contentsazhou/teaching/18w/hw...for each y, the x-integral goes from 4 to 4, so is symmetric about...

MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables

Contents

1 Homework 1 - Solutions 3



1


HOMEWORK 1 - SOLUTIONS

Problem 1 (16.1.15). Use symmetry to evaluate

∫∫Rx3 dA for R = [−4, 4]× [0, 5].

Solution. For each y, the x-integral goes from −4 to 4, so is symmetric about the x-axis, while thefunction x3 is odd, meaning that (−x)3 = −x3. This means that the negative and positive portionsof the integral cancel, so each x-integral is 0, hence the entire area integral is 0.

Problem 2 (16.1.17). Use symmetry to evaluate

∫∫R

sinx dA for R = [0, 2π]× [0, 2π].

Solution. For each y, the x-integral is the integral of sinx over a whole period, so is 0. Thus theentire area integral is 0.

Problem 3 (16.1.24). Evaluate

∫ 1

−1

∫ π

0

x2 sin y dx dy.

Solution. ∫ y=1

y=−1

∫ x=π

x=0

x2 sin y dx dy =

∫ y=1

y=−1

[x3

3sin y

]x=πx=0

dy

=

∫ y=1

y=−1

π3

3sin y dy

=π3

3[− cos y]y=1

y=−1

=π3

3(cos(−1)− cos 1)

= 0.


∫ 2

1

∫ 4

0

dy dx

x+ y.

Solution. ∫ x=2

x=1

∫ y=4

y=0

dy dx

x+ y=

∫ x=2

x=1

[ln(x+ y)]y=4y=0 dx

=

∫ x=2

x=1

(ln(x+ 4)− ln(x)) dx

= [(x+ 4) ln(x+ 4)− (x+ 4)]x=2x=1 − [x lnx− x]x=2

x=1

= [(6 ln 6− 6)− (5 ln 5− 5)]− [(2 ln 2− 2)− (1 ln 1− 1)]

= 6 ln 6− 5 ln 5− 2 ln 2.

3

Homework Solutions MATH 32B-2 (18W)


∫ 8

0

∫ 2

1

x dx dy√x2 + y

.

Solution.∫ y=8

y=0

∫ x=2

x=1

x dx dy√x2 + y

=

∫ y=8

y=0

∫ u=4+y

u=1+y

(1/2) du dy√u

(u = x2 + y)

=

∫ y=8

y=0

[√u]u=4+y

u=1+ydy =

∫ y=8

y=0

(√4 + y −

√1 + y

)dy

=

[2

3(4 + y)3/2 − 2

3(1 + y)3/2

]y=8

y=0

=2

3

[(123/2 − 93/2

)−(

43/2 − 13/2)]

=2

3

[24√

3− 34]

=−68 + 48

√3

3.


∫∫R

y

x+ 1dA for R = [0, 2]× [0, 4].

Solution. ∫∫R

y

x+ 1dA =

∫ x=2

x=0

∫ y=4

y=0

y

x+ 1dy dx (Fubini)

=

∫ x=2

x=0

[1

2

y2

x+ 1

]y=4

y=0

dx =

∫ x=2

x=0

8

x+ 1dx

= 8 [ln(x+ 1)]x=2x=0 = 8 ln 3.


∫∫Re3x+4y dA for R = [0, 1]× [1, 2].

Solution. ∫∫Re3x+4y dA =

∫ y=2

y=1

∫ x=1

x=0

e3x+4y dx dy (Fubini)

=

∫ y=2

y=1

[1

3e3x+4y

]x=1

x=0

dy =1

3

∫ y=2

y=1

e4y+3 − e4y dy

=e3 − 1

3

∫ y=2

y=1

e4y dy =e3 − 1

3

[1

4e4y]y=2

y=1

=(e3 − 1)(e8 − e4)

12.

4



∫ 1

0

∫ 1

0

y

1 + xydy dx.

