define our sphirical-volume differential.First, let’s consider a nearly infinitisimal small sphirical prism, that is,

E = [ρ, ρ+ ∆ρ]× [θ, θ + ∆θ]× [φ, φ+ ∆φ].

If we recall the length of an arc is given by

Rϕ,

we then have the following side lengths of

∆ρ, ρ∆φ, ρ sinφ∆θ.

36


Now, since we’re dealing with an infinitesimal rectangles, the volume is approxi-mately calculated in the same way as Cartersian rectangles, that is,

∆V ≈ (∆ρ)(ρ∆φ)(ρ sinφ∆θ),

and hence taking limits, we see that our spherical-volume differential is given by

dV = ρ2 sinφdρdθdφ.

You can make this more precise, as in the case of polar coordinates, by using thevolume of a spherical cones.

Finally, if E = [a, b]× [α, β]× [γ, δ] is a sphirical prism˚

E

f(x, y, z)dV =

ˆ γ

δ

ˆ α

β

ˆ b

a

f(ρ cos θ sinφ, ρ sin θ sinφ, ρ cosφ)ρ2 sinφdρdθdφ

37

3.3 Change of Variables 3 COORDINATE SYSTEMS

Example 32:Use spherical coordinates to compute the volume of the ball of radius r.

Since a ball of radius R is given by

E = [0, R]× [0, 2π]× [0, π],

we have

V (E) =

˚E

dV

=

ˆ π

0

ˆ 2π

0

ˆ R

0

ρ2 sinφdρdθdφ

=R3

3

ˆ π

0

ˆ 2π

0

sinφdθdφ

=2

3πR3

ˆ π

0

sinφdφ

=4

3πR3.

3.3 Change of Variables

We’ve seen a few examples in this course of changing variables already. By a change ofvariable, we mean a different coordinate system. In R2, we’ve seen polar coordinates,where

(x, y) =⇒ (r, θ),

and in R3, we’ve seen cylindrical coordinates

(x, y, z) =⇒ (r, θ, z)

and spherical coordinates(x, y, z) =⇒ (ρ, θ, φ).

These were all examples of change of variables, and the main goal of introducingthese was to allow us to describe regions in a new way as to facilitate integration. Forintegration, however, we had to use very example-specific methods to determine thearea differential for polar coordinates or the volume differentials for cylindrical andsphereical coordinates. This leads to the question: Is there a more general method ofdetermining these differentials?

To answer this, let’s recall from Math 124, u-substitution:

ˆ b

a

f(x)dx =

ˆ d

c

f(g(u))g′(u)du,

38


where g(c) = a and g(d) = b. This is actually an example of change of variables with

(x) =⇒ (u),

given by the conversionx = g(u).

And the length differential is represented by g′(u)du in complete generality for 1-dimension.

We can generalize this type of change of variable to R2. Indeed, let’s take

(x, y) =⇒ (u, v),

with conversionx = g(u, v), y = h(u, v).

This is sometimes called a transformation and denoted T (u, v) = (x, y).It transforms a region in the uv-plane to a region in the xy-plane.

Example 33:What is the image under T of the region in the uv-plane bounded by u = 0, u =1, v = 0, v = π

2, where T is the transformation given by x = u cos v, y = u sin v.

This is just polar coordinates!

39


Example 34:What is the image of the square R = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 1} under thetransformation x = u2 − v2, y = 2uv?

We see the line u = 0 corresponds x = −v2, y = 0, so we lie on the negative x-axixmoving to 0 as v goes to 0. Similiarly when v = 0, y = 0 and x = u2, so it loes onthe positive x-axis moving to 0 as u goes to 0. When v = 1, x = u2 − 1 and y = 2uand so substituting, x = y2

4− 1, and similar yor u = 1, x = 1− y2

4.

Now, let’s return to using change of variables for integration. Suppose we havea transformation T (u, v) = (x, y) with x = g(u, v), y = h(u, v) which takes someregion R in the uv-plane to the region D in the xy-plane. Then, in similiarity tou-substitution, for double integrals,

¨D

f(x, y)dxdy =

¨R

f(g(u, v), h(u, v))J(u, v)dudv.

What is this function J(u, v) replacing the single derivative in the u-substitution case?Notational Aside: As a function T : R2 → R2 with T (u, v) = (g(u, v), h(u, v)).

We will also write x = x(u, v) and y = y(u, v) supressing the g and h functions. Thisis in similarity to single-variable calculus, where in u-substitution, we write u = u(x)and du = u′(x)dx.

40


Definition 35:The Jacobian of the transformation T (u, v) = (x, y), denoted ∂(x,y)

∂(u,v)is given by

∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ .Its derivation is similar in nature to the surface area derivation, but it’s not

particularly enlightening to go through that again. See the textbook for furtherdetails.

Theorem 36:Suppose that T is a transformation with x = g(u, v) and y = h(u, v) such that g andh are continuous and have continuous first partial derivatives. If the Jacobian is notidentically zero, and T (R) = D, then

¨D

f(x, y)dA =

¨R

f(g(u, v), h(u, v))

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ dudv.Example 37:Compute the Jacobian for polar coordinates.

∂(x, y)

∂(r, θ)=

∣∣∣∣∂x∂r ∂x∂θ

∂y∂r

∂y∂θ

∣∣∣∣= r cos2 θ + r sin2 θ

= r

and so J = r, which we already know since the polar-area differential is given by

dA = rdrdθ.

Example 38:Use the change of variables x = u2−v2 and y = 2uv to compute the integral

˜DydA,

where D is the region bounded by the lines y = 0, x = 1− y2/4 and x = y2/4− 1.From Example 34, we know the transformation T takes the unit sqaure R onto

D. Thus, we need to only compute the Jacobian

∂(x, y)

∂(u, v)=

∣∣∣∣2u −2v2v 2u

∣∣∣∣= 4u2 + 4v2.

41


And so ¨D

ydA =

¨R

(2uv)(4u2 + 4v2)dudv

= 4

ˆ 1

0

ˆ 1

0

(u2 + v2)2uvdudv

= 2

ˆ 1

0

(u2 + v2)2∣∣u=1

u=0vdv

= 2.

So far, we’ve introduced change-of-variables in R2, but an identical process worksfor change-of-variables in R3.

Definition 39:If T is a transformation which maps a region R in uvw-space to a region E in xyz-space, by the equations

x = g(u, v, w), y = h(u, v, w), z = k(u, v, w),

then the Jacobian of T , denoted ∂(x,y,z)∂(u,v,w)

is computed as

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∂x∂u

∂x∂v

∂x∂w

∂y∂u

∂y∂v

∂y∂w

∂z∂u

∂z∂v

∂z∂w

∣∣∣∣∣∣ .Also, in similiarty to before, we have the equality˚

E

f(x, y, z)dV =

˚R

f(g(u, v, w), h(u, v, w), k(u, v, w))

∣∣∣∣ ∂(x, y, z)

∂(u, v, w)

∣∣∣∣ dudvdw.Example 40:Compute the Jacobian for spherical coordinates.

∂(x, y, z)

∂(ρ, θ, φ)=

∣∣∣∣∣∣∣∂ρ cos θ sinφ

∂ρ∂ρ cos θ sinφ

∂θ∂ρ cos θ sinφ

∂φ∂ρ sin θ sinφ

∂ρ∂ρ sin θ sinφ

∂θ∂ρ sin θ sinφ

∂φ∂ρ cosφ∂ρ

∂ρ cosφ∂θ

∂ρ cosφ∂φ

∣∣∣∣∣∣∣=

∣∣∣∣∣∣cos θ sinφ −ρ sin θ sinφ ρ cos θ cosφsin θ sinφ ρ cos θ sinφ ρ sin θ cosφ

cosφ 0 −ρ sinφ

∣∣∣∣∣∣= cosφ(ρ2 cos2 θ cosφ sinφ+ ρ2 sin2 θ cosφ sinφ)

− ρ sinφ(−ρ cos2 θ sin2 φ− ρ sin2 θ sin2 φ)

= ρ2 sinφ cos2 φ+ ρ2 sinφ sin2 φ

= ρ2 sinφ

42


In summary, this might’ve seemed like a fairly convoluted process, however, whendoing these types of problems, it’s best to keep in mind the three change-of-variableswe’re familiar with (polar, cylindrical, spherical), and think about the order in whichwe compute things with them. Let’s try and outline the methodology of using arbi-trary coordinates:

1. Choose a change-of-variable x = g(u, v), y = h(u, v) that will simplify the prob-lem (this is not unique).

2. Determine the bounds on u and v which correspond to the desired bounds onx and y. This depends on your choice in Step 1.

3. Compute the Jacobian of the transformation chosen.

4. Replace the x’s and y’s with u and v in the integration, along with dxdy =J(u, v)dudv.

5. Integrate!

43

4 DERIVATIVES

4 Derivatives

We now turn our attention to revisiting derivatives, which we will then relate tointegration later in the course.

We know how to take partial derivatives from Math 1262, but we need anothertechnique. Recall the single-variable chain rule: If y = f(x) and x = g(t), then

dy

dt=dy

dx

dx

dt.

Definition 41:Suppose z = f(x, y) with x = g(t) and y = h(t), then the multivariable chain ruleis

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt.

Example 42:If z = x3 + xy2 + y, x = e2t and y = sin t, compute dz

dt.

