MAT 10B: Highlights

Fall 2011 - Williams

Latest Update: November 22, 2011

Section 5.1. Introduction: Areas and Volumes

Definition (Standard rectangle in R2). Let a, b, c, d be real numbers such that a < b andc < d. These numbers determine a closed rectangle

R = {(x, y) | a x b , c y d} ⇢ R2

.

We also denote this rectangle as

R = [a, b]⇥ [c, d] .

Computational Tool: Iterated integrals

Let R = [a, b]⇥ [c, d] be a rectangle. Let f : R ! [0,1) be a continuous function. Then theintegral Z

d

c

Zb

a

f(x, y) dx dy

is interpreted as the iterated integralZd

c

✓Zb

a

f(x, y) dx

◆dy .

When computing the “inner integral” Zb

a

f(x, y) dx ,

treat y as a constant.

Similarly, Zb

a

Zd

c

f(x, y) dy dx =

Zb

a

✓Zd

c

f(x, y) dy

◆dx .

1

Example. Z2

0

Z3

1

x

2 + y dy dx =

Z2

0

✓Z3

1

x

2 + y dy

◆dx

=

Z2

0

hx

2

y +y

2

2

iy=3

y=1

dx

=

Z2

0

✓3x2 +

9

2

◆�✓x

2 +1

2

◆dx

=

Z2

0

2x2 + 4 dx

=40

3.

We compute the associated integral in which the order of integration is switched.Z3

1

Z2

0

x

2 + y dx dy =

Z3

1

✓Z2

0

x

2 + y dx

◆dy

=

Z3

1

hx

3

3+ xy

ix=2

x=0

dy

=

Z3

1

8

3+ 2y dy

=40

3.

We happen to get the same answer of 40

3

. This is no coincidence since the integral representsthe same geometric quantity; also see Fubini’s Theorem in the next section.

Proposition. Let R = [a, b]⇥ [c, d] be a rectangle.

Let f : R ! [0,1) be a continuous function.

If V denotes the volume of the region between the graph of f and the xy–plane, then

V =

Zb

a

Zd

c

f(x, y) dy dx =

Zd

c

Zb

a

f(x, y) dx dy .

A simplified notation for the above iterated integral isZZR

f dA .

Example (Volume of a box). Suppose that a box has dimensions l, w, h. Draw a picture inR3. Then the volume of the box is

V =

Zl

0

Zw

0

h dy dx =

Zl

0

wh dx = lwh .

2

Example (Volume of a complicated region). Find the volume of the region bounded by thegraph of

f(x, y) = 9� x

2 � y

2

over the rectangle R = [1, 2]⇥ [0, 1].The volume is

V =

Z2

1

Z1

0

9� x

2 � y

2

dy dx

=

Z2

1

✓26

3� x

2

◆dx

=19

3.

Section 5.2. Double Integrals

Goals: Define the double integral ZZR

f dA

1. for a more general function f : R ! R (not necessarily nonnegative nor continuous)

2. over a more general region R ⇢ R (not necessarily a rectangle).

Part 1: Integral of a function over a rectangle

Definition (Partition of order n). Let R = [a, b] ⇥ [c, d] be a rectangle, and let n be apositive integer. Consider partitions

a = x

0

< x

1

< x

2

< · · · < x

n

= b for [a, b], and

c = y

0

< y

1

< y

2

< · · · < y

n

= d for [c, d].

A partition of order n (of R) is the collection of subrectangles

R

ij

= [xi�1

, x

i

]⇥ [yj�1

, y

j

]

for i, j = 1, . . . , n. Visualize the partition as a tiling of R.Given R

ij

, the base is �x

i

= x

i

� x

i�1

and the height is �y

j

= y

j

� y

j�1

.The area of R

ij

is�A

ij

= �x

i

·�y

j

.

Furthermore, the partition is called regular if �x

i

= b�a

n

and �y

j

= d�c

n

for all i, j =1, . . . , n.

3

Definition (Riemann Sum). Let R = [a, b]⇥ [c, d] be a rectangle.Choose a partition {R

ij

} for R.For each i, j = 1, . . . , n, choose a sample point c

ij

2 R

ij

.Let f : R ! R be a function.The quantity

S =nX

i,j=1

f(cij

) �A

ij

,

is called a Riemann sum.

Now we give the definition of the double integral.

Definition (Double Integral). Let R = [a, b]⇥ [c, d] be a rectangle.Let f : R ! R be a function.The double integral of f over R is defined to beZZ

R

f dA = limn!1

nXi,j=1

f(cij

) �A

ij

over all regular partitions Rij

of R.

Remark. The above definition ofRR

R

f dA is equivalent to the one in the textbook. If theintegral

RRR

f dA exists, we say that f is integrable on R.

Theorem. If R is a rectangle and f : R ! R is continuous, then

RRR

f dA exists.

Theorem (Fubini’s Theorem (Simple Form)). Let R = [a, b] ⇥ [c, d] be a rectangle. If

f : R ! R is continuous, thenZZR

f dA =

Zb

a

Zd

c

f(x, y) dy dx =

Zd

c

Zb

a

f(x, y) dx dy .

Example. Z2

0

Z3

1

x

2 � y dy dx =

Z2

0

✓Z3

1

x

2 � y dy

◆dx

=

Z2

0

hx

2

y � y

2

2

iy=3

y=1

dx

=

Z2

0

✓3x2 � 9

2

◆�✓x

2 � 1

2

◆dx

=

Z2

0

2x2 � 4 dx

= �8

3.

4

Proposition (Properties of the integral). Suppose that f and g are both integrable on the

standard rectangle R. Then

1. f + g is also integrable on R andZZR

f + g dA =

ZZR

f dA+

ZZR

g dA .

2. cf is also integrable on R for any c 2 R andZZR

cf dA = c

ZZR

f dA .

3. If f(x, y) g(x, y) for all (x, y) 2 R, thenZZR

f dA ZZ

R

g dA .

4. |f | is also integrable on R and����ZZR

f dA

���� ZZR

|f | dA .

Examples (Properties of the integral). Let R ⇢ R2 be a standard rectangle.

1. ZZR

xy + y

2

dA =

ZZR

xy dA+

ZZR

y

2

dA .

2. ZZR

�5xy dA = �5

ZZR

xy dA .

3. Suppose that R = [0, 2]⇥ [0, 5]. Let’s prove the inequalityZZR

x

2

y dA 200 .

Since 0 x 2 and 0 y 5, we conclude that x2

y 22 · 5 = 20 for all (x, y) 2 R.By Property 3, we haveZZ

R

x

2

y dA ZZ

R

20 dA = 20 · 2 · 5 = 200 .

4. Suppose that R = [�1, 2]⇥ [�3, 5]. Let’s prove the inequality����ZZR

xy dA

���� 240 .

Since �1 x 2 and �3 y 5, we conclude that |xy| = |x| · |y| 2 · 5 = 10 for all(x, y) 2 R. By Properties 3 and 4, we have����ZZ

R

xy dA

���� ZZR

|xy| dA ZZ

R

10 dA = 10 · 3 · 8 = 240 .

5

Part 2: Integral over a general region

Definition (Elementary Region). A subset D ⇢ R2 is called an elementary region if itadmits a description of one (or both) of the following types.

1. D = {(x, y) | �(x) y �(x) , a x b} for continuous functions �(x) and �(x), or

2. D = {(x, y) | �(y) x �(y) , c y d} for continuous functions �(y) and �(y).

Theorem. Let D ⇢ R2

be an elementary region. Suppose that f : D ! R is a continuous

function. Then f is integrable on D. Furthermore

• If D is of type 1, then ZZD

f dA =

Zb

a

Z�(x)

�(x)

f(x, y) dy dx .

• If D is of type 2, then ZZD

f dA =

Zd

c

Z�(y)

�(y)

f(x, y) dx dy .

Example. Sometimes a region can be represented as both types. Let D be the regionbounded by the curves y = x and y = x

2. As a type 1 region, express D as

D = {(x, y) | x2 y x , 0 x 1} .

Then ZZD

xy dA =

Z1

0

Zx

x

2

xy dy dx

=

Z1

0

✓xy

2

2

iy=x

y=x

2

◆dx

=

Z1

0

✓x

3

2� x

5

2

◆dx

=1

24.

As a type 2 region, express D as

D = {(x, y) | y x py , 0 y 1} .

6

Then ZZD

xy dA =

Z1

0

Z py

y

xy dx dy

=

Z1

0

✓x

2

y

2

ix=

py

x=y

◆dy

=

Z1

0

✓y

2

2� y

3

3

◆dy

=1

24.

Double integrals can also be used to calculate areas of elementary regions.

Theorem (Area of an elementary region in R2). Suppose that D is an elementary region in

R2

. Then

Area(D) =

ZZD

1 dA =

ZZD

dA .

Putting elementary regions together

The main idea is that we can integrate over a union of elementary regions as long as anytwo of the elementary pieces do not overlap too much. More precisely, we have

Theorem. Let D1

and D

2

be elementary regions in R2

. Let D = D

1

[D

2

. If D

1

\D

2

has

zero area, (e.g. D

1

\D

2

is a line segment), thenZZD

f(x, y) dA =

ZZD1

f(x, y) dA+

ZZD2

f(x, y) dA .

See Example 6 from pages 302-303 in the textbook. We will later describe the annulus andany integrals over it in a more appropriate way (using polar coordinates).

Section 5.3. Changing the Order of Integration

Main Idea. Given an iterated integral, it is sometimes helpful to change the order of inte-gration when

• the given integral is impossible to evaluate by hand, or

• the given integral is more convenient to evaluate (by hand or computer) in the otherorder.

7

Example. Consider the iterated integralZ1

0

Z2

2y

e

x

2dx dy .

There is no simple technique to calculate an antiderivative of ex2, so we want to integrate

e

x

2with respect to y. The given integral tells us that the (relevant) domain of f(x, y) = e

x

2

isR = {(x, y) | 2y x 2, 0 y 1} .

Draw the region R in the plane. It is easy to see that R is a triangle with vertices (0, 0), (2, 0)and (2, 1). Changing perspective, re-interpret R as

R = {(x, y) | 0 y x/2, 0 x 2} .

Therefore Z1

0

Z2

2y

e

x

2dx dy =

Z2

0

Zx/2

0

e

x

2dy dx

=

Z2

0

✓ye

x

2iy=x/2

y=0

◆dx

=

Z2

0

x

2e

x

2dx

= (e4 � 1)/4 .

Section 5.4. Triple Integrals

Main Idea. Generalize the double integral for a function of three variables.

Computational tool: iterated integrals

Triple iterated integrals just generalize double iterated integrals.

Example. Z1

0

Z2

�1

Z3

1

3yz2 dx dy dz =

Z1

0

Z2

�1

h3yz2x

ix=3

x=1

dy dz

=

Z1

0

Z2

�1

6yz2 dy dz

=

Z1

0

h3y2z2

iy=2

y=�1

dz

=

Z1

0

9z2 dz

= 3 .

8

Definition (Standard box in R3). Let a, b, c, d, p, q be real numbers such that a < b, c < d,and p < q. These numbers determine a (closed) box

B = {(x, y, z) | a x b , c y d , p z q} ⇢ R3

.

We also denote this box asB = [a, b]⇥ [c, d]⇥ [p, q] .

Goals: Define the triple integral ZZZB

f dV

1. for a function f : B ! R where B is a standard box.

2. over a more general region B ⇢ R3 (not necessarily a standard box).

Part 1: Integral of a function over a box

As with double integrals, we

• partition the box B = [a, b]⇥ [c, d]⇥ [p, q]

• compute f(c) for various sample points c = c

ijk

2 B

• form the Riemann sum

S =nX

i,j,k=1

f(cijk

) �V

ijk

• define the integral ZZZB

f dV = limn!1

S ,

the limit of Riemann sums.

Remark. See the textbook for the precise definition of the triple integral. If the integralRRRB

f dV exists, we say that f is integrable on B.

Theorem. If B is a standard box and f : B ! R is continuous, then

RRRB

f dV exists.

9

Theorem (Fubini’s Theorem (Simple Form)). Let B = [a, b] ⇥ [c, d] ⇥ [p, q] be a box. If

f : B ! R is continuous, thenZZZB

f dV

=

Zb

a

Zd

c

Zq

p

f(x, y, z) dz dy dx =

Zb

a

Zq

p

Zd

c

f(x, y, z) dy dz dx

=

Zd

c

Zb

a

Zq

p

f(x, y, z) dz dx dy =

Zd

c

Zq

p

Zb

a

f(x, y, z) dx dz dy

=

Zq

p

Zb

a

Zd

c

f(x, y, z) dy dx dz =

Zq

p

Zd

c

Zb

a

f(x, y, z) dx dy dz .

Example. The integral Z1

0

Z2

�1

Z3

1

3yz2 dx dy dz

can also be calculated asZ3

1

Z2

�1

Z1

0

3yz2 dz dy dx =

Z3

1

Z2

�1

hyz

3

iz=1

z=0

dy dx

=

Z3

1

Z2

�1

y dy dx

=

Z3

1

3

2dx

= 3 .

Proposition (Properties of the integral). Suppose that f and g are both integrable on the

closed box B. Then

1. f + g is also integrable on B andZZZB

f + g dV =

ZZZB

f dV +

ZZZB

g dV .

2. cf is also integrable on B for any c 2 R andZZZB

cf dV = c

ZZZB

f dV .

3. If f(x, y, z) g(x, y, z) for all (x, y, z) 2 B, thenZZZB

f dV ZZZ

B

g dV .