Solution.∫ x=1

x=0

∫ y=1

y=0

y

1 + xydy dx =

∫ y=1

y=0

∫ x=1

x=0

y

1 + xydx dy (Fubini to change order)

=

∫ y=1

y=0

∫ u=1+y

u=1

1

udu dy =

∫ y=1

y=0

[lnu]u=1+yu=1 dy (u = 1 + xy)

=

∫ y=1

y=0

ln(1 + y) dy = [(1 + y) ln(1 + y)− (1 + y)]y=1y=0

= [2 ln 2− 2]− [1 ln 1− 1] = 2 ln 2− 1.

Problem 9 (16.1.49). Using Fubini’s theorem, argue that the solid in Figure 1 has volume AL,where A is the area of the front face of the solid.

Figure 1

Solution. The surface bounding the solid from above is the graph of a positive function z = f(y)that does not depend on x. (Here a is the largest value that y can take, which is not labeled in thediagram.) The volume of the solid is∫∫

[0,L]×[0,a]f(y) dA =

∫ y=a

y=0

∫ x=L

x=0

f(y) dx dy (Fubini)

=

∫ y=a

y=0

[xf(y)]x=Lx=0 dy =

∫ y=a

y=0

Lf(y) dy

= L

∫ y=a

y=0

f(y) dy = LA,

where in the last step, we have used the fact that A is the area under the graph of z = f(y) in the(y, z)-plane (which is equal to the area of the front face of the solid), which is then given by theintegral of f over the interval [0, a].

5


Problem 10 (16.2.10). Sketch the region D between y = x2 and y = x(1 − x). Express D as asimple region and calculate the integral of f(x, y) = 2y over D.

Solution.

y = x2

y = x(1− x)

(1/2, 1/4)

The region D is both vertically simple and horizontally simple, but the bounds for y in terms of xare simpler than the bounds for x in terms of y, so when we use Fubini’s theorem to evaluate theintegral, we take the y-integral on the inside and the x-integral on the outside. This gives us∫∫

Df(x, y) dA =

∫ x=1/2

x=0

∫ y=x(1−x)

y=x2

2y dy dx (Fubini)

=

∫ x=1/2

x=0

[y2]y=x(1−x)y=x2 dx

=

∫ x=1/2

x=0

x2(1− x)2 − x4 dx

=

∫ x=1/2

x=0

x2(1− 2x) dx

=

[1

3x3 − 1

2x4]x=1/2

x=0

=1

24− 1

32=

1

96.

6


Problem 11 (16.2.14). Integrate f(x, y) = (x + y + 1)−2 over the triangle with vertices (0, 0),(4, 0), and (0, 8).

Solution.

(4, 0)

(0, 8)

y = −2x+ 8

The bounds for y in terms of x are simpler than the bounds for x in terms of y (in particular, nofractions), so we take the y-integral on the inside and the x-integral on the outside. We have∫∫

Df(x, y) dA =

∫ x=4

x=0

∫ y=−2x+8

y=0

(x+ y + 1)−2 dy dx (Fubini)

=

∫ x=4

x=0

[−(x+ y + 1)−1

]y=−2x+8

y=0dx

=

∫ x=4

x=0

((x+ 1)−1 − (−x+ 9)−1

)dx

=

∫ x=4

x=0

((x+ 1)−1 + (x− 9)−1

)dx

= [ln|x+ 1|+ ln|x− 9|]x=4x=0

= (ln 5 + ln 5)− (ln 1 + ln 9)

= ln(25/9).

7


Problem 12 (16.2.16). Integrate f(x, y) = x over the region bounded by y = x, y = 4x− x2, andy = 0 in two ways: as a vertically simple region and as a horizontally simple region.

Solution.

y = x

y = 4x− x2

(3, 3)

(4, 0)

If we regard this region as vertically simple, so the y-integral is inside, then we have to split theintegral into two parts depending on which function bounds y from above. That is,∫∫

Df(x, y) dA =

∫ x=4

x=0

∫ y=ymax(x)

y=0

x dy dx (Fubini)

=

∫ x=3

x=0

∫ y=x

y=0

x dy dx+

∫ x=4

x=3

∫ y=4x−x2

y=0

x dy dx

=

∫ x=3

x=0

x2 dx+

∫ x=4

x=3

∫x(4x− x2) dx

= 9 +

[4

3x3 − 1

4x4]x=4

x=3

=175

12.

