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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
866 SECTION 17.1
CHAPTER 17
SECTION 17.1
1.3∑
i=1
3∑j=1
2i−13j+1 =
(3∑
i=1
2i−1
)⎛⎝ 3∑j=1
3j+1
⎞⎠ = (1 + 2 + 4)(9 + 27 + 81) = 819
2. 2 + 22 + 3 + 32 + 4 + 42 + 5 + 52 = 68
3.4∑
i=1
3∑j=1
(i2 + 3i)(j − 2) =
[4∑
i=1
(i2 + 3i)
]⎡⎣ 3∑j=1
(j − 2)
⎤⎦ = (4 + 10 + 18 + 28)(−1 + 0 + 1) = 0
4.22
+23
+24
+25
+26
+27
+42
+43
+44
+45
+46
+47
+62
+63
+64
+65
+66
+67
= 19435
.
5.m∑i=1
Δxi = Δx1 + Δx2 + · · · + Δxm = (x1 − x0) + (x2 − x1) + · · · + (xm − xm−1)
= xm − x0 = a2 − a1
6. (y1 − y0) + (y2 − y1) + · · · + (yn − yn−1) = yn − y0 = b2 − b1
7.m∑i=1
n∑j=1
Δxi Δyj =
(m∑i=1
Δxi
)⎛⎝ n∑j=1
Δyj
⎞⎠ = (a2 − a1) (b2 − b1)
8.n∑
j=1
q∑k=1
Δyj Δzk =
⎛⎝ n∑
j=1
Δyj
⎞⎠( q∑
k=1
Δzk
)= (b2 − b1) (c2 − c1)
9.m∑i=1
(xi + xi−1)Δxi =m∑i=1
(xi + xi−1)(xi − xi−1) =m∑i=1
(xi2 − x2
i−1)
= xm2 − x0
2 = a22 − a1
2
10.n∑
j=1
12(yj2 + yjyj−1 + yj−1
2)Δyj =12
n∑j=1
(yj3 − yj−13) =
12(b23 − b1
3)
11.m∑i=1
n∑j=1
(xi + xi−1)ΔxiΔyj =
(m∑i=1
(xi + xi−1)Δxi
)( n∑j=1
Δyj
)
(Exercise 9)∧
=(a2
2 − a12)(b2 − b1)
12.m∑i=1
n∑j=1
(yi + yj−1)ΔxiΔyj =
(m∑i=1
Δxi
)⎡⎣ n∑j=1
(yj2 − yj−12)
⎤⎦ = (a2 − a1)(b22 − b1
2)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.2 867
13.m∑i=1
n∑j=1
(2Δxi − 3Δyj) = 2
(m∑i=1
Δxi
)( n∑j=1
1)− 3
(m∑i=1
1
)( n∑j=1
Δyj
)
= 2n(a2 − a1) − 3m(b2 − b1)
14.m∑i=1
n∑j=1
(3Δxi − 2Δyj) = 3m∑i=1
n∑j=1
Δxi − 2m∑i=1
n∑j=1
Δyj = 3n(a2 − a1) − 2m(b2 − b1).
15.m∑i=1
n∑j=1
q∑k=1
Δxi Δyj Δzk =
(m∑i=1
Δxi
)⎛⎝ n∑j=1
Δyj
⎞⎠( q∑
k=1
Δzk
)
= (a2 − a1)(b2 − b1)(c2 − c1)
16.m∑i=1
n∑j=1
q∑k=1
(xi + xi−1)ΔxiΔyjΔzk =
[m∑i=1
(xi2 − xi−1
2)
]⎛⎝ n∑j=1
Δyj
⎞⎠( q∑
k=1
Δzk
)
= (a22 − a1
2)(b2 − b1)(c2 − c1)
17.n∑
i=1
n∑j=1
n∑k=1
δijkaijk = a111 + a222 + · · · + annn =n∑
p=1
appp
18. Start withm∑i=1
n∑j=1
aij . Take all the aij (there are only a finite number of them) and order them
in any order you chose. Call the first one b1, the second b2, and so on. Thenm∑i=1
n∑j=1
aij =r∑
p=1
bp where r = m× n.
