multiple integrals - christian brothers universityfacstaff.cbu.edu/wschrein/media/m232...

CHAPTER 13

Multiple Integrals

1. Double Integrals

Definition (1.1 – Riemann Integral). Suppose f is continuous on [a, b].Partition [a, b] into n subintervals

a = x0 < x1 < x2 < · · · < xn�1 < xn = b.

Let �xi = xi � xi�1 be the width of the i’th subinterval [xi�1, xi] and let thenorm of the partition kPk be the largest of the �xi’s. Then the definite integralof f on [a, b] is Z b

af(x) dx = lim

kPk!0

nXi=1

f(ci)�xi

providd the limit exists and is the same for all choices of the evaluation pointsci 2 [xi�1, xi] for i = 1, 2, . . . , n. In this case, we say f is integrable on [a, b].

112

1. DOUBLE INTEGRALS 113

Double Integrals over a Rectangle

Let f(x, y) be continuous on the rectangle R = {(x, y)|a x b, c y d}.Partition R by laying a rectangular grid on top of R consisting of n smallerrectangles R1, R2, . . . , Rn (not necessarily the same size).

Let �Ai be the area of rectangle Ri and define the norm of the partition kPkto be the largest diagonal of any rectangle in the partition.

Definition. The double integral of f over R isZZR

f(x, y) dA = limkPk!0

nXi=1

f(ui, vi) �Ai

provided the limit exists and is the same for every choice of the evaluation points(ui, vi) in Ri, for i = 1, 2, . . . , n. When this happens, we say f is integrableover R.

114 13. MULTIPLE INTEGRALS

How to Compute

If f(x, y) � 0, the volume below the surface is

V =

ZZR

f(x, y) dA.

Using slices parallel to the y-axis (as in Calculus II),

V =

Z b

aA(x) dx

where A(x) is the area of the slice at x, which is

A(x) =

Z d

cf(x, y) dy

by a partial integration with respect to y. Then,

V =

Z b

aA(x) dx =

Z b

a

"Z d

cf(x, y) dy

#dx =

Z b

a

Z d

cf(x, y) dydx,

an iterated integral.

Symetrically, taking slices parallel to the x-axis,

V =

Z d

cA(y) dy =

Z d

c

"Z b

af(x, y) dx

#dy =

Z d

c

Z b

af(x, y) dxdy.


Theorem (1.1 – Fubini’s Theorem). Suppose f is integrable over therectangle R = {(x, y)|a x b, c y d}. ThenZZ

Rf(x, y) dA =

Z b

a

Z d

cf(x, y) dydx =

Z d

c

Z b

af(x, y) dxdy.

Example. If R = {(x, y)|2 x 4, 0 y 2} = [2, 4] ⇥ [0, 2], findZZR(x2y2 � 17) dA.

ZZR(x2y2 � 17) dA =

Z 4

2

Z 2

0(x2y2 � 17) dydx =

Z 4

2

hx2y3

3� 17y

��20

idx

=

Z 4

2

⇣8x2

3� 34

⌘dx =

8x3

9� 34x)

��42

=⇣512

9� 136

⌘�⇣64

9� 68

⌘=

448

9� 68 = �164

9.

Example.

ZZ[0,ln 2]⇥[0,ln 3]

ex+y dA =

Z ln 3

0

Z ln 2

0ex+y dxdy =

Z ln 3

0

hex+y

��ln 2

0

idy

=

Z ln 3

0

⇣eln 2+y � ey

⌘dy =

Z ln 3

0ey dy

= ey��ln 3

0= 3� 1 = 2


Double Integrals over General Regions

For a general region R,we sum over inner partitions using the rectangles thatlie entirely within the region R.

Definition. For any function f(x, y) defined on a bounded region R 2 R2,the double integral of f over R isZZ

Rf(x, y) dA = lim

kPk!0

nXi=1

f(ui, vi)�Ai

provided the limit exists and is the same for every choice of evaluation points(ui, vi) in Ri, for i = 1, 2, . . . , n. In this case, we say f is integrable over R.


How to Compute

(1) Vertically simple region:

Theorem (1.2). Suppose f(x, y) is continuous on R defined by

R = {(x, y)|a x b, g1(x) y g2(x)}for continuous functions g1 and g2 where g1(x) g2(x) for all x 2 [a, b].Then ZZ

Rf(x, y) dA =

Z b

a

Z g2(x)

g1(x)f(x, y) dydx.