If we regard this region as horizontally simple instead, so the x-integral is inside, then the leftboundary is always given by x = y and the right boundary is always given by the larger value of xfor which y = 4x − x2. Using the quadratic formula for the equivalent equation x2 − 4x + y = 0,this value is x = 2 +

√4− y, so we have∫∫

Df(x, y) dA =

∫ y=3

y=0

∫ x=2+√4−y

x=y

x dx dy

=1

2

∫ y=3

y=0

(2 +√

4− y)2 − y2 dy

=175

12.

8


Problem 13 (16.2.20). Compute the double integral of f(x, y) = cos(2x + y) over the domain1/2 ≤ x ≤ π/2 and 1 ≤ y ≤ 2x.

Solution.∫ x=π/2

x=1/2

∫ y=2x

y=1

cos(2x+ y) dy dx =

∫ x=π/2

x=1/2

sin(4x)− sin(2x+ 1) dx

=

[−1

4cos(4x) +

1

2cos(2x+ 1)

]x=π/2x=1/2

=

[−1

4cos(2π) +

1

2cos(π + 1)

]−[−1

4cos 2 +

1

2cos 2

]= −1

4− 1

2cos 1− 1

4cos 2.

Problem 14 (16.2.21). Compute the double integral of f(x, y) = 2xy over the domain boundedby x = y and x = y2.

Solution.

(1, 1)

x = y

x = y2

As a horizontally simple region, for each value of y, the left boundary is given by x = y2 and theright boundary is given by x = y, so∫ y=1

y=0

∫ x=y

x=y22xy dx dy =

∫ y=1

y=0

(y3 − y5) dy =1

12.

9


Problem 15 (16.2.28). For

∫ 1

0

∫ e

exf(x, y) dy dx, sketch the domain of integration and express as

an iterated integral in the opposite order.

Solution. The domain of integration is the set of points (x, y) for which 0 ≤ x ≤ 1 and ex ≤ y ≤ e,which produces the diagram below.

x = 0 x = 1

y = e

y = ex

(0, 1)

The iterated integral expresses the integral over the domain interpreted as a vertically simple region.It is also a horizontally simple region, with 1 ≤ y ≤ e and 0 ≤ x ≤ ln y, so∫ x=1

x=0

∫ y=e

y=exf(x, y) dy dx =

∫ y=e

y=1

∫ x=ln y

x=0

f(x, y) dx dy.

10



∫ 1

0

∫ 1

y=x

xey3

dy dx, sketch the domain of integration. Then change

the order of integration and compute. Explain the simplification achieved by changing the order.

Solution. The domain of integration is 0 ≤ x ≤ 1 and x ≤ y ≤ 1.

x = 0 x = 1

y = x

y = 1

Changing the order, so the domain of integration is equivalently given by 0 ≤ y ≤ 1 and 0 ≤ x ≤ y,∫ x=1

x=0

∫ y=1

y=x

xey3

dy dx =

∫ y=1

y=0

∫ x=y

x=0

xey3

dx dy

=1

2

∫ y=1

y=0

y2ey3

dy

=1

2

∫ u=1

u=0

eu

3du (u = y3)

=e− 1

6.

The function we were required to integrate can only be exactly integrated in one direction, sochanging the order of integration to use this direction first gives us a chance to obtain somethingwhich can be (exactly) integrated a second time.

11


Problem 17 (16.2.52). Calculate the average height above the x-axis of a point in the region0 ≤ x ≤ 1 and 0 ≤ y ≤ x2.

Solution. For a point in the plane, its height above the y-axis is its y-coordinate, so the averageheight in the given region is

average height =1

area of region

∫ x=1

x=0

∫ y=x2

y=0

y dy dx.

The area of a region is found by integrating 1 over the region, so

average height =

(∫ x=1

x=0

∫ y=x2

y=0

1 dy dx

)−1(∫ x=1

x=0

∫ y=x2

y=0

y dy dx

)

=

(∫ x=1

x=0

x2 dx

)−1(∫ x=1

x=0

x4

2dx

)=

(1

3

)−1(1

10

)=

3

10.