SECTION 17.2
1. Lf (P ) = 2 14 , Uf (P ) = 5 3
4 2. Lf (P ) = 3, Uf (P ) = 5
3. (a) Lf (P ) =m∑i=1
n∑j=1
(xi−1 + 2yj−1) Δxi Δ yj , Uf (P ) =m∑i=1
n∑j=1
(xi + 2yj) Δxi Δyj
(b) Lf (P ) ≤m∑i=1
n∑j=1
[xi−1 + xi
2+ 2(yj−1 + yj
2
)]Δxi Δyj ≤ Uf (P ).
The middle expression can be writtenm∑i=1
n∑j=1
12(xi
2 − x2i−1
)Δyj +
m∑i=1
n∑j=1
(yj
2 − y2j−1
)Δxi.
The first double sum reduces tom∑i=1
n∑j=1
12(xi
2 − x2i−1
)Δyj =
12
(m∑i=1
(xi
2 − x2i−1
))( n∑j=1
Δyj
)=
12
(4 − 0) (1 − 0) = 2.
In like manner the second double sum also reduces to 2. Thus, I = 4; the volume of the prism
bounded above by the plane z = x + 2y and below by R.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
868 SECTION 17.2
4. Lf (P ) = −7/16, Uf (P ) = 7/16 5. Lf (P ) = −7/24, Uf (P ) = 7/24
6. (a) Lf (p) =m∑i=1
n∑j=1
(xi−1 − yj)ΔxiΔyj , Uf (P ) =m∑i=1
n∑j=1
(xi − yj−1)ΔxiΔyj
(b) Lf (P ) ≤m∑i=1
n∑j=1
(xi + xi−1
2− yj + yj−1
2
)ΔxiΔyj ≤ Uf (P )
The middle expression can be writtenm∑i=1
n∑j=1
12(xi
2 − xi−12)Δyj −
m∑i=1
n∑j=1
12(yj2 − yj−1
2)Δxi.
The first sum reduces to
12
(m∑i=1
(xi2 − xi−1
2)
)⎛⎝ n∑j=1
Δyj
⎞⎠ =
12(1 − 0)(1 − 0) =
12.
In like manner the second sum also reduces to 12 . Thus I = 1
2 − 12 = 0.
7. (a) Lf (P ) =m∑i=1
n∑j=1
(4xi−1 yj−1) Δxi Δyj , Uf (P ) =m∑i=1
n∑j=1
(4xi yj) Δxi Δyj
(b) Lf (P ) ≤m∑i=1
n∑j=1
(xi + xi−1) (yj + yj−1) Δx1 Δyj ≤ Uf (P ).
The middle expression can be writtenm∑i=1
n∑j=1
(xi
2 − x2i−1
) (yj
2 − y2j−1
)=
(m∑i=1
xi2 − x2
i−1
)( n∑j=1
yj2 − y2
j−1
)
by (17.1.5)∧
=(b2 − 02
) (d2 − 02
)= b2d2.
It follows that I = b2d2.
8. (a) Lf (P ) =m∑i=1
n∑j=1
3(xi−12 + yj−1
2)ΔxiΔyj , Uf (P ) =m∑i=1
n∑j=1
3(xi2 + yj
2)ΔxiΔyj
(b) Lf (P ) ≤m∑i=1
n∑j=1
[(xi
2 + xixi−1 + xi−12) + (yj2 + yjyj−1 + yj−1
2)]ΔxiΔyj ≤ Uf (P )
Since in general (A2 + AB + B2)(A−B) = A3 −B3, the middle expression can be written
m∑i=1
n∑j=1
(xi3 − xi−1
3)Δyj +m∑i=1
n∑j=1
(yj3 − yj−13)Δxi,
which reduces to(m∑i=1
xi3 − xi−1
3
)⎛⎝ n∑j=1
Δyj
⎞⎠+
(m∑i=1
Δxi
)⎛⎝ n∑j=1
yj3 − yj−1
3
⎞⎠ .