4. |f | is also integrable on B and����ZZZB

f dV

���� ZZZB

|f | dV .

10

Part 2: Integral over a general region

Definition (Elementary Region). A subset W ⇢ R3 is called an elementary region if itadmits a description of at least one of the following types.

1. Type 1 Regions [The most fundamental of regions]

(a) W = {(x, y, z) | '(x, y) z (x, y) , �(x) y �(x) , a x b} , or

(b) W = {(x, y, z) | '(x, y) z (x, y) , ↵(y) x �(y) , c y d} .

Simply put, the region W is the region between the graphs of z = '(x, y) and z = (x, y) over an elementary region in the xy–plane.

2. Type 2 Regions

(a) W = {(x, y, z) | ↵(y, z) x �(y, z) , �(z) y �(z) , p z q} , or

(b) W = {(x, y, z) | ↵(y, z) x �(y, z) , '(y) z (y) , c y d} .

Simply put, the region W is the region between the graphs of x = ↵(y, z) and x =�(y, z) over an elementary region in the yz–plane.

3. Type 3 Regions

(a) W = {(x, y, z) | �(x, z) y �(x, z) , ↵(z) x �(z) , p z q} , or

(b) W = {(x, y, z) | �(x, y) y �(x, y) , '(x) z (x) , a x b} .

Simply put, the regionW is the region between the graphs of y = �(x, z) and y = �(x, z)over an elementary region in the xz–plane.

Theorem. Let W be an elementary region in R3

and let f be a function on W .

1. If W is a type 1 region, thenZZZW

f dV =

Zb

a

Z�(x)

�(x)

Z (x,y)

'(x,y)

f(x, y, z) dz dy dx ,

or ZZZW

f dV =

Zd

c

Z�(y)

↵(y)

Z (x,y)

'(x,y)

f(x, y, z) dz dx dy .

2. If W is a type 2 region, thenZZZW

f dV =

Zq

p

Z�(z)

�(z)

Z�(y,z)

↵(y,z)

f(x, y, z) dx dy dz ,

or ZZZW

f dV =

Zd

c

Z (y)

'(y)

Z�(y,z)

↵(y,z)

f(x, y, z) dx dz dy .

11

3. If W is a type 3 region, thenZZZW

f dV =

Zq

p

Z�(z)

↵(z)

Z�(x,z)

�(x,z)

f(x, y, z) dy dx dz ,

or ZZZW

f dV =

Zb

a

Z (x)

'(x)

Z�(x,z)

�(x,z)

f(x, y, z) dy dz dx .

Triple integrals can also be used to compute volume of elementary regions in R3.

Theorem (Volume of an elementary region in R

3). Suppose that W is an elementary region

in R

3

. Then

Volume(W ) =

ZZZW

1 dV =

ZZZW

dV .

Let’s take a look at some examples of each type.

Examples of type 1 regions

Example. Let W ⇢ R3 be the solid containing (0, 0, 0) in the first octant bounded byz = 2� x

2 � y

2, z = 0 and x+ y = 1. Compute the integralZZZW

x

2

y dV .

Solution: The shadow of W in the xy–plane is the triangle in the first quadrant bounded bythe line x+ y = 1. Hence the shadow can be expressed as

W

shadow

= {(x, y) | 0 y 1� x , 0 x 1} .

It follows that

W = {(x, y, z) | 0 z 2� x

2 � y

2

, 0 y 1� x , 0 x 1} .

Therefore, the iterated integral that computesRRR

W

x

2

y dV isZ1

0

Z1�x

0

Z2�x

2�y

2

0

x

2

y dz dy dx .

12

Now we evaluate the integral:Z1

0

Z1�x

0

Z2�x

2�y

2

0

x

2

y dz dy dx =

Z1

0

Z1�x

0

hx

2

yz

iz=2�x

2�y

2

z=0

dy dx

=

Z1

0

Z1�x

0

x

2

y(2� x

2 � y

2) dy dx

=

Z1

0

hx

2

y

2 � x

4

y

2

2� x

2

y

4

4

iy=1�x

y=0

dx

=

Z1

0

✓x

2(1� x)2 � x

4(1� x)2

2� x

2(1� x)4

4

◆dx

=11

420.

Since the shadow of W can be expressed as

W

shadow

= {(x, y) | 0 x 1� y , 0 y 1} ,

the integral can also be set-up as

Z1

0

Z1�y

0

Z2�x

2�y

2

0

x

2

y dz dx dy .

Example. Let W ⇢ R3 be the solid bounded by z = 3x2 + 2y2 and z = 16 � x

2 � 2y2.Compute the integrals

(a) ZZZW

xz dV

(b) ZZZW

dV =

ZZZW

1 dV

Solution: Find the shadow ofW in the xy–plane by solving for the intersection of the surfaces:

z = 3x2 + 2y2 meets z = 16� x

2 � 2y2 =) 3x2 + 2y2 = 16� x

2 � 2y2

=) 4x2 + 4y2 = 16

=) x

2 + y

2 = 4 .

So the shadow of W in the xy–plane is

W

shadow

= {(x, y) | x2 + y

2 4} = {(x, y) | �p4� x

2 y p4� x

2

, �2 x 2} .

13

(a) An iterated integral that computesRRR

W

xz dV isZ2

�2

Z p4�x

2

�p4�x

2

Z16�x

2�2y

2

3x

2+2y

2

xz dz dy dx .

An easy way to compute this integral is to use the symmetry of the solid W across theplane x = 0. The function f(x, y, z) = xz satisfies f(�x, y, z) = �f(�x, y, z). In asense, the function f is odd with respect to the yz–plane. Therefore, the integral hasthe value 0.

(b) An iterated integral that computesRRR

W

dV isZ2

�2

Z p4�x

2

�p4�x

2

Z16�x

2�2y

2

3x

2+2y

2

dz dy dx .

Of course the integralRRR

W

dV represents the volume of W , so we expect the integralto be positive.

Z2

�2

Z p4�x

2

�p4�x

2

Z16�x

2�2y

2

3x

2+2y

2

dz dy dx =

Z2

�2

Z p4�x

2

�p4�x

2

⇥z

⇤z=16�x

2�2y

2

z=3x

2+2y

2 dy dx

=

Z2

�2

Z p4�x

2

�p4�x

2

�(16� x

2 � 2y2)� (3x2 + 2y2)�dy dx

=

Z2

�2

Z p4�x

2

�p4�x

2

16� 4(x2 + y

2) dy dx .

To finish the calculation, we should use polar coordinates as in Section 5.5 (to bediscussed). It turns out that this transformation will turn our integral into a simplerdouble integral:

Z2⇡

0

Z2

0

(16� 4r2)r dr d✓ = 32⇡ .

Examples of type 2 and type 3 regions

Example. Let W be the solid in the first octant bounded by the plane x+2y+3z = 6. Letf(x, y, z) be a continuous function. Set-up an iterated integral that computesZZZ

W

f dV

by considering W as

14

(a) a type 1 region

(b) a type 2 region

(c) a type 3 region

Solution: First, we should draw the plane x + 2y + 3z = 6 in the first octant. This surfaceis the triangle with vertices (6, 0, 0), (0, 3, 0) and (0, 0, 2).

(a) As a type 1 region, W is the region between z = 0 and z = 6�x�2y

3

; the shadow of Win the xy–plane is the triangle with vertices (0, 0), (6, 0) and (0, 3). It follows that

W = {(x, y, z) | 0 z 6� x� 2y

3, 0 y 6� x

2, 0 x 6} .

Therefore ZZZW

f dV =

Z6

0

Z 6�x

2

0

Z 6�x�2y3

0

f(x, y, z) dz dy dx .

(b) As a type 2 region, W is the region between x = 0 and x = 6� 2y� 3z; the shadow ofW in the yz–plane is the triangle with vertices (0, 0), (3, 0) and (0, 2). It follows that

W = {(x, y, z) | 0 x 6� 2y � 3z , 0 y 6� 3z

2, 0 z 2} .

Therefore ZZZW

f dV =

Z2

0

Z 6�3z2

0

Z6�2y�3z

0

f(x, y, z) dx dy dz .

(c) As a type 3 region, W is the region between y = 0 and y = 6�x�3z

2

; the shadow of Win the xz–plane is the triangle with vertices (0, 0), (6, 0) and (0, 2). It follows that

W = {(x, y, z) | 0 y 6� x� 3z

2, 0 x 6� 3z , 0 z 2} .

Therefore ZZZW

f dV =

Z2

0

Z6�3z

0

Z 6�x�3z2

0

f(x, y, z) dy dx dz .

Putting elementary regions together

The main idea is that we can integrate over a union of elementary regions as long as anytwo of the elementary pieces do not overlap too much. More precisely, we have

Theorem. Let W1

and W

2

be elementary regions in R3

. Let W = W

1

[W

2

. If W

1

\W

2

has

zero volume, (e.g. W

1

\W

2

is a surface), thenZZW

f(x, y, z) dV =

ZZW1

f(x, y, z) dV +

ZZW2

f(x, y, z) dV .

15

Section 5.5. Change of Variables: Double Integrals

Some basic linear algebra

Definition (Matrix). A 2 by 2 matrix is a square array of real numbers a, b, c, d 2 Ra b

c d

�.

Definition (Determinant). The determinant of a matrix is a real number given by theformula

det

a b

c d

�= ad� bc .

Definition (Linear transformation). A linear transformation is a function of the form

T : R2

uv

! R2

xy

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

given by the equations

x = au+ bv

y = cu+ dv

where a, b, c, d 2 R. In this case, T admits a matrix formx

y

�=

a b

c d

� u

v

�.

It is convenient to write

T ⌘a b

c d

�and to define the determinant of T to be the determinant of its matrix.

Remark. The determinant of T tells us how T distorts (or rescales) area. See Proposition5.1 in the textbook.

Definition (Invertible Linear Transformation). A linear transformation T is invertible(a.k.a. nonsingular) if ↵ = detT 6= 0. The inverse of T is denoted by T

�1, and the matrixof T�1 is given by the formula

T

�1 ⌘d/↵ �b/↵

�c/↵ a/↵

�.

In this case, the determinants of T and T

�1 are reciprocals of each other:

det(T�1) = 1�det(T ) .

16

Linear change of variables for integrals

Theorem (Linear change of variables). Suppose that T : R2

uv

! R2

xy

is an invertible

linear transformation. Suppose that D

⇤ ⇢ R2

uv

and D ⇢ R2

xy

are elementary regions

such that T (D⇤) = D. If f : D ! R is continuous, thenZZD

f(x, y) dx dy =

ZZD

⇤f(x(u, v), y(u, v)) | detT | du dv .

Example (Problem 9). Evaluate the integralZ2

0

Z(x/2)+1

x/2

x

5(2y � x)e(2y�x)

2dy dx

by making the substitution u = x and v = 2y � x.

Solution:

The substitution u = x and v = 2y � x gives a linear transformation

T

�1 : R2

xy

! R2

uv

whose matrix is 1 0�1 2

�.

So detT�1 = 2. Therefore, the linear transformation

T : R2

uv

! R2

xy

has detT = 1/2.

A quick sketch of the region D reveals that D is the parallelogram spanned by the vectors(2, 1) and (0, 1). We would like to find D

⇤ so that T (D⇤) = D.

To do this, we notice that

(x, y) = (2, 1) =) (u, v) = (2, 0) and (x, y) = (0, 1) =) (u, v) = (0, 2) .

It follows that D⇤ is the parallelogram spanned by the vectors (2, 0) and (0, 2). Hence

D

⇤ = {(u, v) | 0 u 2 , 0 v 2} = [0, 2]⇥ [0, 2] .

17

Therefore ZZD

⇤f(x(u, v), y(u, v)) | detT | du dv =

Z2

0

Z2

0

u

5

ve

v

2 1

2du dv

=1

2

✓Z2

0

u

5

du

◆✓Z2

0

ve

v

2dv

◆=

1

2

✓64

6

◆✓e

4 � 1

2

◆=

8

3(e4 � 1) .

Example (Problem 10). Determine ZZD

rx+ y

x� 2ydA

where D is the region enclosed by the lines y = x/2, y = 0, and x+ y = 1.

Solution: Let’s try the substitution u = x + y and v = x� 2y. This gives a linear transfor-mation

T

�1 : R2

xy

! R2

uv

whose matrix is 1 11 �2

�.

So detT�1 = �2� 1 = �3 . Therefore, the linear transformation

T : R2

uv

! R2

xy

has detT = �1/3.

A quick sketch of D reveals that it is the triangle with verticies (0, 0), (1, 0) and (2/3, 1/3).We would like to find D

⇤ so that T (D⇤) = D.

To do this, we notice that

(x, y) = (1, 0) =) (u, v) = (1, 1) and (x, y) = (2/3, 1/3) =) (u, v) = (1, 0) .

So D

⇤ is the triangle with verticies (0, 0), (1, 0) and (1, 1):

D

⇤ = {(u, v) | 0 u 1 , 0 v u}

18

It follows that ZZD

rx+ y

x� 2ydA =

ZZD

⇤

pupv

|detT | du dv

=

Z1

0

Zu

0

pupv

(1/3) dv du

= 1/3 .

General change of variables

Recall a couple of definitions about functions. The definitions below generalize to subsets ofRn and Rm.

Definition (C1 function). A function T : Rn ! Rm is of class C1 if its first-order partialderivatives exist and are continuous.