Problem 18 (16.2.55). What is the average value of the linear function

f(x, y) = mx+ ny + p

on the ellipse (x/a)2 + (y/b)2 ≤ 1? Argue by symmetry rather than calculation.

Solution. If (x, y) is a point in the ellipse, then (−x,−y) is a point in the ellipse as well, and if wepair these two points together, they average to

f(x, y) + f(−x,−y)

2=

(mx+ ny + p) + (−mx− ny + p)

2= p.

Doing this for every pair of points, the average of f(x, y) on the ellipse as a whole is p.

Problem 19 (16.2.59). Prove the inequality

∫∫D

dA

4 + x2 + y2≤ π, where D is the disc x2+y2 ≤ 4.

Solution. Since x2, y2 ≥ 0 for all (x, y), we have

1

4 + x2 + y2≤ 1

4

for all (x, y). Thus ∫∫D

dA

4 + x2 + y2≤∫∫D

dA

4=

1

4· (area of D) =

1

4· 4π = π.

12




∫∫∫Bxey−2z dV for the box B = {0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1}.

Solution. ∫∫∫Bxey−2z dV =

∫ z=1

z=0

∫ y=1

y=0

∫ x=2

x=0

xeye−2z dx dy dz

=

∫ z=1

z=0

e−2z dz

∫ y=1

y=0

ey dy

∫ x=2

x=0

x dx

=1

2(1− e−2) · (e− 1) · 2 = (e− 1)(1− e−2).


∫∫∫Wex+y+z dV over the region

W = {0 ≤ z ≤ 1, 0 ≤ y ≤ x, 0 ≤ x ≤ 1}.

Solution.∫∫∫Wex+y+z dV =

∫ z=1

z=0

∫ x=1

x=0

∫ y=x

y=0

exeyez dy dx dz

=

∫ z=1

z=0

ez dz

∫ x=1

x=0

∫ y=x

y=0

exey dy dx = (e− 1)

∫ x=1

x=0

ex(ex − 1) dx

= (e− 1)

∫ u=e−1

u=0

u du =1

2(e− 1)3. (u = ex − 1)

Problem 3 (16.3.15). Calculate the integral of f(x, y, z) = z over the region W below the hemi-sphere of radius 3 and lying over the triangle D in the xy-plane bounded by x = 1, y = 0, andx = y.

Solution. We have D = {0 ≤ x ≤ 1 and 0 ≤ y ≤ x}, and for z to lie below the hemisphere of radius3, we must have x2 + y2 + z2 ≤ 9, so

W = {0 ≤ x ≤ 1 and 0 ≤ y ≤ x and 0 ≤ z ≤√

9− x2 − y2}.

Then ∫∫∫Wf(x, y, z) dV =

∫ x=1

x=0

∫ y=x

y=0

∫ z=√

9−x2−y2

z=0

z dz dy dx

=

∫ x=1

x=0

∫ y=x

y=0

9− x2 − y2 dy dx =

∫ x=1

x=0

9x− x3 − 1

3x3 dx

=9

2− 1

4− 1

12=

25

6.

13


Problem 4 (16.3.17). Integrate f(x, y, z) = x over the region in the first octant (x, y, z ≥ 0) abovez = y2 and below z = 8− 2x2 − y2.

Solution. The bounds on (x, y) are given by the intersection of the two surfaces,

z = y2 and z = 8− 2x2 − y2 =⇒ x2 + y2 = 4.

Hence the region of interation is

W = {0 ≤ y ≤ 2 and 0 ≤ x ≤√

4− y2 and y2 ≤ z ≤ 8− 2x2 − y2},

and ∫∫∫Wf(x, y, z) dV =

∫ y=2

y=0

∫ x=√

4−y2

x=0

∫ z=8−2x2−y2

z=y2x dz dx dy

=

∫ y=2

y=0

∫ x=√

4−y2

x=0

x(8− 2x2 − 2y2) dx dy

=

∫ y=2

y=0

∫ u=0

u=8−2y2−1

4u du dy (u = 8− 2x2 − 2y2)

=

∫ y=2

y=0

1

8(8− 2y2)2 dy

=

∫ y=2

y=0

1

2y4 − 4y2 + 8 dy

=32

10− 32

3+ 16 =

128

15.