This can be evaluated as b3d + bd3 = bd(b2 + d2). It follows that I = bd(b2 + d2).
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.2 869
9. (a) Lf (P ) =m∑i=1
n∑j=1
3(x2i−1 − yj
2)
Δxi Δyj , Uf (P ) =m∑i=1
n∑j=1
3(xi
2 − y2j−1
)Δxi Δyj
(b) Lf (P ) ≤m∑i=1
n∑j=1
[ (xi
2 + xixi−1 + x2i−1
)−(yj
2 + yjyj−1 + y2j−1
) ]Δxi Δyj ≤ Uf (P ).
Since in general(A2 + AB + B2
)(A−B) = A3 −B3, the middle expression can be written
m∑i=1
n∑j=1
(xi
3 − x3i−1
)Δyj −
m∑i=1
n∑j=1
(yj
3 − y3j−1
)Δxi,
which reduces to(m∑i=1
xi3 − x3
i−1
)⎛⎝ n∑j=1
Δyj
⎞⎠−
(m∑i=1
Δxi
)⎛⎝ n∑j=1
yj3 − y3
j−1
⎞⎠ .
This can be evaluated as b3d− bd3 = bd(b2 − d2
). It follows that I = bd (b2 − d2).
10. On each subrectangle, the minimum and the maximum of f are equal, so f is constant on each
subrectangle and therefore (since f is continuous) on the entire rectangle R. Then∫∫R
f(x, y) dxdy = f(a, c)(b− a)(d− c).
11.∫∫Ω
dxdy =∫ b
a
φ(x) dx
12. Suppose that there is a point (x0, y0) on the boundary of Ω at which f is not zero. As (x, y)
tends to (x0, y0) through that part of R which is outside Ω, f(x, y), being zero, tends to zero.
Since f(x0, y0) is not zero, f(x, y) does not tend to f(x0, y0). Thus the extended function f can not
be continuous at (x0, y0).
13. Suppose f(x0, y0) �= 0. Assume f(x0, y0) > 0. Since f is continuous, there exists a disc Ωε with radius
ε centered at (x0, y0) such that f(x, y) > 0 on Ωε. Let R be a rectangle contained in Ωε. Then∫∫R
f(x, y) dxdy > 0, which contradicts the hypothesis.
14.∫∫R
(x + 2y) dx dy = 4; area(R) = (2)(1) = 2, so average value =42
= 2
15. By Exercise 7, Section 17.2,∫∫R
4xy dxdy = 2232 = 36. Thus
favg =1
area (R)
∫∫R
4xy dxdy =16
(36) = 6
16.∫∫R
(x2 + y2) dxdy =bd(b2 + d2)
3; area(R) = bd, so average value =
b2 + d2
3
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
870 SECTION 17.3
17. By Theorem 16.2.10, there exists a point (x1, y1) ∈ Dr such that∫∫Dr
f(x, y) dxdy = f(x1, y1)∫∫R
dxdy = f(x1, y1)πr2 =⇒ f(x1, y1) =1
πr2
∫∫Dr
f(x, y) dxdy
As r → 0, (x1, y1) → (x0, y0) and f(x1, y1) → f(x0, y0) since f is continuous.
The result follows.
18. 0 ≤ sin(x + y) ≤ 1 for all (x, y) ∈ R. Thus, 0 ≤∫∫R
sin(x + y) dxdy ≤∫∫R
dxdy = 1
19. z =√
4 − x2 − y2 on Ω : x2 + y2 ≤ 4, x ≥ 0, y ≥ 0;∫∫Ω
√4 − x2 − y2 dxdy is the volume V of one
quarter of a hemisphere; V = 43π.