(1) Horizontally simple region:

Theorem (1.3). Suppose f(x, y) is continuous on R defined by

R = {(x, y)|c y d, h1(y) x h2(y)}for continuous functions h1 and h2 where h1(y) h2(y) for all y 2 [c, d].Then ZZ

Rf(x, y) dA =

Z d

c

Z h2(y)

h1(y)f(x, y) dxdy.


Theorem (1.4). Let f(x, y) and g(x, y) be integrable over R 2 R2 andlet c 2 R. Then

(1)

ZZR

cf(x, y) dA = c

ZZR

f(x, y) dA

(2)

ZZR

⇥f(x, y) + g(x, y)

⇤dA =

ZZR

f(x, y) dA +

ZZR

g(x, y) dA

(3) If R = R1 [R2 where R1 \R2 = �, thenZZR

f(x, y) dA =

ZZR1

f(x, y) dA +

ZZR2

f(x, y) dA

Note. This means any region can be decomposed into vertically and hori-zontally simple regions.


Example. Z 1

0

Z py

y(x + y) dxdy =

Z 1

0

⇣x2

2+ xy

��p

y

y

⌘dy =

Z 1

0

h⇣y

2+ y3/2

⌘�⇣y2

2+ y2

⌘idy =

Z 1

0

⇣y

2+ y3/2 � 3y2

2

⌘dy =

y2

4+

2y5/2

5� y3

2

��10

=1

4+

2

5� 1

2=

3

20This region is also vertically simple where x = y =) y = x and x =

py =)

y = x2. Thus Z 1

0

Z py

y(x + y) dxdy =

Z 1

0

Z x

x2(x + y) dydx.


Example. Find

ZZR

xy dA where R is bounded by x = y2 and 3x+2y = 8.

Points of intersection:

(substitution) 3y2 + 2y = 8 =) 3y2 + 2y� 8 = 0 =) (3y� 4)(y + 2) = 0 =)y =

4

3or y = �2. Thus

⇣16

9,4

3

⌘and (4,�2) are points of intersection, giviving

us a region both vertically and horizontally simple.

horizontal: h1(y) = y2 and h2(y) = �2

3y +

8

3=)

ZZR

xy dA =

Z 4/3

�2

Z �23 y+8

3

y2xy dxdy =

Z 4/3

�2

⇣x2y

2

��23 y+8

3

y2

⌘dy =

Z 4/3

�2

hy2

⇣� 2

3y +

8

3

⌘2� y5

2

idy =

Z 4/3

�2

⇣2y3

9� 16y2

9+

32y

9� y5

2

⌘dy =

hy4

18� 16y3

27+

16y2

9� y6

12

i4/3

�2=

⇣128

729� 1024

729+

256

81� 1024

2187

⌘�⇣8

9+

128

27+

64

9� 16

3

⌘=

3200

2187� 200

27= �13000

2187.

vertical: ZZR

xy dA =

Z 16/9

0

Z px

�pxxy dydx +

Z 4

16/9

Z �32 x+4

�pxxy dydx.


Example. Find Z 3

0

Z 9

y2y sin(x2) dxdy

Zy sin(x2) dx has no elementary antiderivative.

The region is both horizotally and vertically simple. Also, x = y2 =) y =p

x.Z 3

0

Z 9

y2y sin(x2) dxdy =

Z 9

0

Z px

0y sin(x2) dydx =

Z 9

0

⇣y2

2sin(x2)

��p

x

0

⌘dx =

Z 9

0

x sin(x2)

2dx =

u = x2 du = 2xdxdu

4=

xdx

2

1

4

Z 81

0sin u du =

1

4

⇣� cos u

��81

0

⌘=

1

4

�� cos 81� (�1)

�=

1� cos 81

4.


2. Area, Volume, and Center of Mass

Consider a function f(x, y) � 0 over a region R in the x-y plane and thevolume of the region lying beneath the surface and above R.

For vertically simple, we can think:

V =

Z b

a

Z g2(x)

g1(x)f(x, y)| {z }

height

dy|{z}width

dx)|{z}length

.

For horizontally simple, we can think:

V =

Z d

c

Z h2(y)

h1(y)f(x, y)| {z }

height

dx|{z}width

dy)|{z}length

.