Problem 5 (16.3.20). Find the volume of the solid in R3 bounded by y = x2, x = y2, z = x+y+5,and z = 0.

Solution. Note that the solid is symmetric about the plane x = y (the set of bounding equationsis unchanged upon swapping x and y). Therefore, it suffices to compute the volume of the solidbounded by y = x2, y = x, z = x+ y + 5, and z = 0, then double it. Hence we compute

2

∫ x=1

x=0

∫ y=x

y=x2

∫ z=x+y+5

z=0

1 dz dy dx = 2

∫ x=1

x=0

∫ y=x

y=x2

x+ y + 5 dy dx

= 2

∫ x=1

x=0

1

2(x+ x+ 5)2 − 1

2(x+ x2 + 5)2 dx

= 2

∫ x=1

x=0

−1

2x4 − x3 − 7

2x2 + 5x dx

= 2

(− 1

10− 1

4− 7

6+

5

2

)=

59

30.

14


Problem 6 (16.3.25). Describe the domain of integration of the integral∫ 2

−2

∫ √4−z2

−√4−z2

∫ √5−x2−z2

1

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

Solution. The domain is

D = {−2 ≤ z ≤ 2 and −√

4− z2 ≤ x ≤√

4− z2 and 1 ≤ y ≤√

5− x2 − z2}

= {−2 ≤ z ≤ 2 and x2 + z2 ≤ 4 and 1 ≤ y ≤√

5− x2 − z2}= {x2 + z2 ≤ 4 and y ≥ 1 and x2 + y2 + z2 ≤ 5}= {y ≥ 1 and x2 + y2 + z2 ≤ 5},

where in the last step, we removed the first inequality since it follows from the other two. This isthe smaller region between the sphere of radius

√5 centered at the origin and the plane y = 1.

Problem 7 (16.3.28). Let W be the region bounded by y + z = 2, 2x = y, x = 0, and z = 0.Express and evaluate the triple integral of f(x, y, z) = z by projecting W onto the

(a) xy-plane;

(b) yz-plane;

(c) xz-plane.

Solution. 1. The projection onto the xy-plane is the triangle {0 ≤ x ≤ 1 and 2x ≤ y ≤ 2}, so∫∫∫Wf(x, y, z) dV =

∫ x=1

x=0

∫ x=1

y=2x

∫ z=2−y

z=0

z dz dy dx

=

∫ x=1

x=0

∫ y=2

y=2x

1

2(2− y)2 dy dx

=

∫ x=1

x=0

1

6(2− 2x)3 dx

=

∫ x=1

x=0

4

3(1− x)3 dx =

1

3.

2. The projection onto the yz-plane is the triangle {0 ≤ y ≤ 2 and 0 ≤ z ≤ 2− y}, so∫∫∫Wf(x, y, z) dV =

∫ y=2

y=0

∫ z=2−y

z=0

∫ x=y/2

x=0

z dx dz dy =1

3.

3. The projection onto the xz−plane is the triangle {0 ≤ x ≤ 1 and 0 ≤ z ≤ 2− 2x}, so∫∫∫Wf(x, y, z) dV =

∫ x=1

x=0

∫ z=2−2x

z=0

∫ y=2−z

y=x/2

z dy dz dx =1

3.

15


Problem 8 (16.3.29). Let

W ={

(x, y, z) |√x2 + y2 ≤ z ≤ 1

}.

Express

∫∫∫Wf(x, y, z) dV as an iterated integral in the order dz dy dx.

Solution. We must have x2 + y2 ≤ 1, so −√

1− x2 ≤ y ≤√

1− x2. Thus∫∫∫Wf(x, y, z) dV =

∫ x=1

x=−1

∫ y=√1−x2

y=−√1−x2

∫ z=1

z=√x2+y2

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

Problem 9 (16.3.30). Repeat the previous exercise for the order dx dy dz.