By the chain rule, we have that

dz

dt= (3x2 + y2)(2e2t) + (2xy + 1)(cos t)

= (3e4t + sin2 t)(2e2t) + (2e2t sin t+ 1) cos t.

Example 43:If z = sin(xy) and x = 3t, y = t2, find dz

dt.

By the chain rule,

dz

dt= (y cos(xy))(3) + (x cos(xy))(2t)

= (3t2 + 6t2) cos(3t3)

= 9t2 cos(3t3).

Proposition 44:Suppose z = f(x, y) and x = g(s, t), y = h(s, t), then

∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s,

∂z

∂t=∂z

∂x

∂x

∂t+∂z

∂y

∂y

∂t.

Example 45:Suppose z = e2x cos y and x = s2 + st, y = st2, then compute ∂z

∂sand ∂z

∂t.

2See my 126 notes, Section 5.1 on Partial Derivatives

44

http://sites.math.washington.edu/~mrrobins/324Autumn17/126notes.pdf

4 DERIVATIVES

∂z

∂s= (2e2x cos y)(2s+ t) + (−e2x sin y)(t2)

= 2e2(s2+st) cos(st2)(2s+ t)− e2(s2+st) sin(st2)t2

and

∂z

∂t= ∗2e2(s2+st) cos(st2)(s)− e2(s2+st) sin(st2)(2st)

You can generalize the chain rule to an arbitrary number of variables as well.

Proposition 46:Suppose u = f(x1, ..., xn) and and for each j, xj = g(t1, ..., tm), then for each k,

∂u

∂tk=

∂u

∂x1

∂x1∂tk

+∂u

∂x2

∂x2∂tk

+ · · ·+ ∂u

∂xn

∂xn∂tk

=n∑j=1

∂u

∂xj

∂xj∂tk

.

Example 47:If u = xyz + xz2, where x = rset, y = rse−t, and z = rs, find ∂u

∂s.

∂u

∂s= (yz + z2)(ret) + (xz)(re−t) + (xy + 2xz)(r)

= (r2s2e−t + r2s2)(ret) + (r2s2et)(re−t) + (r2s2 + 2r2s2et)(r)

We can generalize implicit differentiation as well. Suppose we have the equation

F (x, y) = 0.

Then we can consider the function as implicitly defining y as a function of x. To finddydx

, we take the derivative of the above equation with respect to x, and by the chainrule, we see that

0 =∂F

∂x

dx

dx+∂F

∂y

dy

dx,

and hence solving for dydx

and using the fact that dxdx

= 1, we get the following theorem

Theorem 48 (Implicit Function Theorem):Under certain hypotheses (beyond the scope of this course) where F (x, y) = 0 definesy implicitly as a differentiable function of x, then

dy

dx= −

∂F∂x∂F∂y

.

45

4.1 Directional Derivatives 4 DERIVATIVES

Note that one of the hypotheses of this theorem must be tht ∂F∂y

is nonvanishing.Also note that this theorem can generalize to an arbitrary number of variables.

For example if F (x, y, z) = 0, a similar calculation shows that

∂z

∂x= −

∂F∂x∂F∂z

,∂z

∂y= −

∂F∂y

∂F∂z

.

Example 49:Compute dy

dxif x2 + xy = y2.

Let F (x, y) = x2 + xy − y2. Then

dy

dx= −

∂F∂x∂F∂y

= −2x+ y

x− 2y.

4.1 Directional Derivatives

Before defining directional derivatives, the motivation for using them is becausethey’re one of the natural generalization of partial derivatives. To see this, recallthe definition of partial derivaitves: If z = f(x, y), then

∂z

∂x= lim

h→0

f(x+ h, y)− f(x, y)

h∂z

∂y= lim

h→0

f(x, y + h)− f(x, , y)

h.

These represented the rates of change in the directions e1 = i = 〈1, 0〉 and e2 = j =〈0, 1〉, respectively.

Suppose we wish to calculate the rate of change in an arbitrary direction u = 〈a, b〉.This leads to the following defintion:

Definition 50:The directional derivative of the function f(x, y) in the direction u = 〈a, b〉 at thepoint (x0, y0) is given by

Duf(x0, y0) = limh→0

f(x0 + ha, y0 + hb)− f(x, y)

h.

How do we compute directional derivatives in practice?

46

4.1 Directional Derivatives 4 DERIVATIVES

Theorem 51:

Duf(x0, y0) = a∂f

∂x(x0, y0) + b

∂f

∂y(x0, y0).

To see this, let g(h) = f(x0 + ha, y0 + hb). Then on one hand

g′(0) = limh→0

g(h)− g(0)

h

= limh→0

f(x0 + ha, y0 + hb)− f(x0, y0)

h= Duf(x0, y0).

On the other hand, if we let g(h) = f(x, y) where x = x0 + ha, y = y0 + hb, thenby the chain rule

dg

dh=∂f

∂x

dx

dh+∂f

∂y

dy

dh

= a∂f

∂x+ b

∂f

∂y,

and so

g′(0) = a∂f

∂x(x0, y0) + b

∂f

∂y(x0, y0),

thus showing the theorem.

Note that with this theorem, it’s easy to see that partial derivatives have aninterpretation as direction derivatives. Indeed,

∂f

∂x= De1f,

∂f

∂y= De2f.

One caveat with directional derivatives that differs with the book’s definition. Thebook only defines directional derivatives for u being a unit vector. As we saw above,neither the definition nor the theorem requires this, so it’s a superfluous hypothesis.However, since the homework uses this convention, we’ll adopt it as well from thispoint forward.

Furthermore, since we’re now only dealing with the case of u being of unit-length,if θ is the angle the vector u makes with the positive x-axis, then we can write

u = 〈cos θ, sin θ〉 .

47

4.2 The Gradient 4 DERIVATIVES

Example 52:Find the directional derivative of f(x, y) = x3− 3xy+ 4y2 in the direction of the unitvector with angle θ = π

6.

Since u = 〈cos(π/6), sin(π/6〉 = 12

⟨√3, 1⟩, we see the directional derivative

Duf(x, y) =

√3

2(3x2 − 3y) +

1

2(8y − 3x)

Example 53:Find the directional derivative of f(x, y) = ln(x2 + y2) at the point (2, 1) in thedirection 〈−1, 2〉.

Normalizing the vector, we see that

u =1√5〈−1, 2〉 .

Then

Duf(2, 1) =

[− 1√

5

2x

x2 + y2+

2√5

2y

x2 + y2

]∣∣∣∣(2,1)

= − 4

5√

5+

4

5√

5

= 0.

4.2 The Gradient

Recalling the dot product of two vectors u = 〈u1, u2〉 , v = 〈v1, v2〉 is given by

u · v = u1v1 + u2v2.

Then we can rewrite the directional derivative of f(x, y) in the direction u = 〈a, b〉as follows

Duf(x, y) = a∂f

∂x+ b

∂f

∂y= 〈∂xf, ∂yf〉 · 〈a, b〉 .

This motivates the following definition:

Definition 54:The gradient of a function f(x, y) is the vector-valued function

grad (f) (x, y) = ∇f(x, y) = 〈∂xf(x, y), ∂yf(x, y)〉 .

48


Using this notation, we see that the directional derivative can we written as

Duf = ∇f · u.

Note that this definition generealizes to any number of variables in the usual way:If f(x, y, z) is a function, then

∇f = 〈fx, fy, fz〉 .

Example 55:Compute the gradient for f(x, y, z) = x sin(yz).

∇f = 〈sin(yz), xz cos(yz), xy cos(yz)〉 .

What information does this gradient vector encapsulate?Well one natural question one might first ask about a function f(x, y) is in what

direction does f change the fastest? What is the maximum rate of change? Weactually already know this answer.

Let u be any unit-direction. Then

Duf = ∇f · u= |∇f ||u| cos θ

= |∇f | cos θ,

there θ is the angle between the vectors ∇f and u. Thus the directional derivative ismaximized when cos θ is maximized which happens when θ = 0. That is, when ∇fand u are in the same direction.

Theorem 56:If f is a differentiable function of any number of variables, then the maximum valueof the directional derivative Duf is |∇f | and it occurs when u is the same directionas ∇f .

Example 57:Find the maximum rate of change of f(x, y) = 4y

√x at the point (1, 1) and the

direction at which is occurs.

∇f(1, 1) =

⟨2y√x, 4√x

⟩∣∣∣∣(1,1)

= 〈2, 4〉 .

49


Thus the maximum value is|∇f | = 2

√5,

and the direction is1

2√

5〈2, 4〉 .

There is also a close connection between gradients and tangent planes to surfaces.Suppose F (x, y, z) = k is a function of three variables and k is constant. We can thinkof this as an implicitly defined function for z of the two variables x, y or as a levelsurface S of the function w = F (x, y, z). Now imagine that C is a curve on S passingthrough the point P = (x0, y0, z0). So C can be paramterized as the vector-valuedfunction

r(t) = 〈x(t), y(t), z(t)〉 .Now let’s differentiate F (x, y, z) = k with respect to t to see that via the chain rule

0 =∂F

∂x

dx

dt+∂F

∂y

dy

dt+∂F

∂z

dz

dt.

Sincer′(t) = 〈x′(t), y′(t), z′(t)〉 ,

the above equation turns into∇F · r′(t) = 0.