For any R 2 R2, volume under the surface z = 1 is

V =

ZZR

1 dA =

ZZR

dA = (1)(Area of R) = Area of R.

2. AREA, VOLUME, AND CENTER OF MASS 123

Problem (Page 931 #6). Find the area bounded by y = x3 and y = x2.

The curves intersect where x3 = x2 or x3 � x2 = x2(x� 1) = 0, i.e., at (0, 0)and (1, 1). We have a vertically (and horizontally) simple region:

A =

ZZR

dA =

Z 1

0

Z x2

x3dydx =

Z 1

0

hy��x2

x3

idx =

Z 1

0(x2 � x3) dx =

x3

3� x4

4

��10

=1

3� 1

4=

1

12.


Problem (Page 931 #18). Find the volume of the solid bounded by z =2x + y + 1, z = �2x, x = y2, and x = 1.

The last two are cylinders whose cross sections in the x-y plane are shown inthe right half of the right graph. The first two are planes that intersect where2x + y + 1 = �2x or 4x + y = �1, whose projection on the x-y plane isshown by the decreasing line on right graph. On the side of the line where ourcylinders lie, since 2x + y + 1

��(0,0)

= 1 and �2x��(0,0)

= 0, z = 2x + y + 1 liesabove z = �2x. With the region bounded by the cylinders horizontally simple,

V =

Z 1

�1

Z 1

y2

⇥(2x + y + 1)� (�2x)

⇤dxdy =

Z 1

�1

Z 1

y2(4x + y + 1) dxdy =

Z 1

�1

⇣2x2 + yx + x

��1y2

⌘dy =

Z 1

�1

⇥(2 + y + 1)� (2y4 + y3 + y2)

⇤dy =

Z 1

�1(�2y4 � y3 � y2 + y + 3) dy = �2y5

5� y4

4� y3

3+

y2

2+ 3y

��1�1

=

⇣� 2

5� 1

4� 1

3+

1

2+ 3

⌘�⇣2

5� 1

4+

1

3+

1

2� 3

⌘=

151

60�⇣� 121

60

⌘=

272

60=

68

15.

Maple. See solidvolume(13.2).mw or solidvolume(13.2).pdf.


Moments and Center of Mass

Consider a lamina (a thin, flat plate) in the shape of a region R 2 R2 whosedensity varies throughout the plate.

Our goal is to find the center of mass. But first we need to find the total mass.Assume we have a mass density function ⇢(x, y).

We partition the region, and if a subregion Ri is small enough, then ⇢(x, y) isalmost constant on Ri, whose mass is then

mi ⇡ ⇢(ui, vi)| {z }mass/unit area

�Ai|{z}area

where (ui, vi) is an arbitrary point in Ri. Summing

m ⇡nX

i=1

⇢(ui, vi)�Ai.

Then the exact mass is

m = limkPk!0

nXi=1

⇢(ui, vi)�Ai =

ZZR

⇢(x, y) dA.


First Moments

With respect to the y-axis:

My = limkPk!0

nXi=1

ui⇢(ui, vi)�Ai =

ZZR

x⇢(x, y) dA.

With respect to the x-axis:

Mx = limkPk!0

nXi=1

vi⇢(ui, vi)�Ai =

ZZR

y⇢(x, y) dA.

Center of Mass

The center of mass is then (x, y) where

x =My

mand y =

Mx

m.


Problem (Page 932 #32). Find the mass and center of mass of the laminabounded by y = x2 � 4 and y = 5, ⇢(x, y) = the square of the distance fromthe y-axis.

x2 � 4 = 5 =) x2 = 9 =) x = ±3. Also, the region is vertically simple.

m =

Z 3

�3

Z 5

x2�4x2 dydx =

Z 3

�3

⇣x2y

��5x2�4

⌘dx =

Z 3

�3

h5x2 � (x4 � 4x2)

idx =

Z 3

�3

�9x2 � x4

�dx = 3x3 � x5

5

��3�3

=⇣81� 243

5

⌘�⇣� 81 +

243

5

⌘=

324

5

My =

Z 3

�3

Z 5

x2�4x3 dydx =

Z 3

�3

⇣x3y

��5x2�4

⌘dx =

Z 3

�3

h5x3 � (x5 � 4x3)

idx =

Z 3

�3

�9x3 � x5

�dx =

9

4x4 � x6

6

��3�3

=⇣729

4� 243

2

⌘�⇣729

4� 243

2

⌘= 0


Thus x =My

m= 0.