Solution. Here we have 0 ≤ z ≤ 1, then −z ≤ y ≤ z, then −√z2 − y2 ≤ x ≤

√z2 − y2, so

∫∫∫Wf(x, y, z) dV =

∫ z=1

z=0

∫ y=z

y=−z

∫ x=√z2−y2

x=−√z2−y2

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

Problem 10 (16.3.35). Draw the region W bounded by the surfaces given by z = y2, y = z2,x = 0, and x+y+z = 4. Set up, but do not compute, a single triple integral that yields the volumeof W.

Solution. [drawing omitted for now]

The region isW = {0 ≤ z ≤ 1 and z2 ≤ y ≤

√z and 0 ≤ x ≤ 4− y − z},

so the volume of W is ∫∫∫WdV =

∫ z=1

z=0

∫ y=√z

y=z2

∫ x=4−y−z

x=0

dx dy dz.

Problem 11 (12.3.13). Convert the equation r = 2 sin θ to rectangular coordinates.

Solution. Multiply by r to get r2 = 2r sin θ, which gives x2 + y2 = 2y, or x2 + (y − 1)2 = 1.

Problem 12 (12.3.16). Convert the equation r = 1/(2− cos θ) to rectangular coordinates.

Solution. Clearing denominators, 2r − r cos θ = 1, or 2√x2 + y2 = 1 + r cos θ = 1 + x. Squaring,

4x2 + 4y2 = 1 + 2x+ x2, or 3x2 − 2x+ 4y2 = 1.

16


Problem 13 (12.3.18). Convert the equation x = 5 to a polar equation of the form r = f(θ).

Solution. This becomes r cos θ = 5, so r = 5 sec θ.

Problem 14 (12.3.19). Convert the equation y = x2 to a polar equation of the form r = f(θ).

Solution. This becomes r sin θ = r2 cos2 θ, so r = sin θ/ cos2 θ = sec θ tan θ.

Problem 15 (12.3.20). Convert the equation xy = 1 to a polar equation of the form r = f(θ).

Solution. This becomes r2 sin θ cos θ = 1, so r =√

sec θ csc θ.

Problem 16 (12.3.23). Match each equation with its description:

(a) r = 2

(b) θ = 2

(c) r = 2 sec θ

(d) r = 2 csc θ

(i) Vertical line

(ii) Horizontal line

(iii) Circle

(iv) Line through origin

Solution. (a) This is the circle of radius 2 around the origin, which fits (iii).

(b) This is a ray from the origin outward in the direction corresponding to θ = 2, which fits mostclosely (though not quite) with (iv).

(c) This equation becomes r cos θ = 2, or x = 2, so we have a vertical line, which fits (i).

(d) This equation becomes r sin θ = 2, or y = 2, so we have a horizontal line, which fits (ii).

Problem 17 (12.3.37). Show thatr = a cos θ + b sin θ

is the equation of a circle passing through the origin. Express the radius and center (in rectangularcoordinates) in terms of a and b, and write down the equation in rectangular coordinates.

Solution. Multiply through by r to get r2 = ar cos θ + br sin θ, or x2 + y2 = ax + by. Completingthe square gives us (

x− 1

2a

)2

+

(y − 1

2b

)2

=a2 + b2

4,

so the center is (a/2, b/2) and the radius is√a2 + b2/2.

Problem 18 (12.3.47). Show that every line that does not pass through the origin has a polarequation of the form

r =b

sin θ − a cos θ,

where b 6= 0.

Solution. Any line which does not pass through the origin has rectangular equation y = ax+ b forsome a and b 6= 0. Then r sin θ = ar cos θ + b, and solving for r, we get the required form.

17



Problem 1 (16.4.4). Sketch D = {y ≥ 0 and x2 +y2 ≤ 1} and integrate f(x, y) = y(x2 +y2)3 overD using polar coordinates.

Solution.

From the diagram, we can see that D = {0 ≤ θ ≤ π and 0 ≤ r ≤ 1}, so∫∫Df(x, y) dA =

∫ θ=π

θ=0

∫ r=1

r=0

r sin θ(r2)3 · r dr dθ =

∫ θ=π

θ=0

sin θ dθ

∫ r=1

r=0

r8 dr = 2 · 1

9=

2

9.