This tells us that the gradient of F is always orthogonal to any curve C that lies onthe surface S, that is, for any point P on S, ∇F is orthogonal to every tangent vectorat P . Hence ∇F is orthogonal to every tangent plane on S. This means that ∇F canbe used to describe the tangent planes, since a normal vector and a point completelydescribe a plane.

Proposition 58:Suppose S is the level surface defined by the equation F (x, y, z) = k. Then for anyP = (x0, y0, z0) on S, the tangent plane at P is given by the equation

Fx(x0, y0, z0)(x− x0) + Fy(x0, y0, z0)(y − y0) + Fz(x0, y0, z0)(z − z0) = 0.

Example 59:Find the tangent plane to ellipsoid 4x2 + y2 + 2z2 = 4 at the point (1, 0, 0).

Letting F (x, y, z) = 4x2 + y2 + 2z2, we see that

∇F (1, 0, 0) = 〈8x, 2y, 4z〉|(1,0,0) = 〈8, 0, 0〉 ,

and so we get the plane8(x− 1) = 0, x = 1.

50

5 VECTOR FIELDS

5 Vector fields

A vector field is a a function that assigns to each point in R2 or R3 a vector. Thatis, as a function F and can be written in several different ways (in R2)

F(x, y) = P (x, y)i +Q(x, y)j

= P (x, y)e1 +Q(x, y)e2

= 〈P (x, y), Q(x, y)〉= 〈P,Q〉 .

We’ve already seen an important example of vector fields, namely the gradientvector field, that is, if f(x, y) is a function then

∇f = 〈fx, fy〉is a vector field.

Example 60:Sketch the vector field given by F(x, y) = 〈x, x+ y〉.

Example 61:Sketch the vector field given by F(x, y) = −yi + xj.

51

5 VECTOR FIELDS

Example 62:Sketch the vector field given by F(x, y) = ye2.

Example 63:Find and sketch the gradient vector field given by f(x, y) = x2 + y2.

The gradient is given by∇f = 〈2x, 2y〉 .

Note that at each point, the gradient vector points in the direction of maximumincrease of the function, and its length represents how rapidly the function is increas-ing.

Definition 64:A vector field F is called conservative if there is a function f such that

F = ∇f.

If such an f exists, it is called a potential function for F.Note: Potential functions are not unique! Indeed, if f is any potential function,

i.e., ∇f = F, then for any constant c, f + c is another potential function, since∇c = 0, we have that

∇(f + c) = ∇f +∇c = ∇f = F.

52

5 VECTOR FIELDS

Example 65:Is the vector field F = 〈y, x〉 conservative?

Yes. Taking antiderivatives, we see that

f(x, y) = xy

works.

Example 66:Is the vector field F = 〈−y, x〉 conservative?

Taking the antiderivative of both components, we see that f must equal both −xyand xy, which can’t happen. Thus it’s not conservative.

Example 67:Is the vector field F = 〈x, y〉 conservative?

Taking the antiderivatives, we see that

f(x, y) =1

2(x2 + y2)

works.

Conservative vector fields have surprising applications in integration, that we’llcome back to in detail later.

53

6 INTEGRATION OVER CURVES

6 Integration over Curves

So far, we’ve seen double integrals over regions in R2 and triple integrals over solidsin R3. Let’s now return to single integrals, however, instead of integrating along aninterval [a, b], we wish to generalize this notation to integrating over curves C whichlie in either R2 or R3.

Definition 68:Given a smooth curve C in R2, the line integral of f(x, y) over C is given byˆ

C

f(x, y)ds,

where s is the length of the curve C. That is, this integral is being taken with respectto arclength.

This integral can be interpreted as the area of the “wall” under the function f(x, y)and above the curve C. Let’s see how to compute line integrals in practice.

Suppose C is any curve in R2 and suppose we have a parametrization of C givenby

r(t) = 〈x(t), y(t)〉 , a ≤ t ≤ b.

Recall that the length of the curve C from r(a) to r(t) is given by

s(t) =

ˆ t

a

|r′(λ)|dλ =

ˆ t

a

√(dx

dλ

)2

+

(dy

dλ

)2

dλ.

Thus, by the Fundamental Theorem of Calculus

ds

dt=

√(dx

dt

)2

+

(dy

dt

)2

,

or rather, in differential form

ds =

√(dx

dt

)2

+

(dy

dt

)2

dt.

ds is called the arclength line differential, or just llne differential.

Proposition 69:If C is a smooth curve with parametrization r(t) = 〈x(t), y(t)〉, a ≤ t ≤ b, then

ˆC

f(x, y)ds =

ˆ b

a

f(x(t), y(t))

√(dx

dt

)2

+

(dy

dt

)2

dt.

54


Note that this is independent of parametrization provided you you only traversethe curve C once.

A few notes about parametrizations:

i. If C is part of a circle of radius r, then one parametrization of C would ber(t) = 〈r cos t, r sin t〉 with appropriate bounds on t.

ii. If C is the graph of some function y = f(x), then one parametrization of C wouldbe r(t) = 〈t, f(t)〉 with appropriate bounds on t.

iii. If C is a line (in R2 or R3), we can use the equation of a line as in part (ii.), or wecan write r(t) = P + vt, where P is any point on the line C and v is a directionvector.

Example 70:Evaluate

Ć

(4 + 2xy)ds where C is the upper half of the cirve x2 + y2 = 1.First, we parametrize r(t) = 〈cos t, sin t〉 , 0 ≤ t ≤ π. Then

ˆC

(4 + 2xy)ds =

ˆ π

0

(4 + 2 cos t sin t)√

cos2 t+ sin2 tdt

=

ˆ π

0

(4 + sin 2t)dt

= 4π.

We can also do integration over piecewise differentiable curves C. If C = C1 +C2,where C1 and C2 are differentiable curves, thenˆ

C

f(x, y)ds =

ˆC1

f(x, y)ds+

ˆC2

f(x, y)ds.

Example 71:Compute

Ćxds where C is the quarter of the circle x2 + y2 = 4 from (2, 0) to (0, 2),

followed by the straight line connecting (0, 2) to (−2, 0).Let C1 be the circle arc parametrized by r1(t) = 〈2 cos t, 2 sin t〉 with 0 ≤ t ≤ π

2

and let C2 be the straight line with rapametrization r2(t) = 〈t, 2 + t〉 with −2 ≤ t ≤ 0.Then ˆ

C

xds =

ˆC1

xds+

ˆC2

xds

=

ˆ π2

0

cos t4dt+

ˆ 0

−2t√

2dt

= 4− 2√

2

55


Definition 72:If C is a piecewise smooth curve with density ρ(x, y), then the mass of C is

m =

ˆC

f(x, y)ds,

and the center of mass (x, y), where

x =1

m

ˆC

xρ(x, y)ds

y =1

m

ˆC

yρ(x, y)ds.

There are specific examples of line integrals which which will be useful later.

Definition 73:Given a curve C with r(t) = 〈x(t), y(t)〉, a ≤ t ≤ b, we can compute the line integralof f(x, y) over C with respct to x by

ˆC

f(x, y)dx =

ˆ b

a

f(x(t), y(t))x′(t)dt,

and the line integral of f(x, y) over C with respect to y by

ˆC

f(x, y)dy =

ˆ b

a

f(x(t), y(t))y′(t)dt.

In the future, we will see these written in the formˆC

P (x, y)dx+

ˆC

Q(x, y)dy,

or more notationally convenientˆC

P (x, y)dx+Q(x, y)dy.

Example 74:If C is the curve given by x(t) = t3, y(t) = t, 0 ≤ t ≤ 2 then compute

Ćy2dx+ xdy.

ˆC

y2dx+ xdy =

ˆ 2

0

t2(3t2)dt+

ˆ 2

0

t3dt

=3

525 +

24

4

56


When we parametrize C, we naturally choose an orientation of C, meaning a wayat which we travel along C. If we travelled the opposite way, we would get a newparametrization. If we call this curve −C, then

ˆC

f(x, y)ds =

ˆ−C

f(x, y)ds,

but ˆC

f(x, y)dx = −ˆ−C

f(x, y)dx

and ˆC

f(x, y)dy = −ˆ−C

f(x, y)dy.

So far, all of these examples and definitions dealth with curves in R2, but they workjust the same for curves in R3. In summary, C is a curve in R3 with parametrizationr(t) = 〈x(t), y(t), z(t)〉 , a ≤ t ≤ b. Then the line integral of f(x, y, z) over C withrespect to arclength is given by

ˆC

f(x, y, z)ds =

ˆ b

a

f(x(t), y(t), z(t))

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt.

We also have the inegrals with respect to x, y, and z

ˆC

f(x, y, z)dx =

ˆ b

a

f(r(t))x′(t)dt,

ˆC

f(x, y, z)dy =

ˆ b

a

f(r(t))y′(t)dt,

ˆC

f(x, y, z)dz =

ˆ b

a

f(r(t))z′(t)dt.

Example 75:Compute

Ćy sin zds where C is the curve parametrized by r(t) = 〈cos t, sin t, t〉 with

0 ≤ t ≤ 2π.

ˆC

y sin zds =

ˆ 2π

0

sin2 t√

2dt

=√

2π.

Now that we know how to integrate functions over curves, we can actually takethis a bit further, an integrate vector fields over curves!

57

6.1 Integrating Vector Fields over Curves 6 INTEGRATION OVER CURVES

6.1 Integrating Vector Fields over Curves

Let’s now see the importance of vector fields and their relation to integration. First,perhaps a brief motivation: In previous classes, you’ve probably learned various for-mulas for work. That is,

W = F × d,

where W is the work done by some constant force F accross some constant distanced. If F is variable, then from Math 125, we’ve learned that

W =

ˆ b

a

F (x)dx.