Mx =

Z 3

�3

Z 5

x2�4x2y dydx =

Z 3

�3

⇣x2y2

2

��5x2�4

⌘dx =

Z 3

�3

h25x2

2� x2(x2 � 4)2

2

idx =

Z 3

�3

h25x2

2�⇣x6

2� 8x4

2+

16x2

2

⌘idx =

Z 3

�3

⇣� 1

2x6 + 4x4 +

9

2x2⌘i

dx = �x7

14+

4x5

5+

3x3

2

��3�3

=

⇣� 2187

14+

972

5+

81

2

⌘�⇣2187

14� 972

5� 81

2

⌘=

2754

35�⇣� 2754

35

⌘=

5508

35

Thus y =Mx

m=

5508353245

=17

7, and the center of mass is

⇣0, 17

7

⌘.

Maple. See center of mass(13.2).mw or center of mass(13.2).pdf.

3. Double Integrals in Polar Coordinates

Example. Find the volume below the surface f(x, y) = x2 + y2 and abovethe first two quadrants of the circle x2 + y2 = 1.

This is a vertically simple region.

V =

Z 1

�1

Z p1�x2

0(x2 + y2) dydx =

Z 1

�1

⇣x2y +

y3

3

��p

1�x2

0

⌘dx =

Z 1

�1

⇣x2p

1� x2 +1

3(1� x2)3/2

⌘dx = · · · ugly

3. DOUBLE INTEGRALS IN POLAR COORDINATES 129

We try polar coordinates instead.

Suppose we wish to integrate f(r, ✓), converted from f(x, y), over a region Rof the type

R =�(r, ✓)|↵ ✓ � and g1(✓) r g2(✓)

,

where 0 g1(✓) g2(✓) for all ✓ 2 [↵,�], as seen below.

We make a grid of elementary polar regions (shown above on the right) andagain use an inner partition.

Let r =1

2

�r1 + r2

�, the average radius of r1 and r2.

�A = area of outer sector� area of inner sector

=1

2r22�✓ � 1

2r21�✓ =

1

2(r2

2 � r21)�✓

=1

2(r2 + r1)(r2 � r1)�✓ = r�r�✓.


Then the volume above the ith elementary polar region and below the surfaceis

Vi ⇡ f(ri, ✓i)| {z }height

�Ai|{z}area of base

= f(ri, ✓i)ri�ri�✓i

where (ri, ✓i) is a point in Ri and ri is the average radius in Ri.

Summing and taking the limit,

V = limkPk!0

nXi=1

f(ri, ✓i)ri�ri�✓i =

Z �

↵

Z g2(✓)

g1(✓)f(r, ✓) rdrd✓.

Theorem (3.1 – Fubini). Suppose f(r, ✓) is continuous on the region

R =�(r, ✓)|↵ ✓ � and g1(✓) r g2(✓)

where 0 g1(✓) g2(✓) for all ✓ 2 [↵,�]. ThenZZ

Rf(r, ✓) dA =

Z �

↵

Z g2(✓)

g1(✓)f(r, ✓) rdrd✓.

Example (continued). Since x2 + y2 = r2, 0 ✓ ⇡, and 0 r 1,

V =

Z 1

�1

Z p1�x2

0(x2 + y2) dydx =

Z ⇡

0

Z 1

0r2 rdrd✓ =

Z ⇡

0

⇣r4

4

��10

⌘d✓ =

Z ⇡

0

1

4d✓ =

1

4✓��⇡0

=⇡

4.

3. DOUBLE INTEGRALS IN POLAR COORDINATES 131

Problem (Page 939 #6). Find the area inside r = 1 and outside r =2� 2 cos ✓.

2� 2 cos ✓ = 1 =) 2 cos ✓ = 1 =) cos ✓ =1

2=) ✓ = ±⇡

3.