∫ x=4

x=0

∫ y=√16−x2

y=0

tan−1(yx

)dy dx, sketch the region of integration and

evaluate by changing to polar coordinates.

Solution.

In this range, tan−1(y/x) = θ, so∫ x=4

x=0

∫ y=√16−x2

y=0

tan−1(yx

)dy dx =

∫ θ=π/2

θ=0

∫ r=1

r=0

θ · r dr dθ

=

∫ θ=π/2

θ=0

θ dθ

∫ r=1

r=0

r dr

=(π/2)2

2· 1

2=π2

16.

19


Problem 3 (16.4.21). Find the volume of the wedge-shaped region contained in the cylinderx2 + y2 = 9, bounded above by the plane z = x and below by the xy-plane.

Solution. In cylindrical polar coordinates, the region is given by

0 ≤ r ≤ 3, −π/2 ≤ θ ≤ π/2, 0 ≤ z ≤ r cos θ,

so the volume is ∫ θ=π/2

θ=−π/2

∫ r=3

r=0

∫ z=r cos θ

z=0

r dz dr dθ =

∫ θ=π/2

θ=−π/2

∫ r=3

r=0

r2 cos θ dr dθ

=

∫ θ=π/2

θ=−π/29 cos θ dθ = 18.

Figure 1


∫∫D

√x2 + y2 dA, where D is the domain above.

Hint: Find the equation of the inner circle in polar coordinates and treat the right and left parts ofthe region separately.

Problem 5 (16.4.30). Use cylindrical coordinates to compute

∫∫∫Wz√x2 + y2 dV for the region

given by x2 + y2 ≤ z ≤ 8− (x2 + y2).

Problem 6 (16.4.35). Express

∫ 1

−1

∫ y=√1−x2

y=0

∫ x2+y2

z=0

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

Problem 7 (16.4.40). Use cylindrical coordinates to find the volume of a sphere of radius a fromwhich a central cylinder of radius b has been removed, where 0 < b < a.

Problem 8 (16.4.49). Use spherical coordinates to calculate the triple integral of the function

f(x, y, z) =√x2 + y2 + z2 over the region x2 + y2 + z2 ≤ 2z.

Problem 9 (16.4.52). Find the volume of the region lying above the cone φ = φ0 and below thesphere ρ = R.

20


Problem 10 (16.4.56). Let W be the region within the cylinder x2 + y2 = 2 between z = 0 and

the cone z =√x2 + y2. Calculate the integral of f(x, y, z) = x2 + y2 over W, using both spherical

and cylindrical coordinates.

Problem 11 (16.5.8). Compute the total mass of the plate in Figure 2 assuming a mass densityof f(x, y) = x2/(x2 + y2) g · cm−2.

Figure 2: 16.5.8

Problem 12 (16.5.13). Find the centroid of the quarter circle x2 +y2 ≤ R2 with x, y ≥ 0 assumingthe density δ(x, y) = 1.

Problem 13 (16.5.16). Show that the centroid of the sector in Figure 3a has y-coordinate

y =

(2R

3

)(sinα

α

).

(a) 16.5.16 (b) 16.5.20

Figure 3

Problem 14 (16.5.20). Show that the z-coordinate of the centroid of the tetrahedron bounded bythe coordinate planes and the plane (x/a) + (y/b) + (z/c) = 1 in Figure 3b is z = c/4. Concludeby symmetry that the centroid is (a/4, b/4, c/4).

21


Problem 15 (16.5.21). Find the centroid of the region W lying above the sphere x2 + y2 + z2 = 6and below the paraboloid z = 4− x2 − y2 (Figure 4).

Figure 4: 16.5.21

Problem 16 (16.5.24). Find the center of mass of the region bounded by y2 = x + 4 and x = 0with mass density δ(x, y) = |y|.

Problem 17 (16.5.27). Find the z-coordinate of the center of mass of the first octant of the unitsphere with mass density δ(x, y, z) = y (Figure 5).

Figure 5: 16.5.27

22

contentsazhou/teaching/18w/hw...for each y, the x-integral goes from 4 to 4, so is symmetric about...

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