Definitions (One-to-one function, Onto function). A function T : Rn ! Rm is called

• one-to-one ifT (a) = T (b) =) a = b

for every a,b 2 Rn;

• onto if T (Rn) = Rm.

Definition (Jacobian matrix and Jacobian determinant). Let T : R2

uv

! R2

xy

be a C1

function. The Jacobian matrix of T is the matrix

DT =

"@x

@u

@x

@v

@y

@u

@y

@v

#.

The determinant of this matrix is denoted by

@(x, y)

@(u, v)

and is called the Jacobian determinant; in our textbook, this determinant is simply called“the Jacobian”.

19

Theorem (Change of Variables (Double Integrals)). Let D⇤ ⇢ R2

uv

and D ⇢ R2

xy

be elementary regions. Suppose that

T : R2

uv

! R2

xy

is a C1

function such that T (D⇤) = D and T is one-to-one on D

⇤. If f : D ! R is

integrable, thenZZD

f(x, y) dx dy =

ZZD

⇤f(x(u, v), y(u, v))

����@(x, y)@(u, v)

���� du dv .

Remark. In some computations, it will be easier to set-up the Jacobian of T�1. It is a factthat this Jacobian is the reciprocal of the Jacobian of T :

det(DT

�1) = 1/ det(DT ) .

In coordinate notation for T : R2

uv

! R2

xy

, this is expressed as

@(u, v)

@(x, y)= 1.✓

@(x, y)

@(u, v)

◆.

Example (Problem 12). EvaluateZZD

(2x+ y � 3)2

(2y � x+ 6)2dx dy ,

where D is the square with verticies (0, 0), (2, 1), (3,�1), and (1,�2).

Solution: For this problem, we have to come up with our own substitution formulas for u

and v. A good substitution would be

u = 2x+ y � 3

v = 2y � x+ 6 .

This gives the transformation T

�1 : R2

xy

! R2

uv

. The Jacobian matrix of T�1 is"@u

@x

@u

@y

@v

@x

@v

@y

#=

2 1�1 2

�.

The Jacobian of T�1 is@(u, v)

@(x, y)= 5 .

So the Jacobian of T is the reciprocal:

@(x, y)

@(u, v)= 1.✓

@(u, v)

@(x, y)

◆= 1/5 .

20

Since T�1 is an a�ne (linear + translation) transformation, it takes straight-lines to straight-lines. It follows that D

⇤ = T

�1(D) is a parallelogram with vertices T

�1(0, 0) = (�3, 6),T

�1(2, 1) = (2, 6), T�1(3,�1) = (2, 1), and T

�1(1,�2) = (�3, 1). Therefore

D

⇤ = [�3, 2]⇥ [1, 6] .

ThereforeZZD

(2x+ y � 3)2

(2y � x+ 6)2dx dy =

ZZD

⇤

u

2

v

2

· 15du dv =

Z2

�3

Z6

1

u

2

5v2dv du =

35

18.

Example (Simple rescaling). Consider the linear transformation T : R2

uv

! R2

xy

defined by

(x, y) = T (u, v) = (2u, 3v) .

It is easy to see that T transforms the unit square

[0, 1]⇥ [0, 1] ⇢ R2

uv

into the rectangle[0, 2]⇥ [0, 3] ⇢ R2

xy

.

Also, T transforms the unit disk

D

⇤ = {(u, v) | u2 + v

2 1} ⇢ R2

uv

into the elliptical region

D = {(x, y) | (x/2)2 + (y/3)2 1} ⇢ R2

xy

.

The determinant of T isdetT = 6 .

This linear transformation allows us to compute

Area(D) =

ZZD

1 dx dy =

ZZD

⇤|detT | du dv = |detT |

✓ZZD

⇤1 du dv

◆= 6·Area(D⇤) = 6⇡ .

Review of polar coordinates (See section 1.7)

Polar coordinates are also called angular coordinates. These are the best coordinates fordescribing geometric objects that possesses circular symmetry. The coordinates are (r, ✓);here r represents the radial coordinate and ✓ represents the angular (or circular) coordinate.

Example (Disk). Consider the closed disk centered at the origin with radius 2:

D = {(x, y) | x2 + y

2 4} .

We can describe this region in polar coordinates as

D

⇤ = {(r, ✓) | 0 r 2 , 0 ✓ 2⇡} .

21

Example (Annulus). Consider the closed annulus centered at the orign with inner radius1 and outer radius 2:

A = {(x, y) | 1 x

2 + y

2 4} .

We can describe this region in polar coordinates as simply

A

⇤ = {(r, ✓) | 1 r 2 , 0 ✓ 2⇡} .

Conversions between Rectangular and Polar coordinates

For more complicated examples besides the disk and annulus, we need to a systematic wayto go between the two coordinate systems.

The basic (explicit) conversions from Rectangular to Polar coordinates are

x = r cos ✓ , y = r sin ✓

The basic (implicit) conversions from Polar to Rectangular coordinates are

r

2 = x

2 + y

2

, tan ✓ = y/x

The equations above can be solved for explicit values of r and ✓; this can be tricky for ✓, soit is recommended that you use a diagram of the unit circle for help.

Example (Another Disk). Consider the disk D bounded by the circle

(x� 1)2 + y

2 = 1 .

You should be able to easily plot this curve in R2. This disk can be described in rectangularcoordinates as

D = {(x, y) | 0 x 2 , �p

1� (x� 1)2 y p1� (x� 1)2} .

Use the conversion formulas to rewrite the equation of the circle

(x� 1)2 + y

2 = 1

in terms of r and ✓:

(x� 1)2 + y

2 = 1 =) (r cos ✓ � 1)2 + (r sin ✓)2 = 1

=) (r2 cos2 ✓ � 2r cos ✓ + 1) + (r2 sin2

✓) = 1

=) (r2 cos2 ✓ + r

2 sin2

✓)� 2r cos ✓ = 0

=) r

2 = 2r cos ✓

=) r = 2 cos ✓ .

The picture of D, with polar coordinates in mind, allows us to describe D as

D

⇤ = {(r, ✓) | 0 r 2 cos ✓ , �⇡/2 ✓ ⇡/2} .

22

Integrals in polar coordinates

Consider the polar coordinate transformation

T : R2

r✓

! R2

xy

defined by the formula(x, y) = T (r, ✓) = (r cos ✓, r sin ✓) .

The Jacobian matrix for T is

DT =

"@x

@r

@x

@✓

@y

@r

@y

@✓

#=

cos ✓ �r sin ✓sin ✓ r cos ✓

�.

So the Jacobian determinant is

@(x, y)

@(r, ✓)= r cos2 ✓ + r sin2

✓ = r .

When converting to polar coordinates, D⇤ ⇢ R2

r✓

is typically an elementary region, but D =T (D⇤) ⇢ R2

xy

need not be (e.g. D is an annulus). Also, the polar coordinate transformationis not quite one-to-one. Nevertheless, the formula for change of variables applies:

Change to polar coordinates: The rule for transforming a double integral inrectangular coordinates over D to a double integral in polar coordinates over D⇤ isZZ

D

f(x, y) dx dy =

ZZD

⇤f(r cos ✓, r sin ✓) r dr d✓ .

Example. Previously, we encountered the integralZ2

�2

Z p4�x

2

�p4�x

2

16� 4(x2 + y

2) dy dx .

The region D over which we are integrating is the disk centered at the origin with radius 2.In polar coordinates, this disk is simply expressed as

D

⇤ = {(r, ✓) | 0 r 2 , 0 ✓ 2⇡} .

We convert to polar coorinates: x = r cos ✓ , y = r sin ✓, r2 = x

2 + y

2. The integral simpliesto

23

Z2

�2

Z p4�x

2

�p4�x

2

16� 4(x2 + y

2) dy dx =

Z2⇡

0

Z2

0

(16� 4r2) r dr d✓

=

Z2⇡

0

Z2

0

(16r � 4r3) dr d✓

=

Z2⇡

0

h8r2 � r

4

ir=2

r=0

d✓

=

Z2⇡

0

16 d✓

= 32⇡ .

Example (Another Disk (revisited)). Consider (again) the disk D bounded by the circle

(x� 1)2 + y

2 = 1 .

Recall that, in polar coordinates, D corresponds to

D

⇤ = {(r, ✓) | 0 r 2 cos ✓ , �⇡/2 ✓ ⇡/2} .

So given a continuous function f(x, y) on D, we haveZZD

f(x, y) dx dy =

ZZD

⇤f(r cos ✓, r sin ✓) r dr d✓ =

Z⇡/2

�⇡/2

Z2 cos ✓

0

f(r cos ✓, r sin ✓) r dr d✓ .

For example, the area of D can be calculated asZZD

dx dy =

Z⇡/2

�⇡/2

Z2 cos ✓

0

r dr d✓

=

Z⇡/2

�⇡/2

hr

2

2

ir=2 cos ✓

r=0

d✓

=

Z⇡/2

�⇡/2

(2 cos ✓)2

2d✓

=

Z⇡/2

�⇡/22 cos2 ✓ d✓

=

Z⇡/2

�⇡/21 + cos 2✓ d✓

=h✓ +

sin 2✓

2

i⇡/2

�⇡/2

= ⇡ .

24

Example (Polar coordinates with a twist). Consider the curve

4x2 + y

2 � 8x+ 4y � 8 = 0

in R2. This curve is an ellipse:

4x2 + y

2 � 8x+ 4y � 8 = 0 =) 4(x2 � 2x) + (y2 + 4y) = 8

=) 4(x2 � 2x+ 1) + (y2 + 4y + 4) = 8 + 4 + 4

=) 4(x� 1)2 + (y + 2)2 = 16

=)✓x� 1

2

◆2

+

✓y + 2

4

◆2

= 1 .

Suppose that we want to integrate a continuous function f(x, y) over the region D boundedby this ellipse:

D =

((x, y)

�����✓x� 1

2

◆2

+

✓y + 2

4

◆2

1

).

We could use an a�ne (linear + translation) change of variables

u =x� 1

2, v =

y + 2

4and then change to polar coordinates, or we can do everything at once.

Consider the substitution

r cos ✓ =x� 1

2, r sin ✓ =

y + 2

4.

This substitution corresponds to the transformation

T : R2

r✓

! R2

xy

defined by the equations

x = 2r cos ✓ + 1 , y = 4r sin ✓ � 2 .

Just like in ordinary polar coordinates, we have D

⇤ = [0, 1]⇥ [0, 2⇡] ⇢ R2

r✓

.

The Jacobian matrix for T is

DT =

"@x

@r

@x

@✓

@y

@r

@y

@✓

#=

2 cos ✓ �2r sin ✓4 sin ✓ 4r cos ✓

�.

So the Jacobian determinant is

@(x, y)

@(r, ✓)= detDT = 8r cos2 ✓ + 8r sin2

✓ = 8r .

For instance, we can now directly compute the area of D :

Area(D) =

ZZD

1 dx dy =

ZZD

⇤8r dr d✓ =

Z2⇡

0

Z1

0

8r dr d✓ = 8⇡ .

25

Section 5.5. Change of Variables: Triple Integrals

In our class, we will focus only on changes to cylindrical and spherical coordinates for tripleintegrals.

Review of cylindrical coordinates (See section 1.7)

See Figures 1.89 to 1.95 in the textbook.

Cylindrical coordinates (r, ✓, z) arise from simply adding the coordinate z to polar coordi-nates.

Conversions between Rectangular and Cylindrical Coordinates

The basic conversions between Rectangular and Cylindrical coordinates are

x = r cos ✓ , y = r sin ✓ , z = z

andr

2 = x

2 + y

2

, tan ✓ = y/x , z = z

Integrals in cylindrical coordinates

Consider the cylindrical coordinate transformation

T : R3

r✓z

! R3

xyz

defined by the formula

(x, y, z) = T (r, ✓, z) = (r cos ✓, r sin ✓, z) .

Change to cylindrical coordinates: The rule for transforming a triple integralin rectangular coordinates over W to a triple integral in cylindrical coordinates overW

⇤ is ZZZW

f(x, y, z) dx dy dz =

ZZZW

⇤f(r cos ✓, r sin ✓, z) r dr d✓ dz .

26

Example (Region between a Plane and a Paraboloid). Let W be the solid in R3 boundedby the plane z = 9 and the paraboloid z = x

2 + y

2. Use cylindrical coordinates to

(a) compute the volume of W .

(b) integrate the functionf(x, y, z) =

px

2 + y

2 � x

2 � y

2

over W .

Solution:

Notice that the paraboloid meets the plane in the circle

{(x, y, 9) | x2 + y

2 = 9} .

The shadow of W in the xy–plane is given as

W

shadow

= {(x, y) | x2 + y

2 9}

which, in polar coordinates, is given as

W

⇤shadow

= {(r, ✓) | 0 r 3 , 0 ✓ 2⇡} .

Therefore, W can be described, in cylindrical coordinates, as

W

⇤c

= {(r, ✓, z) | 0 r 3 , 0 ✓ 2⇡ , r

2 z 9} .

To compute integrals, we make the substitutions

x = r cos ✓ , y = r sin ✓ , z = z .

(a) The volume of W is the integral of 1 over W isZZZW

1 dx dy dz =

ZZZW

⇤c

r dr d✓ dz

=

Z2⇡

0

Z3

0

Z9

r

2

r dz dr d✓

=

Z2⇡

0

Z3

0

hrz

iz=9

z=r

2dr d✓

=

Z2⇡

0

Z3

0

9r � r

3

dr d✓

= 243⇡ .