A =

Z ⇡/3

�⇡/3

Z 1

2�2 cos ✓r drd✓ =

Z ⇡/3

�⇡/3

⇣r2

2

��12�2 cos ✓

⌘d✓ =

=

Z ⇡/3

�⇡/3

h1

2�⇣4� 8 cos ✓ + 4 cos2 ✓

2

⌘id✓

=1

2

Z ⇡/3

�⇡/3

⇣� 3 + 8 cos ✓ � 4 cos2 ✓

⌘d✓

=1

2

Z ⇡/3

�⇡/3

h� 3 + 8 cos ✓ � 2(1 + cos 2✓

⌘d✓ =

1

2

h� 5✓ + 8 sin ✓ � sin 2✓

i⇡/3

�⇡/3

=1

2

h⇣� 5⇡

3+ 4p

3�p

3

2

⌘�⇣5⇡

3� 4p

3 +

p3

2

⌘i= �5⇡

3+

7p

3

2


Example. Find the volume of the solid that lies under the paraboloid z =x2 + y2, above the xy-plane, and inside the cylinder x2 + y2 = 2x.

Completing the square, (x� 1)2 + y2 = 1 is the shadow of the cylinder in thexy-plane.

Changing to polar coordinates, the shadow of the cylinder is r2 = 2r cos ✓ orr = 2 cos ✓, so

R =n

(r, ✓)�� ⇡

2 ✓ ⇡

2and 0 r 2 cos ✓

o.

V =

ZZR(x2 + y2) dA =

Z ⇡/2

�⇡/2

Z 2 cos ✓

0r2 rdrd✓ =

Z ⇡/2

�⇡/2

⇣r4

4

��2 cos ✓

0

⌘d✓ =

=

Z ⇡/2

�⇡/24 cos4 ✓ d✓ = 4

Z ⇡/2

�⇡/2cos4 ✓ d✓ =|{z}

#60

4h1

4cos3 ✓ sin ✓ +

3

4

Zcos2 ✓ d✓

i⇡/2

�⇡/2

=h

cos3 ✓ sin ✓ + 3⇣1

2cos ✓ sin ✓ +

1

2

Zd✓⌘i⇡/2

�⇡/2

=h

cos3 ✓ sin ✓ +3

2cos ✓ sin ✓ +

3

2✓i⇡/2

�⇡/2

=h3⇡

4�⇣� 3⇡

4

⌘i=

3⇡

2.

Maple. See polarint(13.3).mw or polarint(13.3).pdf.

5. TRIPLE INTEGRALS 133

5. Triple Integrals

As a first step, we wish to integrate f(x, y, z) over a box

Q =�(x, y, z)|a x b, c y d, r z s

.

We partition Q into sub-boxes by slicing it with planes parallel to the xy-, xz-,and yz-planes. The volume of a sub-box Qi is then

�Vi = �xi�yi�zi.

Definition. For any function f(x, y, z) defined on the box Q, the tripleintegral of f over Q isZZZ

Qf(x, y, z) dV = lim

kPk!0

nXi=1

f(ui, vi, wi) �Vi,

provided the limit exists and is the same for every choice of evaluation points(ui, vi, wi) in Qi for i = 1, 2 . . . , n. When this happens, we say f is integrableover Q.

Theorem (5.1–Fubini). Suppose f(x, y, z) is continuous on the box

Q =�(x, y, z)|a x b, c y d, r z s

.

Then ZZZQ

f(x, y, z) dV =

Z r

s

Z d

c

Z b

af(x, y, z) dxdydz.


Note. Any of the 6 possible orderings of dx, dy, and dz may be used, withthe limits of integration changing accordingly.

Example. For

Q =�(x, y, z)|1 x 2,�1 y 1, 2 z 4

= [1, 2]⇥ [�1, 1]⇥ [2, 4]

ZZZQ(x2z � y2z) dV =

Z 4

2

Z 1

�1

Z 2

1(x2z � y2z) dxdydz =

Z 4

2

Z 1

�1

⇣1

3x3z � xy2z

��21

⌘dydz =

Z 4

2

Z 1

�1

h⇣8

3z � 2y2z

⌘�⇣1

3z � y2z

⌘idydz =

Z 4

2

Z 1

�1

⇣7

3z � y2z

⌘dydz =

Z 4

2

⇣7

3yz � 1

3y3z

��1�1

⌘dz =

Z 4

2

h⇣7

3z � 1

3z⌘�⇣� 7

3z +

1

3z⌘i

dz =

Z 4

2(4z) dz = 2z2

��42

= 32� 8 = 24

Note.