20. 8 − 4√x2 + y2 on Ω is a cone with height h = 8 and radius r = 2; V =
32π3
21. z = 6 − 2x− 3y ⇒ x
3+
y
2+
z
6= 1; the solid is the tetrahedron bounded by the coordinate planes
and the plane:x
3+
y
2+
z
6= 1; V = 1
6 (3)(2)(6) = 6
22. (a) Lf (P ) ∼= 35.4603; Uf (P ) ∼= 36.5403 (c)∫∫R
(3y2 − 2x
)dxdy = 36
SECTION 17.3
1.∫ 1
0
∫ 3
0
x2 dy dx =∫ 1
0
3x2 dx = 1
2.∫ 3
0
∫ 1
0
ex+y dx dy =∫ 3
0
(e1+y − ey) dy =[e1+y − ey
]30
= e4 − e3 − e + 1
3.∫ 1
0
∫ 3
0
xy2 dy dx =∫ 1
0
x
[13y3
]30
dx =∫ 1
0
9x dx =92
4.∫ 1
0
∫ x
0
x3y dy dx =∫ 1
0
x3x2
2dx =
112
5.∫ 1
0
∫ x
0
xy3 dy dx =∫ 1
0
x
[14y4
]x0
dx =∫ 1
0
14x5 dx =
124
6.∫ 1
0
∫ x
0
x2y2 dy dx =∫ 1
0
x2x3
3dx =
118
7.∫ π/2
0
∫ π/2
0
sin (x + y) dy dx =∫ π/2
0
[− cos (x + y)]π/20 dx =∫ π/2
0
[cosx− cos
(x +
π
2
)]dx = 2
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.3 871
8.∫ π/2
0
∫ π/2
0
cos(x + y) dx dy =∫ π/2
0
[sin(π
2+ y)− sin y
]dy =
[cos y − cos
(π2
+ y)]π/2
0= 0
9.∫ π/2
0
∫ π/2
0
(1 + xy) dy dx =∫ π/2
0
[y +
12xy2
]π/20
dx =∫ π/2
0
(12π +
18π2x
)dx =
14π2 +
164
π4
10.∫ 1
−1
∫ √1−y2
−√
1−y2(x + 3y3) dx dy =
∫ 1
−1
6y3√
1 − y2 dy = 0 (integrand is odd)
11.∫ 1
0
∫ y
y2
√xy dx dy =
∫ 1
0
√y
[23x3/2
]yy2
dy =∫ 1
0
23
(y2 − y7/2
)dy =
227
12.∫ 1
0
∫ y2
0
yex dx dy =∫ 1
0
y(ey2 − 1) dy =
[ey
2
2− y2
2
]10
=12(e− 2)
13.∫ 2
−2
∫ 4− 12y
2
12y
2
(4 − y2
)dx dy =
∫ 2
−2
(4 − y2
) [(4 − 1
2y2
)−(
12y2
)]dy
= 2∫ 2
0
(16 − 8y2 + y4
)dy =
51215
14. I =∫ 1
0
∫ x2
x3(x4 + y2) dy dx =
∫ 1
0
[x4y +
y3
3
]x2
x3
dx =∫ 1
0
(4x6
3− x7 − x9
3
)dx
=[4x7
21− x8
8− x10
30
]10
=9
280
15. 0 by symmetry (integrand odd in y, Ω symmetric about x-axis)
16.∫ 1
0
∫ 2y
0
e−y2/2 dx dy =∫ 1
0
2ye−y2/2 dy =[−2e−y2/2
]10
= 2(
1 − 1√e
)
17.∫ 2
0
∫ x/2
0
ex2dy dx =
∫ 2
0
12xex
2dx =
[14ex
2]20
=14(e4 − 1
)
18.∫ 0
−1
∫ x4
x3(x + y) dy dx +
∫ 1
0
∫ x3
x4(x + y) dydx =
∫ 0
−1
[xy +
y2
2
]x4
x3
dx +∫ 1
0
[xy +
y2
2
]x3
x4
dx
=∫ 0
−1
(x5 +
x8
2− x4 − x6
2
)dx +
∫ 1
0
(x4 +
x6
2− x5 − x8
2
)dx
=[x6
6+
x9
18− x5
5− x7
14
]0−1
+[x5
5+
x7
14− x6
6− x9
18
]10
= −13
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
872 SECTION 17.3
19.
∫ 1
0
∫ y1/4
y1/2f(x, y) dx dy
20.