27

(b) The integral of f(x, y, z) =p

x

2 + y

2 � x

2 � y

2 over W isZZZW

⇣px

2 + y

2 � x

2 � y

2

⌘dx dy dz =

ZZZW

⇤c

(r � r

2)r dr d✓ dz

=

Z2⇡

0

Z3

0

Z9

r

2

(r � r

2)r dz dr d✓

=

Z2⇡

0

Z3

0

Z9

r

2

(r2 � r

3) dz dr d✓

=

Z2⇡

0

Z3

0

h(r2 � r

3)ziz=9

z=r

2dr d✓

=

Z2⇡

0

Z3

0

(r2 � r

3)(9� r

2) dr d✓

=

Z2⇡

0

Z3

0

(9r2 � 9r3 � r

4 + r

5) dr d✓

= �29889⇡

140.

Review of spherical coordinates (See section 1.7)

See Figures 1.96 to 1.102 in the textbook.

Spherical coordinates (⇢,', ✓) are useful for describing solids or surfaces in R3 which havespherical symmetry. The (usually nonnegative) coordinate ⇢ is the radial coordinate, andthere are two angular coordinates ' (or �) and the same ✓ coordinate from polar or cylindricalcoordinates. The new angular coordinate ' measures the angle between a line through theorigin and the z–axis. See Figure 1.97.

Conversions between Spherical coordinates and Cylindrical coordinates

The basic conversions between Spherical and Cylindrical coordinates are

r = ⇢ sin' , ✓ = ✓ , z = ⇢ cos'

and⇢

2 = r

2 + z

2

, tan' = r/z , ✓ = ✓

Conversions between Spherical coordinates and Rectangular coordinates

The basic conversions between Spherical and Rectangular coordinates are

x = ⇢ sin' cos ✓ , y = ⇢ sin' sin ✓ , z = ⇢ cos'

and⇢

2 = x

2 + y

2 + z

2

, tan' =p

x

2 + y

2

/z , tan ✓ = y/x

28

Integrals in spherical coordinates

Consider the spherical coordinate transformation

T : R3

⇢'✓

! R3

xyz

defined by the formula

(x, y, z) = T (⇢,', ✓) = (⇢ sin' cos ✓, ⇢ sin' sin ✓, ⇢ cos') .

Change to spherical coordinates: The rule for transforming a triple integralin rectangular coordinates over W to a triple integral in spherical coordinates overW

⇤ isZZZW

f(x, y, z) dx dy dz =

ZZZW

⇤f(⇢ sin' cos ✓, ⇢ sin' sin ✓, ⇢ cos') ⇢2 sin' d⇢ d' d✓ .

Example (Ice Cream Cone). Let W ⇢ R3 be the solid bounded by the cone z =px

2 + y

2

and the sphere x

2 + y

2 + z

2 = 9. Use spherical coordinates to

(a) compute the volume of W .

(b) set-up the interated integral which computes the integral of f(x, y, z) = 1p100�x�y

2+z

3

over W .

Solution:

The solid W looks like an “ice cream cone”.

In spherical coordinates, the equation for cone transforms into

⇢ cos' =p

(⇢ sin' cos ✓)2 + (⇢ sin' sin ✓)2 =) ⇢ cos' = ⇢ sin'

=) cos' = sin'

=) ' = ⇡/4 .

The sphere is naturally described by the equation ⇢ = 3.

Therefore W has the following description in spherical coordinates :

W

⇤s

= {(⇢,', ✓) | 0 ⇢ 3 , 0 ' ⇡/4 , 0 ✓ 2⇡} .

(a) It follows that the volume of W is

29

ZZZW

dx dy dz =

ZZZW

⇤s

⇢

2 sin' d⇢ d' d✓

=

Z2⇡

0

Z⇡/4

0

Z3

0

⇢

2 sin' d⇢ d' d✓

=

Z2⇡

0

Z⇡/4

0

h⇢

3

3sin'

i⇢=3

⇢=0

d' d✓

=

Z2⇡

0

Z⇡/4

0

9 sin' d' d✓

=

Z2⇡

0

Z⇡/4

0

h� 9 cos'

i'=⇡/4

'=0

d✓

=

Z2⇡

0

✓9� 9p

2

◆d✓

= 2⇡

✓9� 9p

2

◆.

(b) The integral of f(x, y, z) = 1p100�x�y

2+z

3over W is

ZZZW

1p100� x� y

2 + z

3

dx dy dz

=

ZZZW

⇤s

1p100� (⇢ sin' cos ✓)� (⇢ sin' sin ✓)2 + (⇢ cos')3

⇢

2 sin' d⇢ d' d✓

=

Z2⇡

0

Z⇡/4

0

Z3

0

1p100� (⇢ sin' cos ✓)� (⇢ sin' sin ✓)2 + (⇢ cos')3

⇢

2 sin' d⇢ d' d✓ .

Example (Ice Cream Cone (revisited)). Let W ⇢ R3 be the solid bounded by the conez =

px

2 + y

2 and the sphere x

2 + y

2 + z

2 = 9. Use cylindrical coordinates to

(a) compute the volume of W .

(b) set-up the interated integral which computes the integral of f(x, y, z) = 1p100�x�y

2+z

3

over W .

Solution:

To give a description of W in cylindrical coordinates, first observe that the cone’s equationis z = r, and the upper hemisphere’s equation is

r

2 + z

2 = 9 =) z =p9� r

2

.

30

So the intersection of the cone and the sphere satisfies the equation

r =p9� r

2 =) r =3p2.

Therefore W has the following description in cylindrical coordinates :

W

⇤c

= {(r, ✓, z) | 0 r 3p2, 0 ✓ 2⇡ , r z

p9� r

2} .

(a) The volume of W can be calculated asZZZW

dx dy dz =

ZZZW

⇤c

r dr d✓ dz

=

Z2⇡

0

Z 3p2

0

Z p9�r

2

r

r dz dr d✓

=

Z2⇡

0

Z 3p2

0

hrz

iz=

p9�r

2

z=r

dr d✓

=

Z2⇡

0

Z 3p2

0

⇣r

p9� r

2 � r

⌘dr d✓

=

Z2⇡

0

✓9� 9p

2

◆d✓

= 2⇡

✓9� 9p

2

◆.

(b) The integral of f(x, y, z) = 1p100�x�y

2+z

3over W is

ZZZW

1p100� x� y

2 + z

3

dx dy dz

=

ZZZW

⇤c

1p100� (r cos ✓)� (r sin ✓)2 + z

3

r dr d✓ dz

=

Z2⇡

0

Z 3p2

0

Z p9�r

2

r

1p100� (r cos ✓)� (r sin ✓)2 + z

3

r dz dr d✓ .

Example (Region between a Plane and a Sphere). Let W be the region between the planez = 2 and the (upper) hemisphere

x

2 + y

2 + z

2 = 16 =) ⇢ = 4 .

Compute the volume of W .

31

Solution:

The plane meets the sphere when

z = ⇢ cos' =) 2 = 4 cos' =) 1/2 = cos' =) ' = arccos(1/2) = ⇡/3 .

So far, we have shown that the bounds for ' and ✓ in W must be

0 ' ⇡/3 and 0 ✓ 2⇡ .

The bounds for ⇢ will require a little more work. Consider the equation z = ⇢ cos'. On theplane z = 2, the equation simplifies to ⇢ = 2 sec'. So the bounds for ⇢ (depending on ')are

2 sec' ⇢ 4 .

Therefore W has the following description in spherical coordinates :

W

⇤ = {(⇢,', ✓) | 2 sec' ⇢ 4 , 0 ' ⇡/3 , 0 ✓ 2⇡} .

The volume of W is

ZZZW

dx dy dz =

ZZW

⇤⇢

2 sin' d⇢ d' d✓

=

Z2⇡

0

Z⇡/3

0

Z4

2 sec'

⇢

2 sin' d⇢ d' d✓

=

Z2⇡

0

Z⇡/3

0

h⇢

3

3sin'

i⇢=4

⇢=2 sec'

d' d✓

=

Z2⇡

0

Z⇡/3

0

✓64

3sin'� 8 sec3 '

3sin'

◆d' d✓

=

Z2⇡

0

Z⇡/3

0

✓64

3sin'� 8

3 cos3 'sin'

◆d' d✓

=

Z2⇡

0

h� 64

3cos'� 4

3 cos2 '

i'=⇡/3

'=0

d✓

=

Z2⇡

0

✓�32

3� 16

3+

64

3+

4

3

◆d✓

=

Z2⇡

0

20

3d✓

=40⇡

3.

32

Section 6.1. Scaler and vector line integrals

It will be useful for you to review parts of Chapter 3.

All paths (curves) under consideration will be piecewise C1, unless otherwise specified. Also,all scaler functions f : Rn ! R and vector fields F : Rn ! Rn will be assumed to becontinuous unless otherwise specified.

The following is a useful way to parametrize a straight line connecting two points P and Q.

Straight-line parametrizations: A convenient way to get a parametrization ofa line segment in Rn beginning at a point P and ending at a point Q is to use theformula

x(t) = (1� t)P+ tQ .

This gives a straight-line path

x : [0, 1] ! Rn

with x(0) = P and x(1) = Q.

Scaler line integrals

Let x : [a, b] ! Rn be a path. Let f : Rn ! R be given. Then the scaler line integral off along x is Z

x

f ds =

Zb

a

f(x(t)) ||x0(t)|| dt .

Notice that ||x0(t)|| is the speed. In particular, the length of the path is given by integratingthe speed:

length(x) =

Zx

1 ds =

Zb

a

||x0(t)|| dt .

Example (Problem 4). Let f(x, y, z) = 3x+ xy + z

3. Consider the parametrization

x(t) = (cos 4t, sin 4t, 3t)

for 0 t 2⇡. ComputeRx

f ds.

Solution:f(x(t)) = 3 cos 4t+ (cos 4t)(sin 4t) + (3t)3

33

and

||x0(t)|| = ||(�4 sin 4t, 4 cos 4t, 3)||=p

(�4 sin 4t)2 + (4 cos 4t)2 + 32

=p

16 sin2 4t+ 16 cos2 4t+ 9

=p16 + 9

= 5 .

Therefore Zx

f ds =

Z2⇡

0

f(x(t)) ||x0(t)|| dt

=

Z2⇡

0

�3 cos 4t+ (cos 4t)(sin 4t) + (3t)3

�5 dt

=

Z2⇡

0

✓3 cos 4t+

sin 8t

2+ (3t)3

◆5 dt

= 180⇡4

.

Example (Problem 5). Consider the function f(x, y, z) = 2x�py + z

2 and the path

x(t) =

((t, t2, 0) if 0 t 1

(1, 1, t� 1) if 1 t 3

Compute the line integral Zx

f ds .

Solution: The path x is the concatenation of the paths

x1

(t) = (t, t2, 0) for 0 t 1

andx2

(t) = (1, 1, t� 1) for 1 t 3 .

The integralRx

f ds can be broken up as the sumZx

f ds =

Zx1

f ds+

Zx2

f ds .

34

We see that Zx1

f ds =

Z1

0

f(x1

(t)) ||x01

(t)|| dt

=

Z1

0

(2t� t) ||(1, 2t, 0)|| dt

=

Z1

0

t

p1 + 4t2 dt

=

Z5

1

pu

8du

⇥substitute u = 1 + 4t2 =) du = 8t dt

⇤=

p125� 1

12.

Also Zx2

f ds =

Z3

1

f(x2

(t)) ||x02

(t)|| dt

=

Z3

1

(1 + (t� 1)2) ||(0, 0, 1)|| dt

=

Z3

1

(1 + (t� 1)2) dt

=14

3.

Therefore, Zx

f ds =

p125� 1

12+

14

3.

Vector line integrals

This time, the integrand will be a vector field.

Let x : [a, b] ! Rn be a path. Let F : Rn ! Rn be a vector field. Then the vector lineintegral of F along x is Z

x

F · ds =Z

b

a

F(x(t)) · x0(t) dt .

Example. Let F(x, y, z) = 3xi+ y

2j+6zk = (3, y2, 6z). Consider the path x(t) = (t, 3, 5t2)defined for 0 t 1. Compute

Rx

F · ds .

Solution:F(x(t)) = (3t, 9, 30t2)

35

andx0(t) = (1, 0, 10t) .

Therefore, Zx

F · ds =Z

b

a

F(x(t)) · x0(t) dt

=

Z1

0

(3t, 9, 30t2) · (1, 0, 10t) dt

=

Z1

0

3t+ 300t3 dt

=153

2.

Di↵erential form

Suppose that we are given a vector field F : R2 ! R2 defined by

F(x, y) = M(x, y)i+N(x, y)j

and a path x : [a, b] ! R2

. We can express the formula of the line integralZx

F · ds

as Zx

(M(x, y), N(x, y)) · (x0(t), y0(t)) dt =

Zx

[M(x, y)x0(t) +N(x, y)y0(t)] dt .

If we multiply dt into the rest of integrand and use the di↵erentials

dx = x

0(t) dt , and dy = y

0(t) dt

we get the di↵erential form Zx

F · ds =Zx

M dx+N dy .

The same is true for a vector field on R3

F(x, y, z) = M(x, y, z)i+N(x, y, z)j+ P (x, y, z)k

and a pathx : [a, b] ! R3

.

The vector line integral of F along x has the di↵erential formZx

F · ds =Zx

M dx+N dy + P dz .

36

Example. Consider the pathx : [0, 1] ! R3

defined by the formula x(t) = (t, t2, t3). ComputeZx

2x dx+ z dy + y dz .