(1) If the region Q is not a box, the integral is the limit over all inner boxes.

(2) If Q =�(x, y, z)|(x, y 2 R and g1(x, y) z g2(x, y)

,

ZZZQ

f(x, y, z) dV =

ZZR

Z g2(x,y)

g1(x,y)f(x, y, z) dzdA.


Example. Find

ZZZQ(2x + 3y) dV where Q is the tetrahedron bounded

by 2x + 3y + z = 6 and the coordinate planes. The intercepts of the plane2x + 3y + z = 6 are x = 3, y = 2, and z = 6, as shown in the figure on the leftbelow.

Note that each point of the solid lies over the vertically simple triangular region

R in the xy-plane bounded by the x- and y-axes and the line y = �2

3x + 2.

ZZZQ(2x + 3y) dV =

ZZR

Z 6�2x�3y

0(2x + 3y) dzdA =

ZZR(2xz + 3yz)

��6�2x�3y

0dA =

ZZR

h2x(6� 2x� 3y) + 3y(6� 2x� 3y)

idA =ZZ

R(12x� 4x2 � 12xy + 18y � 9y2)dA =

Z 3

0

Z �23x+2

0(12x� 4x2 � 12xy + 18y � 9y2) dydx =

Z 3

0(12xy � 4x2y � 6xy2 + 9y2 � 3y3)

��23x+2

0dx =

Z 3

0(�8x2+24x+

8

3x3�8x2�8

3x2+16x2�24x+4x2+36+

8

9x3�8x2+24x�24) dx =

Z 3

0

⇣8

9x3 � 4x2 + 12

⌘dx =

2

9x4 � 4

3x3 + 12x

��30

= 18� 36 + 36 = 18.


Example. Find the volume of the solid formed by the intersection of thecylinders x2 + z2 = 1 and y2 + z2 = 1. (Think of the intersection of tunnelsrunning along the y- and x-axes.)

1) Look along x-axis (figure on left below):

The region is horizontally simple =) we have the integral for “length”Z p1�z2

�p

1�z21 dy

2) Look along y-axis (figure on right above):

The region is horizontally simple =) we have the integral for “area”Z p1�z2

�p

1�z2

Z p1�z2

�p

1�z21 dydx

3) Integrate from z = �1 to z = 1.

V =

Z 1

�1

Z p1�z2

�p

1�z2

Z p1�z2

�p

1�z21 dydxdz =

Z 1

�1

Z p1�z2

�p

1�z2

⇣y��p

1�z2

�p

1�z2

⌘dxdz =

Z 1

�1

Z p1�z2

�p

1�z22p

1� z2 dxdz =

Z 1

�1

⇣2xp

1� z2��p

1�z2

�p

1�z2

⌘dz =

Z 1

�14(1� z2) dz = 4z � 4

3z3��1�1

=⇣4� 4

3

⌘�⇣� 4 +

4

3

⌘=

16

3.


Example. Find the solid whose volume is given byZ 1

0

Z p1�z2

0

Z p1�x2�z2

�p

1�x2�z2dydxdz

and rewrite it using a di↵erent innermost variable.

From the limits of integration for y, we get y = ±p

1� x2 � z2, equationsdefining the right and left hemispheres (in our standard view) of the spherex2 + y2 + z2 = 1 of radius 1 centered at the origin.

From the limits for x, x = 0 and x =p

1� z2, we are restricted to thehemisphere with x � 0.

Finally, the limits for z, z = 0 and z = 1, restrict us to the quarter sphere withx � 0 and above the xy-plane.

A di↵erent version of the integral isZ 1

0

Z p1�x2

�p

1�x2

Z p1�x2�y2

0dzdydx.


Mass and Center of Mass

Suppose a solid Q has mass density function ⇢(x, y, z). As before,

m =

ZZZQ

⇢(x, y, z) dV

Myz =

ZZZQ

x⇢(x, y, z) dV

Mxz =

ZZZQ

y⇢(x, y, z) dV

Mxy =

ZZZQ

z⇢(x, y, z) dV

x =Myz

m, y =

Mxz

m, z =

Mxy

m

Maple. See triple integral(13.5).mw or triple integral(13.5).pdf.

6. Cylindrical Coordinates

Just as (x, y, z) can represent each point in space, so can the coordinates(r, ✓, z), with 0 r <1, 0 ✓ 2⇡, �1 < z <1.