∫ 1
0
∫ 1
√x
f(x, y) dydx
21.
∫ 0
−1
∫ 1
−x
f(x, y) dy dx +∫ 1
0
∫ 1
x
f(x, y) dy dx
22.
∫ 1/2
3√
1/2
∫ y1/3
1/2
f(x, y) dxdy +∫ 1
1/2
∫ y1/3
y
f(x, y) dxdy
23.
∫ 2
1
∫ y
1
f(x, y) dx dy +∫ 4
2
∫ y
y/2
f(x, y) dx dy
+∫ 8
4
∫ 4
y/2
f(x, y) dx dy
24.
∫ −1
−3
∫ 3
−y
f(x, y) dx dy +∫ 1
−1
∫ 3
1
f(x, y) dx dy
+∫ 9
1
∫ 3
√y
f(x, y) dx dy
25.∫ 4
−2
∫ 12x+2
1/4x2dy dx =
∫ 4
−2
[12x + 2 − 1
4x2
]dx = 9
26.∫ 3
0
∫ 4y−y2
y
dx dy =∫ 3
0
(3y − y2) dy =92
27.∫ 1/4
0
∫ y
2y3/2dx dy =
∫ 1/4
0
[y − 2y3/2
]dy =
1160
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.3 873
28.∫ 3
2
∫ 5−x
6/x
dy dx =∫ 3
2
(5 − x− 6/x) dx =52
+ 6 ln23
29. ∫ 1
0
∫ y2
0
sin(y3 + 1
2
)dx dy =
∫ 1
0
y2 sin(y3 + 1
2
)dy
=[−2
3cos(y3 + 1
2
)]10
=23
(cos
12− cos 1
)
30.∫ 1
−1
∫ 0
x2−1
x2 dy dx =∫ 1
−1
x2(1 − x2) dx
=415
31. ∫ ln 2
0
∫ 2
exe−x dy dx =
∫ ln 2
0
e−x (2 − ex) dx
=[−2e−x − x
]ln 2
0= 1 − ln 2
32. ∫ 1
0
∫ √y
0
x3√x4 + y2
dx dy =∫ 1
0
12(√
2 − 1)y dy
=14(√
2 − 1)
33.∫ 2
1
∫ 2/y
y−1
dx dy =∫ 2
1
[2y− (y − 1)
]dy = ln 4 − 1
2
34.∫ 1
0
∫ 1−x
0
(x + y) dy dx =∫ 1
0
[x(1 − x) +
(1 − x)2
2
]dx =
13
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
874 SECTION 17.3
35.∫ 2
0
∫ 3− 32x
0
(4 − 2x− 4
3y
)dy dx =
∫ 3
0
∫ 2− 23y
0
(4 − 2x− 4
3y
)dx dy = 4
36.∫ 1
0
∫ 1
0
(2x + 3y) dy dx =∫ 1
0
(2x +32) dx =
52
37.∫ 2
0
∫ 1− 12x
0
x3y dy dx =∫ 2
0
∫ 2−2y
0
x3y dx dy =215
38.∫ 1
−1
∫ √1−x2
−√
1−x2(x2 + y2) dy dx = 2
∫ 1
−1
[x2√
1 − x2 +13(1 − x2)3/2
]dx =
π
2
39.∫ 2
0
∫ √2x−x2
−√
2x−x2(2x + 1) dy dx =
∫ 1
−1
∫ 1+√
1−y2
1−√
1−y2(2x + 1) dx dy
=∫ 1
−1
[x2 + x
]1+√1−y2
1−√
1−y2
dy
= 6∫ 1
−1
√1 − y2 dy = 6
(π2
)= 3π
40.∫ 1
−1
∫ 1+√
1−x2
1−√
1−x2(4 − y2 − 1
4x2) dy dx =
∫ 1
−1
[6√
1 − x2 − 12x2√
1 − x2 − 23(1 − x2)3/2
]dx =
4316
π.