Solution: Compute the di↵erentials

dx = x

0(t) dt = 1 dt , dy = y

0(t) dt = 2t dt , dz = z

0(t) dt = 3t2 dt .

Then substituteZx

2x dx+ z dy + y dz =

Z1

0

2t dt+ (t3)(2t dt) + t

2(3t2 dt) =

Z1

0

(2t+ 2t4 + 3t4) dt = 2 .

Example (Compare with problem 22). CalculateZC

z dx� x dy + 2y dz

where C is the curve obtained by intersecting the surfaces z = x

2 and x

2 + y

2 = 4 andoriented counterclockwise around the z–axis (as seen from the positive z–axis).

Solution:

Since C is the intersection of two surfaces, we want a parametrization

x : [a, b] ! R3

, x(t) = (x(t), y(t), z(t))

which satisfies both equations

x

2 + y

2 = 4 and z = x

2

.

Start with the equation x

2 + y

2 = 4 because the shadow of C in the xy–plane will be thecircle x

2 + y

2 = 4. The standard (counterclockwise) parametrization of that circle in R2 is

(2 cos t, 2 sin t) , 0 t 2⇡ .

Since C also lies on the surface z = x

2, we just set z(t) = (x(t))2 = (2 cos t)2 = 4 cos2 t .

It follows that the parametrization for C should be

x : [0, 2⇡] ! R3

, x(t) = (2 cos t, 2 sin t, 4 cos2 t) .

The di↵erentials are

dx = �2 sin t dt , dy = 2 cos t dt , dz = �8 cos t sin t dt .

37

Therefore,ZC

z dx� x dy + 2y dz =

Z2⇡

0

(4 cos2 t)(�2 sin t) dt� (2 cos t)(2 cos t) dt+ 2(2 sin t)(�8 cos t sin t) dt

=

Z2⇡

0

(�8 cos2 sin t� 4 cos2 t� 32 sin2

t cos t) dt

= �4⇡ .

The e↵ect of reparametrization

Example. The image of each of the parametrizations

x : [0, 2⇡] ! R2

, x(t) = (cos t, sin t)

andy : [0, ⇡] ! R2

, y(t) = (cos 2t, sin 2t)

is the unit circle {(x, y) | x2 + y

2 = 1}.

Definition. A reparametrization of a path

x : [a, b] ! Rn

is another pathy : [c, d] ! Rn

for which there exists a one-to-one and onto function u : [c, d] ! [a, b] with theproperty y(t) = x(u(t)).

Furthermore, if u(c) = a, then y is an orientation-preserving reparametrizationof x; if u(c) = b, then y is an orientation-reversing reparametrization of x.

Two quick and easy way to see if the reparametrization preserves or reverses orientation is:

1. If you know u(t), then u

0(t) is negative if and only if y reverses orientation.

2. y reverses orientation if and only if the tangent (velocity) vectors for y and x point in

di↵erent directions at the same point on the image curve.

Example. In the previous example, we see that y is a reparametrization of x with

u : [0, ⇡] ! [0, 2⇡] , u(t) = 2t .

Notice that y preserves orientation.

38

Definition (Opposite path). Suppose that

x : [a, b] ! Rn

is a path. The opposite path

xopp

: [a, b] ! Rn

is defined byxopp

(t) = x(a+ b� t) .

Example. The opposite path of

x : [0, 2⇡] ! R2

, x(t) = (cos t, sin t)

isxopp

: [0, 2⇡] ! R2

, x(t) = (cos(2⇡ � t), sin(2⇡ � t)) = (cos t,� sin t).

The path x traces the unit circle in the counter-clockwise direction while the opposite pathxopp

traces the unit circle in the clockwise direction.

Theorem (Reparametrization and scaler line integrals). Let

x : [a, b] ! Rn

be a path. Let f : Rn ! R. If

y : [c, d] ! Rn

is a reparametrization of x, thenZx

f ds =

Zy

f ds .

Since scaler line integrals are not a↵ected by reparametrizations, we can define scaler lineintegrals over curves without having to refer to parametrizations until we need them.

Example. Let � be the upper half of the unit circle

� = {(x, y) | x2 + y

2 = 1, y � 0} .

We would like to integrate the function f(x, y) = y along �. Consider the parametrization

x : [0, ⇡] ! R2

, x(t) = (cos t, sin t) .

So x0(t) = (� sin t, cos t), andZx

y ds =

Z⇡

0

sin tp

sin2

t+ cos2 t dt =

Z⇡

0

sin t dt = 2 .

39

Now consider another parametrization

y : [�1, 1] ! R2

, y(t) = (t,p1� t

2) .

So y0(t) = (1,�t/

p1� t

2), andZy

y ds =

Z1

�1

p1� t

2

r1 +

t

2

1� t

2

!dt =

Z1

�1

dt = 2 .

Theorem (Reparametrization and vector line integrals). Let

x : [a, b] ! Rn

be a path. Let F : Rn ! Rn

be a vector field. Suppose that

y : [c, d] ! Rn

is a reparametrization of x.

1. If y preserves orientation, then

Ry

F · ds =Rx

F · ds.

2. If y reverses orientation, then

Ry

F · ds = �Rx

F · ds.

Example. Let F : R2 ! R2 be the vector field F(x, y) = xyi+ y

2j. Consider the path

x : [0, 4] ! R2

, x(t) = (t, t2) .

Then Zx

F · ds =Z

4

0

(t3, t4) · (1, 2t) dt =Z

4

0

t

3 + 2t5 dt =4288

3.

Consider the reparametrization

y : [�4, 0] ! R2

, y(t) = (�t, t

2) .

Notice that y reverses orientation. ThenZy

F · ds =Z

0

�4

(�t

3

, t

4) · (�1, 2t) dt =

Z0

�4

t

3 + 2t5 dt = �4288

3.

40

Section 6.2. Green’s Theorem

All paths (curves) under consideration will be piecewise C1, unless otherwise specified. Also,all scaler functions f : Rn ! R and vector fields F : Rn ! Rn will be assumed to be C1

unless otherwise specified.

Green’s Theorem

Definitions (simple, closed). A curve in Rn is simple in if it has no self-intersections. A curve in Rn is closed if it is a loop.

Definition (positive orientation). Let C be collection of disjoint simple closedcurves in R2. Suppose that C is the boundary of a region D. The orientationfor C in which D lies to the left of C as one traverses C is called the positiveorientation; when C is a single curve, then positive means counter-clockwise. Theopposite orientation is called the negative orientation.

Special Notation: When expressing a line integral over a closed curve C, we use the specialintegral symbol I

C

in place of the ordinary symbolRC

.

Theorem (Green’s Theorem). Let D ⇢ R2

be a closed, bounded region whose

boundary @D consists of simple closed curves. Give @D the positive orientation.

Then I@D

M dx+N dy =

ZZD

✓@N

@x

� @M

@y

◆dx dy .

Example (Problem 2). Consider the vector field F : R2 ! R2 given by

F(x, y) = M i+N j

where M = x

2� y and N = x+ y

2. Let D be the rectangle with verticies (0, 0), (2, 0), (2, 1),and (0, 1). Verify the formula in Green’s Theorem by calculating each of the integralsI

@D

M dx+N dy and

ZZD

✓@N

@x

� @M

@y

◆dx dy .

Solution:

We start with the double integralZZD

✓@N

@x

� @M

@y

◆dx dy .

41

D is just the rectangle [0, 2]⇥ [0, 1], soZZD

✓@N

@x

� @M

@y

◆dx dy =

ZZD

(1� (�1)) dx dy

=

Z1

0

Z2

0

2 dx dy

= 4 .

Now take care of the line integral I@D

M dx+N dy .

This is more labor-intensive:

The boundary @D consists of four line-segments �1

, �

2

, �

3

, and �4

where

• �

1

connects (0, 0) to (2, 0)

• �

2

connects (2, 0) to (2, 1)

• �

3

connects (2, 1) to (0, 1)

• �

4

connects (0, 1) to (0, 0).

The simplest parametrizations for these segments (with the appropriate orientations) are thefollowing.

• �

1

is given by x1

: [0, 2] ! R2, x1

(t) = (t, 0)

• �

2

is given by x2

: [0, 1] ! R2, x2

(t) = (2, t)

• �

3

is given by x3

: [0, 1] ! R2, x3

(t) =�(1� t)2, 1

�• �

4

is given by x4

: [0, 1] ! R2, x4

(t) = (0, 1� t)

Then I@D

M dx+N dy =4X

i=1

✓Zx

i

M dx+N dy

◆= 4 .

Example (Problem 8). Find the work done by the force field

F = (4y � 3x)i+ (x� 4y)j

along the positively oriented ellipse

x

2 + 4y2 = 4 .

42

Solution: For this problem, let D be the region bounded by the ellipse:

D = {(x, y) | x2 + 4y2 4} .

The work done by F along @D is the line integralI@D

F · ds .

By Green’s Theorem, we haveI@D

F · ds =ZZ

D

✓@N

@x

� @M

@y

◆dx dy

=

ZZD

(1� 4) dx dy

=

ZZD

�3 dx dy

= �3 Area(D) .

The standard form for the equation of @D is⇣x

2

⌘2

+ y

2 = 1 .

Recall that the area of the region D enclosed by the ellipse⇣x

a

⌘2

+⇣y

b

⌘2

= 1

is ab⇡. This is most easily seen by using the polar-like substitution:

x

a

= r cos ✓ ,

y

b

= r sin ✓

with Jacobian@(x, y)

@(r, ✓)= abr .

So

Area(D) =

ZZD

dx dy =

Z2⇡

0

Z1

0

abr dr d✓ = ab⇡ .

It follows that the area of D is 2⇡.

Therefore, the work done by F along @D is �6⇡.

43

Area formula

The following is an easy consequence of Green’s Theorem.

If D is a region to which Green’s Theorem applies, then

Area(D) =

I@D

x dy =

I@D

�y dx =1

2

I@D

x dy � y dx .

The Divergence Theorem

The following theorem is useful for studying the flux of a vector field across a closed curvein R2.

Theorem (Divergence Theorem in the plane). If D is a region to which Green’s

Theorem applies, n is the outward unit normal vector to D, and

F = M(x, y)i+N(x, y)j

is a vector field on D, thenI@D

F · n ds =

ZZD

r · F dA .

Recall that the outward unit normal vector n at a point (x(t), y(t)) can be given by theformula

n = n(t) =y

0(t)i� x

0(t)jp(x0(t))2 + (y0(t))2

.

The quantityr · F

is also known as the divergence of the vector field F, and is usually denoted by div F.

In practice, the equivalent formulaI@D

F · n ds =

ZZD

✓@M

@x

+@N

@y

◆dA

in the Divergence Theorem will be more useful.

Example (Problem 17 (a)). Use the Divergence Theorem to show thatIC

F · n ds = 0

where F = 2yi� 3xj and C is the (positively oriented) unit circle.

44

Solution: By the Divergence Theorem, it is enough to show thatZZD

✓@M

@x

+@N

@y

◆dA = 0 .

Since M = 2y and N = �3x, we see that

@M

@x

+@N

@y

= 0 .

Therefore I@D

F · n ds =

ZZD

✓@M

@x

+@N

@y

◆dA = 0 .

Without the Divergence Theorem, we would have compute the integral directly. Parameter-ize the unit circle in the standard way:

x(t) = (cos t, sin t) , 0 t 2⇡ .

Sox0(t) = (� sin t, cos t) =) ||x0(t)|| = 1 .

To compute the integral, put F and n in terms of x(t):

F(x(t)) = (2 sin t,�3 cos t) , n(t) = (cos t, sin t) .

Therefore I@D

F · n ds =

Z2⇡

0

�F(t) · n(t)

�||x0(t)|| dt

=

Z2⇡

0

�(2 sin t,�3 cos t) · (cos t, sin t)

�dt

=

Z2⇡

0

� cos t sin t dt

= 0 .

45

Section 6.3. Conservative vector fields

All paths (curves) under consideration will be piecewise C1, unless otherwise specified. Also,all scaler functions f : Rn ! R and vector fields F : Rn ! Rn will be assumed to be C1

unless otherwise specified. The same assumptions apply if we restrict the domains to regionsR ⇢ Rn.

Definition. Let R ⇢ Rn be a region. We say that R is connected if every pair ofpoints in R can be joined by a continuous path � ⇢ R. The idea is that R consistsof only one “piece”.

Definition (Path independent line integrals). Let F : R ! Rn be a vector field onR. We say that F has path independent line integrals on R ifZ

C1

F · ds =ZC2

F · ds

for any paths C1

and C

2

in R with the same initial and final points.

Example. We consider two paths C1

and C

2

in R2 starting at (1, 0) and ending at (0, 1):

Let C1

be quarter-circle given by formula x1

(t) = (cos t, sin t) for 0 t ⇡/2.

Let C2

be the line-segment given by the formula x2

(t) = (1� t, t) for 0 t 1.

Consider the vector field F(x, y) = �yi+ xj. ThenZC1

F · ds =Z

⇡/2

0

(� sin t, cos t) · (� sin t, cos t) dt =

Z⇡/2

0

dt = ⇡/2 ,

while ZC2

F · ds =Z

1

0

(�t, 1� t) · (�1, 1) dt =

Z1

0

dt = 1 .

We see that F is does not have path-independent line integrals.

Theorem. Let F : R ! Rn

be a vector field. Then F has path independent line

integrals on R if and only if IC

F · ds = 0

for any simple closed curve C in R.