We still have x = r cos(✓), y = r sin(✓), and r =p

x2 + y2. These arecylindrical coordinates.

6. CYLINDRICAL COORDINATES 139

Surfaces obtained by setting one coordinate equal to a constant:

(1, ✓, z) (r,⇡

4, z) (r, ✓,�1)

(2, ✓, z) (r,3⇡

4, z) (r, ✓, 3)

The volume element in cylindrical coordinates.

�V = �A�z = rdrd✓dz

Other orders of integration are also possible.


Example. Find the volume of the region R bounded by the paraboloidsz = x2 + y2 and z = 36� 3x2 � 3y2, or z = r2 and z = 36� 3r2.

Solving r2 = 36� 3r2 to get r = 3, the paraboloids intersect at z = 9 and theshadow of R on the r✓-plane is r = 3. Thus

V =

ZZZR

1 dV =

Z 2⇡

0

Z 3

0

Z 36�3r2

r2r dzdrd✓ =

Z 2⇡

0

Z 3

0

hrz��36�3r2

r2

idrd✓ =

Z 2⇡

0

Z 3

0(36r � 3r3 � r3) drd✓ =

Z 2⇡

0

Z 3

0(36r � 4r3) drd✓ =

Z 2⇡

0(18r2 � r4)

��30d✓ =

Z 2⇡

0(162� 81) d✓ = 81✓

��2⇡0

= 162⇡.

Problem (Page 962 #17). Set up

ZZZQ

f(x, y, z) dV in cylindrical coor-

dinates if Q is the region bounded by x = y2 + z2 and x = 2� y2 � z2.

Here, because of the y2 +z2, we will use cylindrical coordinates in the yz-plane,i.e., y = r cos ✓, z = r sin ✓, x = x.

Then we have x = r2 and x = 2�r2. Solving r2 = 2�r2, we get r = 1. So theparaboloids intersect at x = 1 and the shadow of Q on the yz-plane is r = 1.

ThenZZZQ

f(x, y, z) dV =

Z 2⇡

0

Z 1

0

Z 2�r2

r2f(x, r cos ✓, r sin ✓)r dxdrd✓.

6. CYLINDRICAL COORDINATES 141

Problem (Page 962 #32). Find

ZZZQ(2x� y) dV , where Q is the tetra-

hedron bounded by 3x + y + 2z = 6 and the coordinate planes.

We get no help from cylindrical coordinates here.

ZZZQ(2x� y) dV =

Z 2

0

Z 6�3x

0

Z 3�32x�1

2y

0(2x� y) dzdydx =

Z 2

0

Z 6�3x

0(2xz � yz)

��3�32x�1

2y

0dydx =

Z 2

0

Z 6�3x

0

h⇣6x� 3x2 � xy

⌘�⇣3y � 3

2xy � 1

2y2⌘i

dydx =

Z 2

0

Z 6�3x

0

⇣6x� 3x2 +

1

2xy � 3y +

1

2y2⌘

dydx =

Z 2

0

⇣12xy � 3x2y +

1

4xy2 � 3y2 +

1

6y3��6�3x

0

⌘dx =

Z 2

0

⇣27x3

4� 63x2 + 135x� 72

⌘dx =

27x4

16� 21x3 +

135x2

2� 72x

��20

=

27� 168 + 270� 144 = �15


Problem (Page 963 #44). Change to cylindrical coordiates and evaluate:Z 1

0

Z p1�x2

0

Z 4

1�x2�y2

px2 + y2 dzdydx =

Z ⇡2

0

Z 1

0

Z 4

1�r2rr dzdrd✓ =

Z ⇡2

0

Z 1

0

⇣r2z

��41�r2

⌘drd✓ =

Z ⇡2

0

Z 1

0

h4r2 � (r2 � r4)

idrd✓ =

Z ⇡2

0

Z 1

0(r4 + 3r2) drd✓ =

Z ⇡2

0

⇣r5

5+ r3

��10

⌘d✓ =

Z ⇡2

0

⇣1

5+ 1

⌘d✓ =

6

5✓��⇡

2

0=

3⇡

5.

Maple. See cylindrical(13.6).mw or cylindrical(13.6).pdf.

multiple integrals - christian brothers universityfacstaff.cbu.edu/wschrein/media/m232...

Documents