Moreover, in vectorial based physics, if F is a constant force vector and d is a dis-placement vector, then the scalar quantity work is given by

W = F · d.

A vector field F = 〈P,Q,R〉 can be thought of as a force field, where the vector ateach point can be considered as the direction and magnitude of force acting at thatpoint. If we have a curve C in some force field, we may be interested in computingthe work required to move a particle along this curve.

To find an integral formula for this, let C be parametrized by r(t), and we splitC up to very tiny pieces, each of acrlength ∆s. But on these little curves ∆s, theparticle travels in the direction of

T(t) =r′(t)

|r′(t)|,

the unit-tangent vector to C at r(t). Therefore, the work to move the particle on thatsmall piece ∆s is given by

W = F ·T(t)∆s.

Summing over all the little pieces, and taking the limit over partitions as in Riemannsums, we get the following formula for work:

W =

ˆC

F(x, y, z) ·T(x, y, z)ds =

ˆC

F ·Tds.

Recalling the formula for ds with paremetrization r(t), a ≤ t ≤ b we also get the

58


following integral

W =

ˆ b

a

F(r(t)) · r(t)

|r′(t)|r′(t)|dt

=

ˆ b

a

F(r(t)) · r′(t)dt

=:

ˆC

F · dr.

Summarizing the aforementioned in a definition:

Definition 76:If F is vector field defined along some curve C with parametrization r(t), a ≤ t ≤ b,the line integral of F over C is defined to be

ˆC

F · dr =

ˆ b

a

F(r(t)) · r′(t)dt =

ˆC

F ·Tds.

Example 77:Find the work done by the force field F = 〈x2,−xy〉 in moving a particle along thequarter circle r(t) = 〈cos t, sin t〉 , 0 ≤ t ≤ π/2.

By the above formula

W =

ˆ b

a


=

ˆ π/2

0

⟨cos2 t,− cos t sin t

⟩· 〈− sin t, cos t〉 dt

= −2

ˆ π/2

0

cos2 t sin tdt

=2

3cos3 t

∣∣t=π/2t=0

= −2

3.

Example 78:Find the work done by the force field F = 〈x2,−xy〉 in moving a particle along thethree-quarter circle r(t) = 〈cos t,− sin t〉 , 0 ≤ t ≤ 3π/2.

59


As before,

W =

ˆ b

a


=

ˆ π/2

0

⟨cos2 t, cos t sin t

⟩· 〈− sin t,− cos t〉 dt

= −2

ˆ π/2

0

cos2 t sin tdt

=2

3cos3 t

∣∣t=π/2t=0

= −2

3.

Note in the previous two examples, we computed the work along two differentpaths both connecting the points (1, 0) to (1, 0) and arrived the same answer. Thisis not always true. But it is true for certain vector fields as we’ll see shortly.

Example 79:Show that a constant force field does zero work on a particle that traverses the unitcircle once.

Let F = 〈c1, c2〉 denote this constant force field and let r(t) = 〈cos t, sin t〉 , 0 ≤t ≤ 2π be the usual parametrization. Then

W =

ˆ 2π

0

(−c1 sin t+ c2 cos t)dt = 0.

We can also relate these notions in terms of line integrals with respect to x, y,and z, namely, if F = 〈P,Q,R〉 and r(t) = 〈x(t), y(t), z(t)〉, a ≤ t ≤ b is some curveC, then

ˆC

F · dr =

ˆ b

a


=

ˆ b

a

(P (r(t))x′(t)dt+Q(r(t))y′(t)dt+R(r(t))z′(t)dt)

=

ˆC

(Pdx+Qdy +Rdz).

Besides just computing various line integrals for force fields, there is another sig-nificant reason for introducing vector fields and their various integrals, which relatethe previous examples.

60


Theorem 80 (Fundamental Theorem for Line Integrals):Let C be a smooth curve parametrized by r(t), a ≤ t ≤ b. Let f be a differentiablefunction of any number of variables. Then

ˆC

∇f · dr = f(r(b))− f(r(a)).

This is an analogue to the usual Fundamental Theorem of Calculus that we’reused to (

´ baF ′(x)dx = F (a)− F (b)), and in fact is a Corollary of it. Indeed,

Proof:

ˆC

∇f · dr =

ˆ b

a

(∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt

)dt

=

ˆ b

a

d

dt(f(r(t))dt by chain rule,

= f(r(b))− f(r(a)) by FTC.

What does this theorem mean for us? Since for gradient fields ∇f , the integralonly depends on the endpoints of the curve C and not the actual path taken! Thatis, the work done along any path connecting two points of a gradient force field is thesame.

This idea leads to a new notion:

Definition 81:We say that the integral

Ć

F · dr is independent of path if

ˆC1

F · dr =

ˆC2

F · dr

for any two paths C1 and C2 in the domain of F with the same start and end points.

This definition leads to a corollary of our FTLI:

Corollary 82:If F is a conservative vector field, then

Ć

F · dr is independent of path.

Remark: We saw in line integrals of scalar functions with respect to arclength,that the result was independent of paramtrization (as long as we didn’t traversethe curve multiple times). This is no longer true when dealing with vector fields.Intuitively, this makes sense in the concept of work: Suppose we want to move a ball

61


from a height of h to the ground in the presence of a gravitation force field. Thenthe work should be positive, since the motion of the ball in the same direction as thefield. However, if we were to traverse the path backwards, that is, taking the ballfrom the ground straight up to a height of h, we should have a negative work doneby the gravitiational field. Formally, if C is any curve and f a scalar function, thenˆ

−Cfds =

ˆC

fds,

however if F is a vector field, thenˆ−C

F · dr = −ˆC

F · dr.

Definition 83:A curve is called closed if its endpoint coincides with the starting point.

We can use closed curves to determine if certain integrals are path independent.Indeed, given two paths C1 and C2 with the same starting and ending points, thenthe path C1 + (−C2) is a closed curve. Conversely, given any closed path C, we canbreak it up (anywhere) to obtain two paths C1 and C2 with the same starting andending points. If for every closed curve C,

Ć

F · dr = 0, then

0 =

ˆC

F · dr

=

ˆC1

F · dr +

ˆ−C2

F · dr

=

ˆC1

F · dr−ˆC2

F · dr,

that is, ˆC1

F · dr =

ˆC2

F · dr.

Conversely, ifĆ

F ·dr is independent of path, then let C be any closed curve, andconsider C = C1 + C2 for any composition of curves with the opposite starting andending points. Then ˆ

C

F · dr =

ˆC1

F · dr +

ˆC2

F · dr

=

ˆC1

F · dr−ˆ−C2

F · dr

= 0.

Thus, we’ve shown the following theorem

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Theorem 84:Ć

F · dr is path independent if and only ifĆ

F · dr = 0 for every closed curve C.

We can go even further with path independence, which is a partial converse toCorollary 82.

Theorem 85:Suppose F is a continuous vector field on a “nice” (open and connected, we’ll revisitthis shortly) domain D. If

Ć

F · dr is path independent in D, then F is conservative.

See the text for the reasoning of the theorem. It’s not particularly enlighteningin telling us if a vector field F is conservative. That is, by a combination of Theorem84 and Theorem 85, we have to show that

Ć

F · dr = 0 for all closed curves C, whichis computationly an unreasonable criteria, since this included an infinite number ofintegrations. So the question remains, how do we quickly know if a vector field F isconvervative?

Suppose F = 〈P,Q〉 is conservative. That is, we have a potential function f suchthat ∇f = F, and so

P =∂f

∂x, Q =

∂f

∂y.

And then, since mixed partials are equal, we have that

∂P

∂y=

∂2f

∂y∂x=

∂2f

∂x∂y=∂Q

∂x.

In turns out that this equality is not coincidental, and that if the domain D of avector field is “sufficiently nice”, then the converse to the above holds.

Theorem 86:Let F = 〈P,Q〉 be a vector field on an open, simply-connected domain D such thatP and Q have continuous first partial derivatives and

∂P

∂y=∂Q

∂x

throughout D. Then F is convervative.Note that this is only applicable in R2. We’ll see criteria for conservative vector

fields a bit later in the course.

The theorem is a pretty straightforward criteria for determining if a vector field isconservative, however, this notion of “simply-connected” domain cannot be ignored,and so we need to explain it in a bit more detail. Moreover, as the technical aspectsof the implications of the soon-to-be defined term “simply-connected” are suprisingquite deep and well beyond the scope of this course, we shall not attempt to explainwhy this theorem is true at this time. Come back to this after

Green.

63


Definition 87:Let D be a region. We say D is open if every point P of D, there is a disk (ofappropraite dimension) centered at P completely contained in D. That is D doesn’tcontain any boundary points (e.g., (a, b) in R, x2 + y2 < 1 in R2, and r < 1 in R3).

We say D is connected if any two points in D can be connected by a path thatcompletely lies in D.

A simple curve is a curve that contains no self-intersection with the possibleexception of its endpoints, that is, for any a < t1 < t2 < b, we have that r(t1) 6= r(t2).

We say D is simply-connected if it’s connected (not necessarily open) and thatevery simple closed curve in D encloses only points that are in D. Intuitively, thismeans D contains no holes.