46

Gradient fields and line integrals

Definition (gradient (or conservative) field, scaler potential). Let R ⇢ Rn be aregion. Suppose that F : R ! Rn is a vector field. We say that F is a gradientfield on R (or conservative field) on R if

F = rf

for some scaler function f : R ! R (of class C2). In this case, we call the functionf a scaler potential function for F. In particular, we realize the component

functions of F as the partial derivatives of f .

Example. A simple example of a conservative vector field in R ⇢ R2 is anything of the form

F(x, y) = M(x)i+N(y)j .

where M and N are C1 functions. Set f(x, y) =RM(x) dx +

RN(y) dy. It is easy to see

thatF = rf .

This construction can be applied in R3 in the obvious way. In particular, F(x, y, z) =2x i+ 3y2 j� 3 sin 3z k and f(x, y, z) = x

2 + y

3 + cos(3z) satisfy F = rf .

Theorem (Fundamental Theorem of Line Integrals). Let R ⇢ R

n

be an open

connected region. Suppose that F : R ! Rn

is a continuous vector field. Then F is

conservative on R if and only if F has path independent line integrals on R.

Furthermore, if F = rf and the path C ⇢ R has initial point A and final point B,

then ZC

F · ds = f(B)� f(A) .

Example. Let C be a continuous path in R2 beginning at A = (1, 2) and ending at B =(�4, 0). Consider the vector field F(x, y) = 2xi+ 3y2j. Compute the vector line integralZ

C

F · ds .

Solution:

Notice that F = rf where f(x, y) = x

2 + y

3; compute f(A) = 9 and f(B) = 16. By theFundamental Theorem of Line Integrals, we haveZ

C

F · ds = f(B)� f(A) = 16� 9 = 7 .

47

Criterion for conservative vector fields

Definition (Simply-connected region). A region R ⇢ Rn is simply-connected ifit is connected and every simple closed curve in R can be continuously shrunk inR to a point.

Example. The following regions are simply-connected.

• Rn; to see this, shrink along straight lines to any desired point P 2 Rn.

• The closed (or open) unit disk in R2; again, shrink along straight lines.

• The closed (or open) unit ball in R3; again, shrink along straight lines.

• The unit sphere in R3; this is not obvious.

• Any “continuously deformed” version of the above examples. For more informationand to see a co↵ee cup deformed into a doughnut, see the website

http://en.wikipedia.org/wiki/Topology

Recall the following definition (see Chapter 3).

Definition (Curl of a vector field). Suppose that F(x, y, z) = M(x, y, z) i +N(x, y, z) j + P (x, y, z) k is a vector field on R3. The curl of F is the vectorfield

r⇥F = det

264 i j k@

@x

@

@y

@

@z

M N P

375 =

✓@P

@y

� @N

@z

◆i�✓@P

@x

� @M

@z

◆j+

✓@N

@x

� @M

@y

◆k .

If F(x, y) = M(x, y)i + N(x, y)j is a vector field on R2, then extend it to a vectorfield F̄(x, y, z) on R3 by the formula

F̄(x, y, z) = M(x, y)i+N(x, y)j+ 0k

and define the curl of F as

r⇥ F = r⇥ F̄ =

✓@N

@x

� @M

@y

◆k .

This is the formula that shows up in the vector reformulation of Green’s Theorem.

Theorem (Conservative = zero curl). Let R ⇢ Rn

be a simply-connected region.

Let F : R ! Rn

be a vector field. Then F is conservative on R if and only if

r⇥ F = 0 on R.

48

Remark. When solving problems, if the region R ⇢ Rn is not specified, then assume that Ris the largest simply-connected region (usually Rn itself) such that F is defined and of classC1.

Example. Consider the vector field

F(x, y) = 2x sin(7y) i+�7x2 cos(7y) + 2y

�j .

Is F conservative?

Solution: We set M = 2x sin(7y) and N = 7x2 cos(7y) + 2y. Then Compute

@N

@x

= 14x cos(7y) and@M

@y

= 14x cos(7y) .

Therefore F is conservative.

Example. Consider the vector field

F(x, y, z) = 3x2

y i + (x3 + 2y) j + 6z k .

Is F conservative?

Solution: We set M = 3x2, N = x

3 + 2y and P = 6z and compute

r⇥F = det

264 i j k@

@x

@

@y

@

@z

M N P

375 = det

264 i j k@

@x

@

@y

@

@z

3x2

x

3 + 2y 6z

375 = (0�0)i�(0�0)j+(3x2�3x2)k = 0 .

Therefore F is conservative.

Example. Consider the vector field

F(x, y, z) = y i� x j+ x

2

yz k .

Is F conservative?

Solution: We set M = y, N = �x and P = x

2

yz and compute

r⇥F = det

264 i j k@

@x

@

@y

@

@z

M N P

375 = det

264 i j k@

@x

@

@y

@

@z

y �x x

2

yz

375 = (x2

z�0)i�(2xyz�0)j+(�1�1)k 6= 0 .

Therefore F is not conservative.

49

Finding scaler potentials on R2

Suppose that F = M i + N j is a conservative vector field on a simply-connectedregion R ⇢ R2. We want to find a scaler potential function f for F. This amountsto solving the system of partial di↵erential equations

@f

@x

= M and@f

@y

= N .

The general procedure for solving the system is to

• Integrate the equation @f

@x

= M to get f(x, y) = (RM dx) + g(y)

• Di↵erentiate this f(x, y) with respect to y to get another formula for @f

@y

;compare to N and deduce a formula for g0(y).

• Integrate one more time: g(y) =Rg

0(y) dy

• Combine the previous formulas to obtain an explicit formula for f(x, y).

There is a variation of this procedure.

Example. Earlier, we showed that

F(x, y) = 2x sin(7y) i+ 7x2 cos(7y) + 2y j

is a conservative vector field on R2. Find a scaler potential for F.

Solution: We follow the standard procedure with M = 2x sin(7y) and N = 7x2 cos(7y) + 2y.

• Integrate @f

@x

= 2x sin(7y) to get

f(x, y) = x

2 sin(7y) + g(y) .

• Di↵erentiate f with respect to y:

@f

@y

=@

@y

�x

2 sin(7y) + g(y)�= 7x2 cos(7y) + g

0(y) ,

and set this equal toN = 7x2 cos(7y) + 2y .

Deduce that g0(y) = 2y.

• Integrate g

0(y) to get g(y):

g(y) =

Zg

0(y) dy = y

2

.

• Conclude thatf(x, y) = x

2 sin(7y) + y

2

.

50

Finding scaler potentials on R3

The method here is a generalization of the method in R2.

Suppose that F = M i + N j + P k is a conservative vector field on a simply-connected region R ⇢ R3. We want to find a scaler potential function f for F. Thisamounts to solving the system of partial di↵erential equations

@f

@x

= M ,

@f

@y

= N , and@f

@z

= P .

The general procedure for solving the system is to

• Integrate the equation @f

@x

= M to get f(x, y, z) = (RM dx) + g(y, z)

• Di↵erentiate this f(x, y, z) with respect to y to get another formula for @f

@y

;

compare to N . Deduce a formula for @g

@y

.

• Integrate @g

@y

with respect to y to get g(y, z) = (R

@g

@y

dy) + h(z); substitutethis g(y, z) into the formula for f(x, y, z) and simplify.

• Di↵erentiate this f(x, y, z) with respect to z to get another formula for @f

@z

;compare to P and deduce a formula for h0(z).

• Integrate one more time: h(z) =Rh

0(z) dz

• Combine the previous formulas to obtain an explicit formula for f(x, y, z).

There are variations of this procedure.

Example. Earlier, we showed that F(x, y, z) = 3x2

y i+(x3+2y) j+6z k is a conservativevector field on R3. Find a scaler potential for F.

Solution: We follow the standard procedure with M = 3x2

y, N = x

3 + 2y and P = 6z.

• Integrate the equation @f

@x

= 3x2

y to get f(x, y, z) = x

3

y + g(y, z)

• Di↵erentiate this f(x, y, z) with respect to y:

@

@y

�x

3

y + g(y, z)�= x

3 +@g

@y

,

and set this equal to N = x

3 + 2y. Deduce that @g

@y

= 2y.

• Integrate @g

@y

= 2y with respect to y to get

g(y, z) = y

2 + h(z) .

51

It follows thatf(x, y, z) = x

3

y + y

2 + h(z) .

• Di↵erentiate this f(x, y, z) with respect to z to get

@f

@z

= 0 + 0 + h

0(z) ;

compare to P = 6z and deduce that h0(z) = 6z.

• Integrate h

0(z) = 6z:

h(z) =

Z6z dz = 3z2 .

• Combine the previous formulas to obtain an explicit formula for f(x, y, z):

f(x, y, z) = x

3

y + y

2 + 3z2 .

52

Section 7.1. Parametrized surfaces

Basic definitions and examples

Definition (Parametrized surface in R3). Let D be a connected region in R2. Aparametrized surface in R3 is a continuous function

X : D ! R3

that is one-to-one on the interior of D. The image

S = X(D) ⇢ R3

is called the underlying surface of X. We can also refer to the function X as aparametrization of the surface S.

See Figure 7.1 in the textbook.

We can write X(s, t) asX(s, t) =

�x(s, t), y(s, t), z(s, t)

�where x(s, t), y(s, t) and z(s, t) are the coordinate functions or component functionsof X.

It be helpful to express the parametrization X : D ! R3 as

X : Dst

! R3

xyz

to emphasize the names of all of the variables involved.

Remark. For the purposes of integration, we will generally want D to be an elementaryregion (as in chapter 5).

Examples (Graphs of functions z = f(x, y)). A major example of surfaces arise fromfunctions of two variables. See Figure 7.5 in the textbook.

1. The graph of z = x

2 + y

2 is the image of the parametrization

X(s, t) = (s, t, s2 + t

2)

where D is any connected region in R2.

2. The graph of z = x

2 is the image of the parametrization

X(s, t) = (s, t, s2)

where D is any connected region in R2.

53

3. In general, the graph of a continuous function z = f(x, y) over a connected regionD ⇢ R2 is the image of the parametrization

X : D ! R3

, X(s, t) = (s, t, f(s, t)) .

Example (Cylinder). The cylinder x2 + y

2 = 4 has the parametrization

X : D✓z

! R3

xyz

defined byX(✓, z) = (2 cos ✓, 2 sin ✓, z)

whereD

✓z

= {(✓, z) | 0 ✓ 2⇡} .

Here, we are using the more appropriate variables ✓ instead of s and z instead of t. SeeFigure 7.4 in the textbook.

Example (Sphere). The sphere x

2 + y

2 + z

2 = 25 has the parametrization

X : D✓'

! R3

xyz

defined byX(✓,') = (5 sin' cos ✓, 5 sin' sin ✓, 5 cos')

whereD

✓'

= {(✓,') | 0 ✓ 2⇡ , 0 ' ⇡} .

Here, we are using the more appropriate variables ✓ instead of s and ' instead of t. SeeFigure 7.3 in the textbook.

Coordinate curves

The main idea here is that we can fix one of variables, s or t, of X(s, t) to produce specialcurves on the surface S = X(D).

54

Definition (Coordinate curves). Let X : D ! R3 be a parametrized surface. Let(s

0

, t

0

) 2 D.

• The s–coordinate curve through X(s0

, t

0

) is the curve

s 7! X(s, t0

)

defined for s–values close to s

0

.

• The t–coordinate curve through X(s0

, t

0

) is the curve

t 7! X(s0

, t)

defined for t–values close to t

0

.

Of course, the largest domain for these curves depends on the region D.

See Figure 7.6 in the textbook.

Tangent vector, tangent plane and normal vector

Let X : Dst

! R3 be a parametrization for the surface S = X(D). Let (s0

, t

0

) 2 D. Considerthe coordinate curves

X(s, t0

) and X(s0

, t)

near the point X(s0

, t

0

) 2 S. The velocity vectors of these coordinate curves at the pointX(s

0

, t

0

) are

Ts

(s0

, t

0

) =@X

@s

(s0

, t

0

) and Tt

(s0

, t

0

) =@X

@t

(s0

, t

0

) .

These vectors are tangent vectors to the surface S. The plane determined by Ts

(s0

, t

0

)and T

t

(s0

, t

0

) is the tangent plane to S at X(s0

, t0

). See Figure 7.10 in the textbook.

The vectors Ts

and Tt

determine a standard normal vector to the tangent plane:

N(s0

, t

0

) = Ts

(s0

, t

0

)⇥Tt

(s0

, t

0

) .

Then the unit normal vector is defined as

n(s0

, t

0

) =N(s

0

, t

0

)

||N(s0

, t

0

)|| .

If N(s0

, t

0

) 6= 0, then the tangent plane to S at the point X(s0

, t

0

) = (x0

, y

0

, z

0

) has theequation

N(s0

, t

0

) · (x� x

0

, y � y

0

, z � z

0

) = 0 .

55

Definition (Smooth point). Let X : D ! R3 is a parametrized surface that isC1 near the point X(s

0

, t

0

) and N(s0

, t

0

) 6= 0, then X(s0

, t

0

) is called a smoothpoint. If S = X(D) consists of all smooth points, then S is called a smoothparametrized surface.

Example. Consider the surface S ⇢ R3 given by the parametrization

X(s, t) = (4es, t2e2s, 3e�s + t)

for (s, t) 2 R2. Find an equation for the tangent plane to S at the point (4, 9, 0).

Solution: First find the point (s, t) 2 R2 that corresponds to the point (4, 9, 0). To do this,solve the equation

X(s, t) = (4, 9, 0) .

