contentsazhou/teaching/18w/hw...for each y, the x-integral goes from 4 to 4, so is symmetric about...
TRANSCRIPT
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
Contents
1 Homework 1 - Solutions 3
2 Homework 2 - Solutions 13
3 Homework 3 - Solutions 19
1
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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.
Problem 4 (16.1.31). Evaluate
∫ 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)
Problem 5 (16.1.34). Evaluate
∫ 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.
Problem 6 (16.1.40). Evaluate
∫∫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.
Problem 7 (16.1.42). Evaluate
∫∫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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
Problem 8 (16.1.47). Evaluate
∫ 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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
Problem 16 (16.2.35). For
∫ 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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
HOMEWORK 2 - SOLUTIONS
Problem 1 (16.3.3). Evaluate
∫∫∫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).
Problem 2 (16.3.10). Evaluate
∫∫∫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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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
Homework Solutions MATH 32B-2 (18W)
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
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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.
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MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
HOMEWORK 3 - SOLUTIONS
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.
Problem 2 (16.4.10). For
∫ 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.
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Homework Solutions MATH 32B-2 (18W)
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
Problem 4 (16.4.23). Evaluate
∫∫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.
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MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
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).
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Homework Solutions MATH 32B-2 (18W)
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
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