Example 88:Is F = 〈x− y, x− 2〉 conservative?

This vector field is defined everywhere, however,

−1 6= 1,

and hence can’t be conservative.

Example 89:Is F = 〈3 + 2xy, x2 − 3y2〉 conservative?

64


This vector field is defined everywhere, and we see that

2x = 2x,

so Theorem 92 implies F is conservative. One potential function would be

f = 3x+ x2y − y3.

Example 90:Show that

Ć

(2xe−ydx+(2y−x2e−ydy) is path independent and evaluate the integralalong any path from (1, 0) to (2, 1).

By Corollary 88, let’s show that the vector field F = 〈2xe−y, 2y − x2e−y〉 is con-servative to imply path independence. Checking partials, we have that

−2xe−y = −2xe−y,

so it’s conservative and hence path independent. Moreover, we have a potentialfunction

f(x, y) = x2e−y + y2.

Thus by the FTLI, let C be any curve connecting the two points and soˆC

∇fdr = f(2, 1)− f(1, 0) =4

e.

Example 91:

If F =⟨−y

x2+y2, xx2+y2

⟩, show that ∂P

∂y= ∂Q

∂x, yet F is not conservative.

We see that both derivatives equal

y2 − x2

(x2 + y2)2.

However, this vector field F is only defined on R2 \ {(0, 0)}, which is not a simply-connected region, so the theorem doesn’t apply. If we compute the line integral alongthe upper circle from (1, 0) to (−1, 0)

ˆ π

0

〈− sin t, cos t〉 · 〈− sin t, cos t〉 dt = π,

and along the lower circle from (1, 0) to (−1, 0),ˆ π

0

〈sin t, cos t〉 · 〈− sin t,− cos t〉 dt = −π,

showing thatĆ

F · dr is not path independent, and hence can’t be conservative.

65

6.2 Green’s Theorem 6 INTEGRATION OVER CURVES

6.2 Green’s Theorem

So far we’ve seen line integrals of scalar functionsĆfds, and then we related vector

fields to line integrals in the form ofĆ

F · dr in order to utilize the fundamentaltheorem for line integrals when dealing with conservative vector fields ∇f = F,ˆ

C

∇f · dr = f(r(b))− f(r(a)).

There’s even more structure to be seen with line integrals besides the FTLI. Infact, they relate closely to double integrals! But first, a quick definition.

Definition 92:Let C be a simple (no self-intersection) closed curve in R2. We say that C haspositive orientation if we refer to a single, counterclockwise traversal of C.

That is, if D is the region enclosed by the curve C, then from the point-of-viewof travelling along C via a parametrization r(t), D will always lie to our left.

Theorem 93 (Green’s Theorem):Let C be a positively oriented, piecewise-differentiable, simple closed curve in R2 andlet D be the region bounded by C. If P and Q have continuous partial derivatives onan open region that contains D, thenˆ

C

Pdx+Qdy =

¨D

(∂Q

∂x− ∂P

∂y

)dA.

Notational Aside: When dealing with line integrals over closed, positively orien-tated curves C, we’ll use the notation˛

C

fds,

˛C

F · dr.

Also, we’ll use ∂D denote the curve C that bounds D (called the boundary of D) andcan write ˆ

∂D

F · dr,˛∂D

F · dr,

and similarly for scalar line integrals.We’re not going to prove Green’s theorem as it’s rather tedious, and the whole

proof is beyond the scope of the course. See the text for a proof of theorem withrather harsh restrictions on D. Let’s now focus on the various applications of thistheorem.

Remark: If F = 〈P,Q〉, then Green’s Theorem holds for all any vector field F,whether it’s conservative or not. However, if F is conservative, then our calculationusing the equality of mixed partials always yields 0. This makes sense, since the FTLIwould have the same endpoints for a closed curve, so it’s nothing new.

66


Example 94:Evaluate

¸Cx4dx+ xydy, where C is the triangular curve with vertices (0, 0), (0, 1),

(1, 0).We could compute the line integral using usual methods, but it would require

three seperate line integrals, so that doesn’t seem optimal.Moreover, based on our previous remark, we should check to see if the vector field

F = 〈x4, xy〉 is conservative, because then we know integral has to be zero (notethat F is defined on all of R2 so it’s open and simply-connected). Checking partialderivatives

∂P

∂y= 0,

∂Q

∂x= y

which are not equal on the whole domain, so it’s not conservative.Now we apply Green’s Theorem

˛C

x4dx+ xydy =

¨D

(y − 0)dA

=

ˆ 1

0

ˆ 1−x

0

ydydx

=

ˆ 1

0

(1− x)2

2dx

= − (1− x)3

6

∣∣∣∣x=1

x=0

=1

6.

Example 95:Evaluate

¸C

F · dr, where F = 〈y4, 2xy3〉 and C is the circle x2 + y2 = 4.This line integral wouldn’t be so bad of a direct computation, but let’s avoid it

anyway.Is F conservative? The domain is again all of R2 and

∂P

∂y= 4y3,

∂Q

∂x= 2y3,

which aren’t equal, so it’s not immediately zero.

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Now, let’s use Green’s theorem˛C

F · dr =

¨D

(2y3 − 4y3)dA

= −2

ˆ 2π

0

ˆ 2

0

r3 sin3 θrdrdθ

= −26

5

ˆ 2π

0

sin3 θdθ

= −26

5

ˆ 2π

0

(1− cos2 θ) sin θdθ

=26

5

(cos θ − 1

3cos3 θ

)∣∣∣∣2π0

= 0.

Example 96:Evaluate

¸Cxdy where C is the upper half of the circle x2 + y2 = 4 followed by the

straight line from (−2, 0) to (0, 2).This is not conservative, so by Green’s theorem

˛C

xdy =

¨D

1dA

= Area(D)

= 2π.

Example 96 leads us to a suprisingly useful application of Green’s theorem, andthat’s using it in the opposite direction when dealing with areas. Since

Area(D) =

¨D

1dA,

we could apply Green’s theorem in reverse with the restriction that F = 〈P,Q〉saitisfies

∂Q

∂x− ∂P

∂y= 1.

There are obviously many vector fields that could satisfy this, but a few importantones are as follows:

P Q F0 x 〈0, x〉−y 0 〈−y, 0〉−y

2x2

12〈−y, x〉

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Using these, we get the following formulas for the Area(D)

Area(D) =

˛C

xdy = −˛C

ydx =1

2

˛C

xdy − ydx.

Example 97:Find the area inside of the ellipse x2

a2+ y2

b2= 1.

Computing the area as usual would require a change of variables (the modifiedpolar coordinates), which isn’t too bad, but i’ts even easier using Green’s theorem.The boundary ellipse has nice parametrization given by x(t) = a cos t, y(t) = b sin t,0 ≤ t ≤ 2π. Then

Area(D) =1

2

˛C

xdy − ydx

=1

2

ˆ 2π

0

(ab cos2 t+ ab sin2 t)dt

= abπ.

69

7 OPERATIONS ON VECTOR FIELDS

7 Operations on Vector Fields

So far when dealing with the important line integral theorems regarding vector fields,we’ve mostly restricted ourselves to working in R2, specifically in determing when avector field is conservative. Recall that F = 〈P,Q〉 on an open, simply-connected Dis conservative if

∂P

∂y=∂Q

∂x.

However, the notion of F being conservative, just means that F = ∇f for somepotential function f . This definition is independent of dimension, so why did we needto restrict ourselves as in Theorem 86?

Suppose F = 〈P,Q,R〉 is conservative, that is, ∇f = F. So

∂f

∂x= P,

∂f

∂y= Q,

∂f

∂z= R,

and again, since mixed partials are equal, we this time have

∂P

∂y=

∂2f

∂x∂y=∂Q

∂x

∂P

∂z=

∂2f

∂x∂x=∂R

∂x∂Q

∂z=

∂2f

∂z∂y=∂R

∂y.

Is it true that if F satisfies the above three equations, then F is conservative?In similar fashion to the two-dimensional case, there are certain restrictions. Also,unlike the two-dimensional case, it’s a bit more tedius to take antiderivatives to finda potential function. This leads to wonder, if there’s a better way, and if there’s anyhidden structure in the above computation?

7.1 The Del Operator

Throughout, let F = 〈P,Q,R〉.

Definition 98:Let ∇ represent the vectorial differential operator

∇ :=

⟨∂

∂x,∂

∂y,∂

∂z

⟩.

It is called the del operator.

70

7.1 The Del Operator 7 OPERATIONS ON VECTOR FIELDS

This may seem strange at first, but it makes some sense from a notational aspect.Indeed, let f(x, y, z) be a scalar function. Then

∇f =

⟨∂

∂x,∂

∂y,∂

∂z

⟩f

=

⟨∂f

∂x,∂f

∂y,∂f

∂z

⟩,

so that certainly lines up with usual notation of gradient, and how vectors and scalarswork with each other.

We also have other operations which act with vectors, namely, the dot productand the cross product.

Definition 99:The curl of a vector field F is

curl (F) := ∇× F.

Note, that by expanding the cross product, we can get the following formula forthe curl:

curl (F) = ∇× F

=

∣∣∣∣∣∣i j j∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣=

(∂R

∂y− ∂Q

∂z

)i +

(∂P

∂z− ∂R

∂x

)j +

(∂Q

∂x− ∂P

∂y

)k.

The above is the actual definition of curl, but it’s harder to recognize the crossproduct in that form.