This amounts to solving the system of equations

4es = 4

t

2

e

2s = 9

3e�s + t = 0 .

We see that the first equation 4es = 4 implies s = 0. Then plug this into the third equation3e�s + t = 0 to get t = �3. The tangent vectors to S have the formulas

Ts

=@X

@s

= (4es, 2t2e2s,�3e�s) and Tt

=@X

@t

= (0, 2te2s, 1) .

At the point X(0,�3), the tangent vectors are

Ts

(0,�3) = (4, 18,�3) and Tt

(0,�3) = (0,�6, 1) .

Now compute the normal vector at X(0,�3):

N(0,�3) = Ts

(0,�3)⇥Tt

(0,�3) = (4, 18,�3)⇥ (0,�6, 1) = (0,�4,�24) .

Therefore, the tangent plane to S at the point X(0,�3) = (4, 9, 0) is

(0,�4,�24) · (x� 4, y � 9, z � 0) = 0 =) 4(y � 9) + 24z = 0 .

Example (Graphs of z = f(x, y)). In the special case of a graph of a C1 function f(x, y)with domain D ⇢ R2, we can make the special parametrization

X : D ! R3

, X(s, t) = (s, t, f(s, t)) .

With this parametrization, the tangent vector formulas are

Ts

=

✓1, 0,

@f

@s

◆and T

t

=

✓0, 1,

@f

@t

◆.

56

So the normal vector is

N = Ts

⇥Tt

=

✓�@f@s

, �@f@t

, 1

◆.

For example, consider the graph of the function f(x, y) = x

2 + xe

y on the domain D = R2.Give the graph of f(x, y) the standard parametrization

X(s, t) = (s, t, s2 + se

t) .

Then

N =

✓�@f@s

, �@f@t

, 1

◆=��(2s+ e

t),�se

t

, 1�.

At the point (1, 0, 2) on the graph of f , the equation of the tangent plane is

N(1, 0)·(x�1, y, z�2) = 0 =) (�3,�1, 1)·(x�1, y, z�2) = 0 =) �3(x�1)�y+(z+2) = 0 .

This simplifies to3x+ y � z = 1 .

Piecewise smooth parametrized surfaces

Definition (Piecewise parametrized surface). A piecewise parametrized sur-face is the union of parametrized surfaces X : D

i

! R3, i = 1, . . . ,m where

• Each Xi

is of class C1, except possibly along @Di

.

• Each S

i

= X(Di

) is smooth, except possibly at finitely many points.

Examples. The following are examples of piecewise smooth parametrized surfaces.

• The boundary of the solid cube [0, 1] ⇥ [0, 1] ⇥ [0, 1] ⇢ R3. See Figure 7.12 in thetextbook.

• The boundary of nearly every solid that appears in section 5.4. In particular, seeFigures 5.47, 5.56, 5.63, 5.65 and 5.67 in the textbook.

Example. Consider the cylindrical solid

{(x, y, z) | x2 + y

2 1, 0 z 1} ⇢ R3

.

Let S denote the boundary of the solid. Then S consists of three smooth pieces S1

, S

2

, S

3

:

• S

1

= {(x, y, 0) | x2 + y

2 1} is the bottom of the cylinder.

57

• S

2

= {(x, y, 1) | x2 + y

2 1} is the top of the cylinder.

• S

3

= {(x, y, z) | x2 + y

2 = 1, 0 z 1} is the lateral portion of S.

See Figure 7.29 in the textbook. Notice that each S

i

is a smooth parametrized surface. Sowhen you need to calculate something over S, you can use parametrizations such as thefollowing.

• For S1

, use X1

(r, ✓) = (r cos ✓, r sin ✓, 0) , (r, ✓) 2 [0, 1]⇥ [0, 2⇡].

• For S2

, use X2

(r, ✓) = (r cos ✓, r sin ✓, 1) , (r, ✓) 2 [0, 1]⇥ [0, 2⇡].

• For S3

, use X3

(✓, z) = (cos ✓, sin ✓, z) , (✓, z) 2 [0, 2⇡]⇥ [0, 1].

Area of a parametrized surface

Let X : D ! R3 be a parametrized surface. The surface area of S = X(D) isdefined to be ZZ

D

||Ts

⇥Tt

|| ds dt =

ZZD

||N(s, t)|| ds dt .

Example. Let S be the portion of the paraboloid z = 25�x

2�y

2 that lies over the xy–plane.Find the area of S.

Solution 1: We can realize S as the graph of f(x, y) = 25� x

2 � y

2 on the domain D ⇢ R2

where D is the shadow of S in the xy–plane. It is easy to see that D is the disk

D = {(x, y) | x2 + y

2 25} .

So consider the standard parametrization

X(s, t) = (s, t, 25� s

2 � t

2) .

ThenT

s

= (1, 0,�2s) and Tt

= (0, 1,�2t) .

So||T

s

⇥Tt

|| = ||(2s, 2t, 1)|| =p

(2s)2 + (2t)2 + 1 .

The area of S is ZZD

||Ts

⇥Tt

|| ds dt =

ZZD

p(2s)2 + (2t)2 + 1 ds dt .

To evaluate this integral, we should use polar coordinates for D. Make the substitutions

s = r cos ✓ , t = r sin ✓ ( 0 r 5 , 0 ✓ 2⇡ )

58

to get ZZD

p(2s)2 + (2t)2 + 1 ds dt =

Z2⇡

0

Z5

0

p(2r cos ✓)2 + (2r sin ✓)2 + 1 r dr d✓

=

Z2⇡

0

Z5

0

p4r2 + 1 r dr d✓

=⇡

6

⇣101

p101� 1

⌘.

Solution 2: Consider the following (cylindrical) parametrization of S:

X : [0, 5]⇥ [0, 2⇡] ! R3

, X(r, ✓) = (r cos ✓, r sin ✓, 25� r

2) .

This time, D is just the rectangle [0, 5]⇥ [0, 2⇡]. Then

Tr

= (cos ✓, sin ✓,�2r) and T✓

= (�r sin ✓, r cos ✓, 0) .

SoT

r

⇥T✓

= (2r2 cos ✓,�2r2 sin ✓, r cos2 ✓ + r sin2

✓) = (2r2 cos ✓,�2r2 sin ✓, r) .

The area of S isZZD

||Tr

⇥T✓

|| dr d✓ =

Z2⇡

0

Z5

0

p4r4 cos2 ✓ + 4r4 sin2

✓ + r

2

dr d✓

=

Z2⇡

0

Z5

0

p4r4 + r

2

dr d✓

=

Z2⇡

0

Z5

0

r

p4r2 + 1 dr d✓

=⇡

6

⇣101

p101� 1

⌘.

Remark. If S is a piecewise smooth parametrized surface where the smooth pieces S1

, . . . , S

m

of S meet along curves, then calculate the area of S in a piecewise way:

Area(S) = Area(S1

) + · · ·+Area(Sm

) .

59

Section 7.2. Surface integrals

Scaler surface integrals

Definition. Let X : D ! R3 be a smooth parametrized surface, where D ⇢ R2 isa bounded region. Let f be a continuous scaler function on S = X(D). The scalersurface integral of f along X isZZ

X

f dS =

ZZD

f

�X(s, t)

�||T

s

⇥Tt

|| ds dt =

ZZD

f

�X(s, t)

�||N(s, t)|| ds dt .

The definition extends to piecewise smooth parametrized surfaces.

Example. Let D = [0, 1]⇥ [0, 2] and consider the parametrized surface

X(s, t) = (s, s+ t, t) .

ComputeRR

X

x

2 + y

2 + z

2

dS.

Solution: The coordinate tangent vectors are

Ts

= (1, 1, 0) and Tt

= (0, 1, 1) ,

so the normal vector Ts

⇥Tt

is

Ts

⇥Tt

= (1,�1, 1) ; so ||Ts

⇥Tt

|| =p3 .

Therefore ZZX

x

2 + y

2 + z

2

dS =

Z2

0

Z1

0

(s2 + (s+ t)2 + t

2)p3 ds dt =

26p3.

Vector surface integrals

Definition. Let X : D ! R3 be a smooth parametrized surface, where D ⇢ R2 isa bounded region. Let F(x, y, z) be a continuous vector field on S = X(D). Thenthe vector surface integral of F along X isZZ

X

F · dS =

ZZD

F�X(s, t)

�·N(s, t) ds dt .

This integral is also known as a flux integral because it computes the flux of avector field F in R3 though the surface S.

60

Example. Consider the vector field F(x, y, z) = 2x i+y j�z k on the parametrized surface

X : [0, 1]⇥ [0, 2] ! R3

, X(s, t) = (s+ t, t, st) .

Compute ZZX

F · dS .

Solution: The coordinate tangent vectors are

Ts

= (1, 0, t) and Tt

= (1, 1, s) ,

so the normal vector isN(s, t) = T

s

⇥Tt

= (�t, t� s, 1) .

Therefore ZZX

F · dS =

Z2

0

Z1

0

�2(s+ t), t,�st

�·�� t, t� s, 1

�ds dt

=

Z2

0

Z1

0

�� 2t(s+ t) + t(t� s)� st

�ds dt

=

Z2

0

Z1

0

�� t

2 � 4st�ds dt

= �20

3.

The e↵ect of reparametrization

The situation is similar to the one for line integrals, the scaler integrals do not depend onreparametrization, and the same is true for vector integrals (up to ± sign).

Definition. A reparametrization of a surface

X : D1

! R3

is another parametrizationY : D

2

! R3

for which there exists a di↵erentiable coordinate transformation H : D2

! D

1

thatis one-to-one and onto, and Y(s, t) = X(H(s, t)).

61

The e↵ect of reparametrization for scaler surface integrals

Theorem (Reparametrization and scaler surface integrals). Let

X : D1

! R3

be a parametrized surface. Let f be a continuous scaler function on S = X(D1

). If

Y : D2

! R3

is a reparametrization of X, thenZZX

f dS =

ZZY

f dS .

Since scaler surface integrals are not a↵ected by reparametrizations, we can define scalersurface integrals ZZ

S

f dS

without having to refer to parametrizations until we need to calculate them.

Orientation

Definition (Orientable surface). Let S ⇢ R3 be a smooth parametrized surface.We say that S is orientable if it is possible to define a continously varying unitnormal vector n to every point on S. Otherwise, S is called nonorientable.

Remark. A surface S ⇢ R3 is orientable if and only if it has two sides. See Figure 7.21 inthe textbook. The only surfaces that we will care about in this class will be orientable.

Definition (Orientation-preserving reparametrizations). Let X : D1

! R3 be aparametrization for an orientable surface S ⇢ R3. Suppose that Y : D

2

! R3 is areparametrization of X. Then Y is an orientation-preserving reparametriza-tion if the normal vectors from X and Y point in the same direction at correspond-ing points on S. Otherwise Y is an orientation-reversing reparametrization.

More precisely, if Y(s, t) = X(H(s, t)) where H(s, t) is a coordinate transformation, then Yis orientation-preserving precisely when the Jacobian determinant of H is positive.

Example (Variable swap). Suppose that X : D1

! R3 is a parametrization of the surfaceS = X(D

1

). Let D2

be the region obtained by swapping the variables in D

1

:

D

2

= {(s, t) | (t, s) 2 D

1

} .

62

Then defineY : D

2

! R3

, Y(s, t) = X(t, s) .

The Jacobian matrix ofH(s, t) = (t, s)

is 0 11 0

�;

the determinant of this matrix is �1. Therefore Y is a orientation-reversing reparametriza-tion of X. You should pictorially compare the tangent vectors and normal vectors of X andY. You can also see Figure 7.20 in the textbook.

The e↵ect of reparametrization for vector surface integrals

Theorem (Reparametrization and vector surface integrals). Let

X : D1

! R3

be a parametrized surface. Let F be a continuous vector field on S = X(D1

). If

Y : D2

! R3

is a reparametrization of X, then

•RR

X

F · dS =RR

Y

F · dS if Y is orientation-preserving.

•RR

X

F · dS = �RR

Y

F · dS if Y is orientation-reversing.

So if we have an oriented surface S, then then the integralZZS

F · dS

is defined to be ZZX

F · dS

where X(s, t) is a parametrization whose orientation agrees with the prescribed orientation.

Outward (pointing) normal

The orientation that we usually prescribe for a surface will one that agrees with the standardnormal vectors of the graph of a function z = f(x, y). The main idea is that we want to definethe “top-side” of the surface to be what we would observe from high above the xy–plane.

63

Definition (Outward (pointing) normal vector). Let S ⇢ R3 be an orientablesurface that bounds a solid W with finite volume. Then the outward (pointing)normal vector can be defined in the obvious way, that is, the normal vector shouldpoint out of the solid.

Remark. In the case that S is the graph of a function z = f(x, y), the upward (pointing)normal will be ✓

�@f@x

,�@f@y

, 1

◆.

In addition, if f(x, y) is nonnegative, then this upward (pointing) normal is the same as theoutward (pointing) normal for the solid between S and the xy–plane.

Example (Cylinder). Define S ⇢ R3 to be the boundary of the solid cylindrical region

{(x, y, z) | x2 + y

2 1 , 0 z 1} .

Give S the outward normal orientation. So S consists of three smooth pieces S1

, S

2

, S

3

:

• S

1

= {(x, y, 1) | x2 + y

2 1} is the bottom of the cylinder.

• S

2

= {(x, y, 0) | x2 + y

2 1} is the top of the cylinder.