41.∫ 1
0
∫ 1−x
0
(x2 + y2) dy dx =∫ 1
0
(2x2 − 4
3x3 − x +
13
)dx =
16
42.∫ 1
−1
∫ √1−x2
−√
1−x2(1 − x) dy dx =
∫ 1
−1
(2√
1 − x2 − 2x√
1 − x2) dx = π
43.∫ 1
0
∫ x
x2(x2 + 3y2) dy dx =
∫ 1
0
(2x3 − x4 − x6
)dx =
1170
44.∫ 2
1
∫ 5−y
2y−1
(1 + xy) dx dy =∫ 2
1
(6 + 9y − 3y2 − 32y3) dy =
558
45.∫ a
0
∫ √a2−x2
0
√a2 − x2 dy dx =
∫ a
0
(a2 − x2) dx =23a3
46.∫ a
0
∫ b(1−x/a)
0
c(1 − x
a− y
b
)dy dx =
∫ a
0
bc
2
(1 − x
a
)2dx =
abc
6
47.∫ 1
0
∫ 1
y
ey/xdx dy =∫ 1
0
∫ x
0
ey/xdy dx =∫ 1
0
[xey/x
]x0dx =
∫ 1
0
x(e− 1) dx =12
(e− 1)
48.∫ 1
0
∫ cos−1 y
0
esinx dx dy =∫ π/2
0
∫ cosx
0
esinx dy dx =∫ π/2
0
cosxesinx dx = e− 1
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.3 875
49.∫ 1
0
∫ 1
x
x2ey4dy dx =
∫ 1
0
∫ y
0
x2ey4dx dy =
∫ 1
0
[13x3ey
4]y0
dy =13
∫ 1
0
y3ey4dy =
112
(e− 1)
50.∫ 1
0
∫ y
0
ey2dx dy =
∫ 1
0
yey2dy =
12(e− 1)
51. favg =18
∫ 1
−1
∫ 4
0
x2y dy dx =18
∫ 1
−1
8x2 dx =∫ 1
−1
x2 dx =23
52. area of Ω = 14π (quarter circle of radius 1)
Average value =4π
∫ 1
0
∫ √1−x2
0
xy dy dx =2π
∫ 1
0
x(1 − x2) dx =12π
53. favg =1
(ln 2)2
∫ 2 ln 2
ln 2
∫ 2 ln 2
ln 2
1xy
dy dx =1
(ln 2)2
∫ 2 ln 2
ln 2
1x
ln 2 dx = 1
54. area of Ω = 1 (parallelogram)
Average value =∫ 1
0
∫ x+1
x−1
ex+y dy dx =∫ 1
0
(e2x+1 − e2x−1
)dx =
12(e3 − 2e + e−1).
55.∫∫R
f(x)g(y) dxdy =∫ d
c
∫ b
a
f(x)g(y) dx dy =∫ d
c
(∫ b
a
f(x)g(y) dx)
dy
=∫ d
c
g(y)(∫ b
a
f(x) dx)
dy =(∫ b
a
f(x) dx) (∫ d
c
g(y) dy)
56. Symmetry about the origin [ we want Ω to contain (−x,−y) whenever it contains (x, y)].
57. Note that Ω = { (x, y) : 0 ≤ x ≤ y, 0 ≤ y ≤ 1}.Set Ω′ = { (x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}.∫∫
Ω
f(x)f(y) dxdy =∫ 1
0
∫ y
0
f(x)f(y) dx dy
=∫ 1
0
∫ x
0
f(y)f(x) dy dx
x and y are dummy variables
=∫ 1
0
∫ x
0
f(x)f(y) dy dx =∫∫Ω′
f(x)f(y) dxdy.
Note that Ω and Ω′ don’t overlap and their union is the unit square
R = { (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
876 SECTION 17.3
If∫ 1
0
f(x) dx = 0, then
0 =(∫ 1
0
f(x) dx) (∫ 1
0
f(y) dy)
=∫∫R
f(x)f(y) dxdy
by Exercise 55
=∫∫Ω
f(x)f(y) dxdy +∫∫Ω′
f(x)f(y) dxdy
= 2∫∫Ω
f(x)f(y) dxdy
and therefore∫∫Ω
f(x)f(y) dxdy = 0.