Notice from our previous calculation, we immediately get the following proposi-tion.

Proposition 100:If F is conservative, then curl (F) = 0, that is, given any function f(x, y, z),

∇×∇f = 0.

What is the physical meaning of curl? If F represents the velocity flow of a fluid,then at a point P in the fluid curl (F) represents the the direction of the axis thatfluid tends to rotate around at P , and the magnitude is how quickly the particlesrotates around said axis. Note that if curl (F) (P ) = 0, then the fluid is free of

71


rotation at that point, or irrotational at P . If curl (F) = 0 everywhere, we say thatF is irrotational. All conservative vector fields are irrotational. We give furtherexplanations later on, once we have developed more tools pertaining to ∇.

It turns out the curl is the answer to our conservative question as well.

Theorem 101:If F is defined on an open, simply-connected domain, and curl (F) = 0, then F isconservative. That is, on open, simply-connected domains, conservative is equivalentto irrotational.

Aside: As simply-connected is much more difficult to characterize in three-dimensions,we shall restrict out attention to only globally defined vector fields, that is, vectorfields with domain R3 which is well known to be open and simply-connected.

Now for our other vector operation:

Definition 102:The divergence of a vector field F is

div (F) := ∇ · F.

Expanding the dot product, we see that

div (F) = ∇ · F

=

⟨∂

∂x,∂

∂y,∂

∂z

⟩· 〈P,Q,R〉

=∂P

∂x+∂Q

∂y+∂R

∂z.

Again, since mixed partials are equal, an identical calculation to the ∇×∇F, weget the following:

Proposition 103:For any vector field F, div (curl (F)) = 0, that is,

∇ · ∇ × F = 0.

What is the physical meaning of divergence? Again, let F represent the velocityflow of a fluid, the div (F) (P ) measures the net rate (with respect to time) of themass of fluid to diverge from the point P . If div (F) = 0, then F is said to beincompressible.

Let’s now do a few computational examples.

72


Example 104:Compute curl (F) and div (F), where F = 〈xz, xyz,−y2〉.

Let’s compute the curl first:

curl (F) =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

xz xyz −y2

∣∣∣∣∣∣= 〈−2y − xy, x− 0, yz − 0〉= 〈−y(2 + x), x, yz〉

And now the divergence:

div (F) =

⟨∂

∂x,∂

∂y,∂

∂z

⟩·⟨xz, xyz,−y2

⟩= z + xz + 0

= (1 + x)z.

Example 105:Is F = 〈y2z3, 2xyz3, 3xy2z2〉 conservative?

Let’s compute the curl:

curl (F) =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

y2z3 2xyz3 3xy2z2

∣∣∣∣∣∣=⟨6xyz2 − 6xyz2, 3y2z2 − 3y2z2, 2yz3 − 2yz3

⟩= 0,

and since F is defined on all of R3, we conclude that it’s conservative.

There’s one final combination of vector operations of ∇, which is probably themost important, yet is beyond the scope of course:

Definition 106:The Laplacian of a function f is

∆f = ∇ · ∇f = div (∇f) .

∆ is called the Laplace operator, and it induces the very basis of most basic physics ∆ = ∇2

equations, Laplace’s equation∆f = 0,

73


which is a partial differental equation, and is valid in every dimension (since there’sno curl), e.g., in R2

∂2f

∂x2+∂2f

∂y2= 0.

Its solutions describe static electromagnetic potentials, as well as, gravitationalpotentials. If ∆f = 0, the function f is called harmonic.

Finally, we remark that we can actually express Green’s theorem a couple of newways using our new operators.

First, let F = 〈P,Q〉 be a vector field on R2. We can consider it as a vector fieldon R3, by writing

F = 〈P,Q, 0〉

(the natural extension). Under this consideration, we then compute

curl (F) =

⟨0, 0,

∂Q

∂x− ∂P

∂y

⟩.

Then,

curl (F) · k =∂Q

∂x− ∂P

∂y,

and so ˛C

F · dr =

¨D

curl (F) · kdA.

Second, let C be given by r(t) = 〈x(t), y(t)〉, a ≤ t ≤ b. Then the unit-tangent toC is given by

T(t) =1

|r′(t)|〈x′(t), y′(t)〉 ,

and the outward unit-normal to C is given by

n(t) =1

|r′(t)|〈y′(t),−x′(t)〉

74


(the other normal, is the usual one defined in Math 126). Now consider

˛C

F · nds =

ˆ b

a

F · n|r′(t)|dt

=

ˆ b

a

F · 〈y′(t),−x′(t)〉|r′(t)|

|r′(t)|dt

=

ˆ b

a

(Py′(t)−Qx′(t))dt

=

ˆC

Pdy −Qdx

=

ˆC

−Qdx+ Pdy

=

¨D

(∂P

∂x+∂Q

∂y

)dA

=

¨D

div (F) dA.

Thus we have the two vector forms of Green’s theorem, summarized as follows

Proposition 107:

˛C

F · dr =

¨D

curl (F) · kdA

and ˛C

F · nds =

¨D

div (F) dA.

75

8 INTEGRATION OVER SURFACES

8 Integration over Surfaces

We’ve now seen various ways to integrate over curves C in either R2 or R3, and thismethodology leads to explore integration over surfaces S in R3. However, as we sawwith line integrals, we needed to parametrize said curves, and in similarity, we’regoing to have to extend this notion, and parametrize surfaces as well.

8.1 Parametric Surfaces

Why did we need to parametrize the curves for line integrals? In our definition for lineintegrals we needed a way to characterize the arclength differential ds. In preview ofwhat’s to come, we’ll similiarly need a way to characterize the surface area differentialdS in generality.

Recall for surfaces defined via z = f(x, y), we had our surface area differentialdefined by

dS =

√1 +

(∂f

∂x

)2

+

(∂f

∂y

)2

dA.

We derived this formula from parametrizing curves on the surface and approximatingvia the tangent plane. We wish to try something similiar here.

In similarity to parametrizing curves via the vector-valued function r(t) of a singlevariable t, this leads to the following definition:

Definition 108:The set of all points (x, y, z) in R3 such that

x = x(u, v), y = y(u, v), z = z(u, v)

and (u, v) varies in some region D in R2 is called a parametric surface S, and thevector-valued function

r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉

are called the parametric equations for the surface S.

Let’s first practice dealing with parametric surfaces before going further.

Example 109:Identify the surface r(u, v) = 〈u, v, 2u+ 2v2〉, D = R2.

Since x = u and y = v, we see that z = 2x + 2y2. Thus the surface is the graphof the function

z = f(x, y) = 2x+ 2y2.

76

8.1 Parametric Surfaces 8 INTEGRATION OVER SURFACES

Example 110:Identify the surface r(u, v) = 〈2 cosu, v, 2 sinu〉, D = R2.

Since y = v is not in the other two, y can be anything. Moreover, note that

x2 + z2 = 4 cos2 u+ 4 sin2 u

= 4.

Thus the surface is an infinite cylinder along the y-axis of radius 2.

Example 111:Find a parametrization of the cylinder x2 + y2 = 9.

As an almost converse to the previous example, we could describe this in cylin-drical coordinates via r = 3, where θ and z could be anything, so letting θ = u andz = v, we get the parametrization

r(u, v) = 〈3 cosu, 3 cosu, v〉

and D = [0, 2π]× R.

Example 112:Find a parametrization of the surface z = 2xy + x2.

When dealing with graphs, the parametrization is given to us after relabling u = xand v = y. That is,

r(u, v) =⟨u, v, 2uv + y2

⟩,

and D = R2.

Example 113:Find a parametrization of the sphere x2 + y2 + z2 = 4.

In spherical coordinates, the sphere is written at ρ = 2, so letting θ and φ be ouru and v, we get

r(u, v) = 〈2 cosu sin v, 2 sinu sin v, 2 cos v〉with D = [0, 2π]× [0, π].

Example 114:Find a parametrization of the surface of revolution given by revolving y = f(x) aroundthe x-axis, a ≤ x ≤ b.

Note that point on a surface of revolution can be expressed as (x, f(x) cos θ, f(x) sin θ),where θ is the angle from the point to the x-axis with the positive xy-plane. Thusletting u = x and v = θ, we get

r(u, v) = 〈u, f(u) cos v, f(u) sin v〉 ,

with D = [a, b]× [0, 2π].

77


Now, let’s see how to compute tangent planes to parametric surfaces, and asbefore, use tangent plane approximations to define the surface area differential.

Recall, to find the tangent plane to the surface S at the point P , we just needa normal vector to the plane, or rather via the cross product, two vectors that aretangent to the plane at P . Let r(u, v) denote the parametric equations to S and letP0 = r(u0, v0). For the fixed u0, letting v vary, we see that r(u0, v) is curve Cv on Spassing through P0, and the tangent vector along Cv at P0 is given by

rv(u0, v0) =∂r

∂v(u0, v0) =

⟨∂x

∂v(u0, v0),

∂y

∂v(u0, v0),

∂z

∂v(u0, v0)

⟩.

Similarly,

ru(u0, v0) =∂r

∂u(u0, v0)

is another tangent vector at P0.If

ru × rv 6= 0,

we say that the surface is smooth, i.e., the parametric equations are differentiablemany (infinitely many) times, and so there are no sharp corners on the surface. Thismeans that ru × rv is the desired normal vector for describing the tangent planes.