• S

3

= {(x, y, z) | x2 + y

2 = 1, 0 z 1} is the lateral portion of S.

We want to parametrize each of the Si

in a way that is consistent with the orientation of S.Let D ⇢ R2 be the closed unit disk. See Figure 7.29 in the textbook.

• A suitable parametrization for S1

is

X1

: D ! R3

, X1

(s, t) = (t, s, 0) .

Notice that

@X1

@s

= (0, 1, 0) and@X

1

@t

= (1, 0, 0) =) N1

=@X

1

@s

⇥ @X1

@t

= (0, 0, 1) .

The outward normal vector is �N1

= (0, 0,�1).

• A suitable parametrization for S2

is

X2

: D ! R3

, X2

(s, t) = (s, t, 1) .

Notice that

@X2

@s

= (1, 0, 0) and@X

2

@t

= (0, 1, 0) =) N2

=@X

2

@s

⇥ @X2

@t

= (0, 0, 1) .

The outward normal vector is N2

.

64

• A suitable parametrization for S3

is

X3

: [0, 2⇡]⇥ [0, 1] ! R3

, X3

(✓, z) = (cos ✓, sin ✓, z) .

Notice that

@X3

@✓

= (� sin ✓, cos ✓, 0) and@X

3

@z

= (0, 0, 1) =) N3

=@X

3

@✓

⇥ @X3

@z

= (cos ✓, sin ✓, 0) .

The outward normal vector is N3

.

Now consider the vector field

F(x, y, z) = x i� 2y j+ (x2 + z) k

on R3. Then the flux of F through S isZZS

F · dS =

ZZS1

F · dS +

ZZS2

F · dS +

ZZS3

F · dS

=

ZZD

(s,�2t, s2) · (0, 0,�1) ds dt +

ZZD

(s,�2t, s2) · (0, 0, 1) ds dt

+

Z2⇡

0

Z1

0

(cos ✓,�2 sin ✓, cos2 ✓ + z) · (cos ✓, sin ✓, 0) dz d✓

=

ZZD

�s

2

ds dt +

ZZD

s

2

ds dt +

Z2⇡

0

Z1

0

(cos2 ✓ � 2 sin2

✓) dz d✓

= 0� ⇡/2

= �⇡/2 .

65

Section 7.3. Stokes’s and Gauss’s Theorems

Stokes’s Theorem

Definition (Boundary orientation). Let S be a piecewise smooth oriented surface.Suppose that @S consists of simple closed curves. Let C be one of the one of theseboundary circles. The boundary orientation (a.k.a. induced orientation) onC is the orientation such that the “top of S” lies to the left of C as one traversesC. See Figures 7.30 and 7.31 in the textbook.

Theorem (Stokes’s Theorem). Let S be a bounded piecewise smooth oriented sur-

face. Suppose that the boundary @S consists of simple closed curves with the bound-

ary orientation. Let F be a C1

vector field on S. ThenZZS

(r⇥ F) · dS =

I@S

F · ds .

Remark: Stokes’s Theorem is a generalization of (the vector reformulation of) Green’sTheorem since

r⇥ (M i+N j+ 0k) =

✓@N

@x

� @M

@y

◆k .

Example (Compare with Example 1 in the textbook). Verify Stokes’s Theorem for thevector field

F(x, y, z) = (z2, 2x, y)

and S the portion of the paraboloid z = 1 � (x2 + y

2) with z � 0; orient S with upwardpointing normal vectors.

Solution: First we calculate ZZS

(r⇥ F) · dS .

The curl of F isr⇥ F = (1, 2z, 2) .

Notice that S is the graph of the function

f(x, y) = 1� (x2 + y

2)

with domain the unit disk in D ⇢ R2. Give S the standard parametriztion

X(s, t) = (s, t, 1� (s2 + t

2)) , (s, t) 2 D .

Then calculate the standard normal vector of X:

N = (2s, 2t, 1) .

66

Then evaluateZZS

(r⇥ F) · dS =

ZZD

(r⇥ F) ·N ds dt

=

ZZD

(1, 2(1� (s2 + t

2)), 2) · (2s, 2t, 1) ds dt

=

ZZD

�2s+ 4t(1� (s2 + t

2)) + 2�ds dt

=

Z2⇡

0

Z1

0

�2r cos ✓ + 4r sin ✓(1� r

2) + 2�r dr d✓

=

Z2⇡

0

2

3r

3 cos ✓ +4

3r

3 sin ✓ � 4

5r

5 sin ✓ + r

2

�r=1

r=0

d✓

=

Z2⇡

0

✓2

3cos ✓ +

4

3sin ✓ � 4

5sin ✓ + 1

◆d✓

= 2⇡ .

Next we calculateH@S

F·ds. Notice that @S is just the unit circle in the xy–plane, and the ap-propriate orientation for @S is the counterclockwise orientation. So give @S the parametriza-tion

x(✓) = (cos ✓, sin ✓, 0) , 0 ✓ 2⇡ .

The velocity (tangent) vector is

x0(✓) = (� sin ✓, cos ✓, 0) .

Therefore I@S

F · ds =Z

2⇡

0

F(x(✓)) · x0(✓) d✓

=

Z2⇡

0

(0, 2 cos ✓, sin ✓) · (� sin ✓, cos ✓, 0) d✓

=

Z⇡

0

2 cos2 ✓ d✓

= 2⇡ .

Example (Compare with Example 2 in the textbook). Consider the vector field

F(x, y, z) = (2y, ez2, e

y

2) .

67

Let S be the portion of the paraboloid z = 1� (x2 + y

2) with z � 0; orient S with upwardpointing normal vectors. Compute ZZ

S

(r⇥ F) · dS .

Solution: First compute the curl of F:

r⇥ F = (2yey2 � 2zez

2, 0,�2) .

To computeRR

S

(r⇥ F) · dS directly, we would parametrize S by

X(s, t) = (s, t, 1� (s2 + t

2)) , (s, t) 2 D = {(s, t) | s2 + t

2 1} .

The normal vector would beN = (2s, 2t, 1) .

Then the integral would beZZS

(r⇥ F) · dS =

ZZD

(2tet2 � 2(1� (s2 + t

2))e(1�(s

2+t

2))

2, 0,�2) · (2s, 2t, 1) ds dt

=

ZZD

�(2tet

2 � 2(1� (s2 + t

2))e(1�(s

2+t

2))

2)(2s)� 2

�ds dt

Then to evaluate the integral, use polar coordinates on D. This computation would beextremely unpleasant to do by-hand or even with a computer.

The remedy for this complication is to pass to a simpler surface S

0 that has the same orientedboundary as S, that is, @S 0 = @S as oriented curves. By applying Stokes’s Theorem to S

and S

0 we haveZZS

(r⇥ F) · dS =

I@S

F · dS =

I@S

0F · dS =

ZZS

0(r⇥ F) · dS .

It is clear that the simpler surface S

0 should be the disk {(x, y, 0) | x2 + y

2 1} with thenatural parametrization (s, t, 0). The normal vector for S 0 is N = (0, 0, 1), soZZ

S

0(r⇥F) ·dS =

ZZS

0(2tet

2, 0,�2) · (0, 0, 1) ds dt =

ZZS

0�2 ds dt = �2 Area(S 0) = �2⇡ .

Gauss’s Theorem (Divergence Theorem)

Definition (Closed surface). Let S be a union of finitely many boundedparametrized surfaces. We call S closed if it has no boundary points.

68

Examples. The following surfaces are closed.

• The sphere; more generally, an ellipsoid.

• The torus.

• The boundary of the solid region bounded by the sphere x

2 + y

2 + z

2 = 25 and thecone z =

px

2 + y

2; this generalizes to many regions that we see in this clas.

Theorem (Gauss’s Theorem or the Divergence Theorem). Let W ⇢ R3

be a

bounded solid whose boundary @W consists of finitely many piecewise smooth, closed

surfaces with outward normal orientations. Let F be a C1

vector field on W . ThenZZZW

r · F dV =

ZZ@W

F · n dS .

Notice that the integral on the right calculates the flux of F through @W .

Keep in mind that the integral ZZ@W

F · n dS

that appears in the Divergence Theorem is the same as the integralZZ@W

F · dS .

Example. Let W be the solid between the paraboloid z = 9 � x

2 � y

2 and the xy–plane.Verify the Divergence Theorem for the vector field

F(x, y, z) = (2x, 5y, 3z) .

Solution: First, we calculateRRR

W

r · F dV :ZZZW

r · F dV =

ZZZW

(2 + 5 + 3) dV =

ZZZW

10 dV .

Then evaluate the integral in cylindrical coordinatesZZZW

10 dV =

Z2⇡

0

Z3

0

Z9�r

2

0

10r dz dr d✓ = 405⇡ .

Now we calculateRR

@W

F · n dS. Keep in mind that this is just the usual flux integralZZ@W

F · dS .

69

The boundary of W consists of two pieces

S

1

= {(x, y, 0) | x2 + y

2 9} and S

2

= {(x, y, z) | z = 9� x

2 � y

2

, z � 0}

With S

1

and S

2

properly oriented, we can computeRR

@W

F · dS as the sumZZ@W

F · dS =

ZZS1

F · dS +

ZZS2

F · dS .

Let D dentote the closed disk {(s, t) | s2 + t

2 9} ⇢ R2.

For S1

, we can use the parametrization

X1

: D ! R3

, X1

(s, t) = (s, t, 0)

as long as we use the downward pointing normal vector

N1

= (0, 0,�1) .

Thus ZZS1

F · dS =

ZZD

(2s, 5t, 0) · (0, 0,�1) ds dt =

ZZD

0 ds dt = 0 .

For S2

, we can use the parametrization

X2

: D ! R3

, X2

(s, t) = (s, t, 9� s

2 � t

2)

with the upward pointing normal vector

N2

= (2s, 2t, 1) .

This yields ZZS2

F · dS =

ZZS2

�2s, 5t, 3(9� s

2 � t

2)�· (2s, 2t, 1) ds dt

=

ZZD

(s2 + 7t2 + 27) ds dt

Then change to polar coordinates

s = r cos ✓ , t = r sin ✓

for (r, ✓) 2 [0, 3]⇥ [0, 2⇡].

70

The integral is nowZZS2

F · dS =

ZZD

(s2 + 7t2 + 27) ds dt

=

Z2⇡

0

Z3

0

(r2 cos2 ✓ + 7r2 sin2

✓ + 27) r dr d✓

=

Z2⇡

0

Z3

0

(r2 + 6r2 sin2

✓ + 27) r dr d✓

= 405⇡ .

Therefore ZZ@W

F · dS = 0 + 405⇡ = 405⇡ .

Example. Let S be the boundary of the cube C = [�1, 1]3; orient S with normal vectorsthat point into C. Consider the vector field

F(x, y, z) =

✓x

⇢

3

,

y

⇢

3

,

z

⇢

3

◆where ⇢ = ⇢(x, y, z) =

px

2 + y

2 + z

2. Use the Divergence Theorem to compute the surfaceintegral ZZ

S

F · dS .

Solution:

The denominator ⇢3 in the components of F is problematic. It makes F is not defined at theorigin, so we can’t just set

RRS

F · dS equal toRRR

C

r · F dV .

The good news is that the Divergence Theorem allows us to integrate F over large spheresinstead of over S:

First, we compute the divergence of F: Use the quotient rule to compute the necessarypartial derivatives:

@

@x

✓x

⇢

3

◆=⇢

2 � 3x2

⇢

5

@

@y

✓y

⇢

3

◆=⇢

2 � 3y2

⇢

5

@

@z

✓z

⇢

3

◆=⇢

2 � 3z2

⇢

5

.

71

It follows that

r · F =@

@x

✓x

⇢

3

◆+

@

@y

✓y

⇢

3

◆+

@

@z

✓z

⇢

3

◆=

3⇢2 � 3⇢2

⇢

5

= 0 .

Let S 0 ⇢ R3 be the sphereS

0 = {(x, y, z) | ⇢ = 2} .

Let W ⇢ R3 be the solid region between S and S

0 along with the surfaces S and S

0. GiveS

0 the outward normal orientation.

Since the origin is not contained in W , the Divergence Theorem givesZZ@W

F · dS =

ZZZW

r · F dV =

ZZZW

0 dV = 0 .

This implies the equalityZZ@W

F · dS =

ZZS

F · dS +

ZZS

0F · dS = 0 .

Hence ZZS

F · dS = �ZZ

S

0F · dS .

It is actually possible to avoid parametrizing S

0 to calculate the integral because

the unit normal to S

0 at a point (x, y, z) 2 S

0 is just

n =

✓x

⇢

,

y

⇢

,

z

⇢

◆.

Then integral is

72

ZZS

0F · dS =

ZZS

0(F · n) dS

=

ZZS

0

✓x

⇢

3

,

y

⇢

3

,

z

⇢

3

◆·✓x

⇢

,

y

⇢

,

z

⇢

◆dS

=

ZZS

0

✓x

2 + y

2 + z

2

⇢

4

◆dS

=

ZZS

0

✓⇢

2

⇢

4

◆dS

=

✓1

⇢

2

◆ZZS

0dS

=

✓1

⇢

2

◆Area(S 0)

=

✓1

⇢

2

◆4⇡⇢2

= 4⇡ .

Therefore ZZS

F · dS = �4⇡ .

Remark: We know that ⇢ = 2 on S

0, but if we want to increase ⇢ to do another problem,the calculations are completely similar. The point here is that ⇢ is a constant function on asphere centered at the origin.

73