Note: ru × rv can and will be zero at certain places where there is not a corner.This is due to the choice of parametrization and is called an artificial singularity. Theusual parametrization for the sphere has this at the North and South pole (check foryourself), but the sphere is obviously smooth. Therefore, the parametrization is nottypically ideal for computing tangent planes, but for integration purposes, we canignore these isolated discrete singularities without affecting the calculation.

As before in describing surface area, the infinitismal area of the tangent plane isthen described by

|ru × rv|∆A,and since this area approximates the surface area, taking limits in the form of Rie-mann sums, we’ve derived the surface area differential in terms of parametric surfaceequations r(u, v):

Definition 115:Let r(u, v) be parametric equations to a surface S defined on D. Then the surfacearea differential is given by

dS = |ru × rv|dA,and so the surface area is given by

Area(S) =

¨S

dS =

¨D

|ru × rv|dA.

78


Example 116:Find the surface area of the sphere of radius 2.

Let’s look at the sphere centered at the origin x2 + y2 + z2 = 4, and recall it’sparametrized by

r(u, v) = 〈2 cosu sin v, 2 sinu sin v, 2 cos v〉 ,with D = [0, 2π]× [0, π]. Moreover, we see that

ru × rv =

∣∣∣∣∣∣i j j

−2 sinu sin v 2 cosu sin v 02 cosu cos v 2 sinu cos v −2 sin v

∣∣∣∣∣∣=⟨−4 cosu sin2 v,−4 sinu sin2 v,−4 sin2 u sin v cos v − 4 cos2 u sin v cos v

⟩= −4

⟨cosu sin2 v, sinu sin2 v, sin v cos v

⟩,

and so

dS = |ru × rv|dA

= 4√

cos2 u sin4 v + sin2 u sin4 v + sin2 v cos2 vdA

= 4√

sin2 v sin2 v + sin2 v cos2 vdA

= 4 sin vdA.

Thus

Area(S) =

¨D

dS

=

ˆ π

0

ˆ 2π

0

4 sin vdudv

= 8π

ˆ π

0

sin vdv

= 16π.

Notice that this is equivalent to the definition we learned for the surface area ofof the graph of a function z = f(x, y). Indeed, parametric equations are defined tobe

r(x, y) = 〈x, y, f(x, y)〉 ,and so

rx × ry =

∣∣∣∣∣∣i j k

1 0 ∂f∂x

0 1 ∂f∂y

∣∣∣∣∣∣=

⟨−∂f∂x− ∂f

∂y,−1

⟩

79

8.2 Surface Integrals 8 INTEGRATION OVER SURFACES

and so

|rx × ry| =

√1 +

(∂f

∂x

)2

+

(∂f

∂y

)2

,

which is exactly what we found before in this case.

8.2 Surface Integrals

We’re now ready to define surface integral of scalar functions f(x, y, z).

Definition 117:Given a function f(x, y, z) on the a surface S defined by the parametrization r(u, v)over a region D, the surface integral of f over the surface S is defined to be¨

S

f(x, y, z)dS =

¨D

f(r(u, v))|ru × rv|dA.

Example 118:Evaluate

˜SydS, where S is the surface z = x+ y2 defined above [0, 1]× [0, 2].

We parametrize the surface as usual

r(u, v) =⟨u, v, u+ v2

⟩with D = [0, 1]× [0, 2]. Then

ru = 〈1, 0, 1〉 , rv = 〈0, 1, 2v〉

and soru × rv = 〈−1,−2v, 1〉 ,

with|ru × rv| =

√2 + 4v2.

Finally, ¨S

ydS =

¨D

v√

2 + 4v2dA

=

ˆ 2

0

ˆ 1

0

v√

2 + 4v2dudv

=

ˆ 2

0

v√

2 + 4v2dv

=1

12(2 + 4v2)3/2

∣∣v=2

v=0

=1

12(183/2 − 23/2)

80



˜SzdS, where S is the cylinder x2 + y2 = 1 between the planes x = 0 and

z = 1 + x.First, we parametrize the cylinder in the usual way

r(u, v) = 〈cosu, sinu, v〉

and now we use the intersecting planes to find the bounds. As u is playing the roleof θ it should range from 0 to 2π, and since v is playing the role of z it should rangefrom 0 to 1 + x = 1 + cosu. Therefore our region D is given by

D = {(u, v) : 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1 + cosu}.

Computingru = 〈− sinu, cosu, 0〉 , rv = 〈0, 0, 1〉 ,

we get|ru × rv| = | 〈cosu, sinu, 0〉 | = 1.

Finally,¨S

zdS =

¨D

vdA

=

ˆ 2π

0

ˆ 1+cosu

0

vdvdu

=1

2

ˆ 2π

0

(1 + cosu)2du

=3π

2.

As usual, we have the standard applications of mass

m =

¨S

ρ(x, y, z)dS

and center of mass

x =1

m

¨S

xρ(x, y, z)dS, y =1

m

¨S

yρ(x, y, z)dS, z =1

m

¨S

zρ(x, y, z)dS.

In similiarity to line integrals, we would like to extend this notion of surfaceintegrals from scalar fields to vector fields. Since line integrals depended on theorientation of the curve, it makes sense that surface integrals should depend on asimilar notion. Let’s extend our notion of orientation to two dimensions.

81


Definition 120:Given a surface S, if it is possible choose a unit normal n at every point P of S,so that n varies continuously, we say that S is an oriented surface, and the givenchoice of n is called the orientation.

Remark: We will only deal with orientable surfaces in the course, which roughlymeans, the surface has an inside and an outside. There are two possible orientationsfor every orientable surface. The Mobius strip is an example of a non-orientablesurface.

To see how we should define surface integrals for vector fields, let’s again use somephysical intuition. Let S be a surface with orientation n. Imagine a fluid with densityρ(x, y, z) flowing through the surface S with velocity field V(x, y, z). Then the rateof flow (mass per unit time) per unit area is given by the vector field

ρV.

Dividing S into tiny pieces of (surface) area ∆S, then we can approximate the massof fluid passing through the region of area ∆S in the direction n per unit time, whichwe denote by Φ, as

Φ = ρV · n∆S.

Making the subdivisions smaller, summing up and taking limits as in a Riemann sum,we obtain the following integral

Φ =

¨S

ρV · ndS.

Definition 121:If F is a continuous vector field defined on an oriented surface S with orientation n,the surface integral of F over S or the flux integral of F is¨

S

F · dS :=

¨S

F · ndS.

This Φ represents the flux of a vector field F through a surface S. It can bethought of as an analogue of work in the following regard: The work W can looselybe thought of a way to take into account all forces acting along a path. The flux φtakes into an account all forces acting along a surface.

Let’s analyze this definition to see if there’s an easier way to compute these surfaceintegrals. To this end, let S be a surface with parametrization r(u, v), (u, v) in someregion D. Then we’ve seen that ru × rv is a normal vector on S, and hence oneunit-normal vector on S can be given by

ru × rv|ru × rv|

,

82


that is,

n = ± ru × rv|ru × rv|

.

Recalling our formula for dS, we see that

¨S

F · dS =

¨S

F · ndS

=

¨D

F ·(± ru × rv|ru × rv|

)|ru × rv|dA

= ±¨D

F · ru × rvdA.

There is no natural choice of n unfortunately, however, once we’ve chosen aparametrization r(u, v), we can use the following convention:

Definition 122:If F is a continuous vector field defined on an oriented surface S with parametrizationr(u, v), (u, v) in D, and orientation n = ru×rv

|ru×rv | the flux integral of F is given by

¨S

F · dS =

¨S

F · ndS

=

¨D

F · ru × rvdA.


˜S

F · dS, where F = 〈y, x, z〉 and S is the surface z = 1 − x2 − y2 abovethe plane z = 0, oriented downwards.

Using the usual parametrization

r(u, v) =⟨u, v, 1− u2 − v2

⟩and D is the unit disk u2 + v2 ≤ 1. Then

ru = 〈1, 0,−2u〉 , rv = 〈0, 1,−2v〉

andru × rv = 〈2u, 2v, 1〉 .

Since the z-coordinate is always positive, this gives the wrong orientation (we should’ve

83


switched u and v), so we’ll just take the negative of this 〈−2u,−2v,−1〉 . Finally¨S

F · dS =

¨D

⟨v, u, 1− u2 − v2

⟩· 〈−2u,−2v,−1〉 dA

=

¨D

(−2uv − 2uv + u2 + v2 − 1)dA

=

¨D

(u2 + v2 − 4uv − 1)dA

=

ˆ 2π

0

ˆ 1

0

(r2 − 4r2 cos θ sin θ − 1)rdrdθ

= −π2

Example 124:Find the flux of the vector field F = 〈z, y, x〉 across the unit sphere x2 + y2 + z2 = 1with outward orientation.

Using the standard parametrization

r(u, v) = 〈cosu sin v, sinu sin v, cos v〉 ,with D = [0, 2π]× [0, π]. Then

ru × rv = 〈− sinu sin v, cosu sin v, 0〉 × 〈cosu cos v, sinu cos v,− sin v〉=⟨− cosu sin2 v,− sinu sin2 v,− sin v cos v

⟩.

Note that this vector does not point outward: take u = 0, v = π2, we get 〈−1, 0, 0〉.

Thus, we take the negative of this to match our orientation. Now,¨S

F · dS =

¨D

〈cos v, sinu sin v, cosu sin v〉 ·⟨cosu sin2 v, sinu sin2 v, sin v cos v

⟩dA

=

¨D

(cosu cos v sin2 v + sin2 u sin3 v + cosu cos v sin2 v)dA

=

ˆ 2π

0

ˆ π

0

(2 cosu cos v sin2 v + sin2 u sin3 v)dvdu

...

=4π

3.

In relation to the previous example, there’s a special case of orientation we wouldlike to introduce. Suppose S is the surface that is the boundary to a 3-dimensionalregion E (like the sphere being the boundary to a ball), the positive orientationon S is the orientation n which is always outward-pointing from the solid E. Thenegative orientation is the orientation n that is always inward-pointing into thesolid. Thus, in the previous example, the prescribed orientation was the positive one.

84

8.3 Stokes’ Theorem 8 INTEGRATION OVER SURFACES

8.3 Stokes’ Theorem

Similiar to Green’s theorem which related the double integral over a region to theintegral over its boundary, we can do a similar thing with surfaces.

Definition 125:Let S be an oriented surface with orientation n with boundary curve C = ∂S. Then ninduces the positive orientation on C. That is, if you walk in the positive directionon C with your head pointing in the direction of n, then the surface lies to your left.

Theorem 126 (Stokes’ Theorem):Let S be an oriented, piecewise-smooth surface, that is bounded by a simple, closed,piecewise-smooth boundary curve C with positive orientation. Let F be a vector fieldwith whose components have continuous first partial derivatives on an open region inR3 containing S. Then ˛

C

F · dr =

¨S

curl (F) · dS.

Stokes’ Theorem is an immediate analogue to Green’s theorem in a higher dimen-sion. Recall Green’s theorem was for regions D in R2 with boundary curve C, so theonly difference is now dealing with a more general surface S in R3 with boundarycurve. Moreover, recall that when finding whether a vector field is conservative

∂Q

∂x− ∂P

∂y,

was the desired quantity to analyze and it also happens to be the integrand in Green’stheorem over the region D. Then when generalizing the question to R3, we realizedthe curl was the higher dimensional analogue, which is again preciesely the integrandover the surface S in Stokes’ theorem. In fact, Green’s theorem is a special case ofStokes’ theorem! If S is a region in the xy-plane, we can consider as a surface in R3

with orientation n = k. If F = 〈P,Q〉 is a vector field defined on the plane, then wecan consider the natural extension F = 〈P,Q, 0〉, and (since P and Q are independentof z)

curl (F) =

⟨0, 0,

∂Q

∂x− ∂P

∂y

⟩.

Thus by Stokes’ theorem˛∂S

F · dr =

¨S

curl (F) · dS

=

¨S

⟨0, 0,

∂Q

∂x− ∂P

∂y

⟩· 〈0, 0, 1〉 dA

=

¨S

(∂Q

∂x− ∂P

∂x

)dA,

85

8.3 Stokes’ Theorem 8 INTEGRATION OVER SURFACES

which is precisely Green’s theorem!Now let’s do some examples.


¸C

F · dr, where F = 〈−y2, x, z2〉 and C is the curve of intersection of theplane y+ z = 2 and the cylinder x2 + y2 = 1, oriented counterclockwise when viewedfrom above.

Note that C is an elllipse when viewed in the plane y + z = 2. Rather thancomputing this line integral, let’s use Stokes’ theorem and compute a surface integral.There are many surfaces with this boundary, but the easiest one is the surface in theplane y + z = 2 constrained by the curve C. Let’s call this surface S. Then S is thegraph of the z = 2− y, and as such has parametrization

r(u, v) = 〈u, v, 2− v〉 ,

where u and v are bounded by u2 + v2 = 1, that is,

D = {(u, v) : u2 + v2 ≤ 1}.

Thenru = 〈1, 0, 0〉 , rv = 〈0, 1,−1〉 ,

and soru × rv = 〈0, 1, 1〉 .

Note that this gives the desired orientation. Morever, we see that

curl (F) = 〈0, 0, 1 + 2y〉 .

Thus, by Stokes’ theorem˛C

F · dr =

¨S

curl (F) · dS

=

¨D

〈0, 0, 1 + 2v〉 · 〈0, 1, 1〉 dA

=

¨D

(1 + 2v)dA

=

ˆ 2π

0

ˆ 1

0

(1 + 2r sin θ)rdrdθ

=

ˆ 2π

0

(1

2+

2

3sin θ

)dθ

= π.

86

8.4 Divergence Theorem 8 INTEGRATION OVER SURFACES

Note that Stokes’ theorem would apply to any surface with boundary curve C.This is useful when relating different surfaces with similar boundaries.


˜S

curl (F) · dS, where F = 〈xz, yz, xy〉 where S is part of the sphere

x2 + y2 + z2 = 4 above the plane z =√

3, oriented upwards.The boundary curve C is the intersection of the plane z =

√3 and the sphere

x2 + y2 + z2 = 4, so substituting, yields the equations x2 + y2 = 1 and z =√

3, wecan then parametrize the this curve via

r(t) =⟨

cos t, sin t,√

3⟩, 0 ≤ t ≤ 2π.

Note that this has the correct positive orientation. Then by Stokes’ theorem¨S

curl (F) · dS =

˛C

F · dr

=

ˆ 2π

0

⟨√3 cos t,

√3 sin t, cos t sin t

⟩· 〈− sin t, cos t, 0〉 dt

=

ˆ 2π

0

(−√

3 cos t sin t+√

3 sin t cos t+ 0)dt

=

ˆ 2π

0

0dt

= 0.

Note that this vector field was not conservative.

8.4 Divergence Theorem

Recall one of our various vector forms of Green’s theorem was given by˛C

F · nds =

¨D

div (F) dA.

Similiar in how Stokes’ theorem is a generalization of Green’s theorem to higherdimensions, we might guess that something of the form¨

S

F · ndS =

˚E

div (F) dV

holds, where S is the boundary of the solid E with positive orientation. It turnsout that this is true, but we’ll only use it with a special type of region E. Solids E,that are simultaneously regions of type I, type II, and type III will be called simplesolids.

87


Theorem 129 (Divergence Theorem):Let E be a simple solid with boundary S = ∂E and positive orientation. Let F bea vector field whose components have continuous first partial derivatives on an openregion containing E. Then

¨S

F · dS =

˚E

div (F) dV.

Notational Aside: In similar fashion to simply closed curves with positive orien-tation, when we have the boundary of a simple region E with positive orientation,we’ll use the notation "

S

F · dS,"∂E

F · dS.

Example 130:Find the flux of the vector field F = 〈z, y, x〉 over the unit sphere x2 + y2 + z2 = 1.

Using the divergence theorem, we see that

‹S

F · dS =

˚E

div (F) dV

=

˚E

(0 + 1 + 0)dV

= Volume(E)

=4

3π.

Example 131:

Evaluate‚S

F · dS, where F =⟨xy, y2 + exz

2, sin(xy)

⟩and S is the boundary of

the region E which is bounded by the parabolic cylinder z = 1 − x2 and the planesz = 0, y = 0, and y + z = 2.

88


We’ll use the divergence theorem because otherwise we would have to do foursurface integrals here. When characterizing this region, we see that

0 ≤ y ≤ 2− z, 0 ≤ z ≤ 1− x2, −1 ≤ x ≤ 1,

and so by the divergence theorem

‹S

F · dS =

˚E

div (F) dV

=

˚E

(y + 2y + 0) dV

=

ˆ 1

−1

ˆ 1−x2

0

ˆ 2−z

0

(3y)dydzdx

=3

2

ˆ 1

−1

ˆ 1−x2

0

(2− z)2dzdx

= −1

2

ˆ 1

−1((1 + x2)3 − 8)dx

=184

35.

89

8.5 The Fundamental Theorems of Vector Calculus 8 INTEGRATION OVER SURFACES

8.5 The Fundamental Theorems of Vector Calculus

In closing, let’s tie all of these important fundamental theorems together, that is, thefundamental theorem for line integrals, Stokes’ theorem (and hence Green’s theorem),and the divergence theorem.

Recall, the divergence theorem

¨∂E

F · dS =

˚E

div (F) dV,

Stokes’ theorem ˆ∂S

F · dr =

¨S

curl (F) · dS,

and the FTLI

f(r(b))− f(r(a)) =

ˆC

∇f · dr.

Note that given a curve C, if it connects the point r(a) to r(b), then we can considerthe boundary of C to be these endpoints, that is, ∂C = {r(a), r(b)}, and if we thinkof integration over a discrete set as just evaluation of the points, we can loosely thinkof the LHS of the FTLI rewritten as

f(r(b))− f(r(a)) =

ˆ∂C

f,

(no differential needed, since it’s a discrete set, the´

really just represents a signedsum) and so the FTLI can be rewritten as

ˆ∂C

f =

ˆC

∇f · dr.

Therefore, all of these theorems can be seen as analgoes to each other based on thedimension of the object we’re integrating over, wether it’s a bounded curve C, abounded surface S, or a bounded solid E with the derivatives being taken on thehigher dimensional side.

In general, if we let M be any-dimensional region with boundary ∂M , and let ddenote the varius forms of differentiation, depending on dimension, then all of theabove theorems state that ˆ

∂M

f =

ˆM

df.

Thus, all of these can be thought of as the same theorem, just stated for differentdimensions. This theorem in generality is called Stokes’ Theorem. It is one of themost fundamental and important results in all of calculus.

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