58.∫ x
0
∫ b
a
∂f
∂x(t, y) dy dt =
∫ b
a
∫ x
0
∂f
∂x(t, y) dt dy
=∫ b
a
[f(x, y) − f(0, y)] dy
=∫ b
a
f(x, y) dy −∫ b
a
f(0, y) dy.
Thus ∫ b
a
f(x, y) dy =∫ x
0
∫ b
a
∂f
∂x(t, y) dy dt +
∫ b
a
f(0, y) dy
and
d
dx
[∫ b
a
f(x, y) dy]
=d
dx
[∫ x
0
∫ b
a
∂f
∂x(t, y) dy dt
]+
d
dx
[∫ b
a
f(0, y) dy]
=d
dx
[∫ x
0
H(t) dt]
+ 0 = H(x) =∫ b
a
∂f
∂x(x, y) dy.
59. Let M be the maximum value of | f(x, y) | on Ω.∫ φ2(x+h)
φ1(x+h)
=∫ φ1(x)
φ1(x+h)
+∫ φ2(x)
φ1(x)
+∫ φ2(x+h)
φ2(x)
|F (x + h) − F (x) | =∣∣∣∣∫ φ2(x+h)
φ1(x+h)
f(x, y) dy −∫ φ2(x)
φ1(x)
f(x, y) dy∣∣∣∣
=∣∣∣∣∫ φ1(x)
φ1(x+h)
f(x, y) dy +∫ φ2(x+h)
φ2(x)
f(x, y) dy∣∣∣∣
≤∣∣∣∣∫ φ1(x)
φ1(x+h)
f(x, y) dy∣∣∣∣ +∣∣∣∣∫ φ2(x+h)
φ2(x)
f(x, y) dy∣∣∣∣
≤∣∣φ1(x) − φ1 (x + h)
∣∣M + |φ2 (x + h) − φ2(x)∣∣M.
The expression on the right tends to 0 as h tends to 0 since φ1 and φ2 are continuous.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.4 877
60. (a)∫ 3
−1
∫ 5
2
xe−xy dy dx ∼= −25.9893 (b)∫ 7
3
∫ 4
1
xy
x2 + y2dy dx ∼= 4.5720
61. (a)∫ 2
1
∫ 1+√x−1
x2−2x+2
1 dy dx = 13 (b)
∫ 2
1
∫ 1+√y−1
y2−2y+2
1 dx dy = 13
62. (a)∫ 1
0
∫ 3−y
2y
√2x + y dx dy ∼= 2.8133 (b)
∫ 2
0
∫ x/2
0
√2x + y dy dx+
∫ 3
2
∫ 3−x
0
√2x + y dy dx
∼= 2.8133
SECTION 17.4
1.∫ π/2
0
∫ sin θ
0
r cos θ dr dθ =∫ π/2
0
12
sin2 θ cos θ dθ =[16
sin3 θ
]π/20
=16
2.∫ π/4
0
∫ cos 2θ
0
r dr dθ =∫ π/4
0
cos2 2θ2
dθ =π
16
3.∫ π/2
0
∫ 3 sin θ
0
r2 dr dθ =∫ π/2
0
9 sin3 θ dθ = 9∫ π/2
0
(1 − cos2 θ) sin θ dθ = 9[− cos θ +
13
cos3 θ]π/20
= 6
4.∫ 2π/3
−π/3
∫ 2 cos θ
0
r sin θ dr dθ =∫ 2π/3
−π/3
2 cos2 θ sin θ dθ =16
5. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1∫∫Γ
(cos r2
)r drdθ =
∫ 2π
0
∫ 1
0
(cos r2
)r dr dθ = 2π
∫ 1
0
r cos r2 dr = π sin 1
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2∫∫Γ
(cos r2
)r drdθ =
∫ 2π
0
∫ 2
1
(cos r2
)r dr dθ = 2π
∫ 2
1
r cos r2 dr = π(sin 4 − sin 1)
6. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1.∫∫Γ
(sin r)r dr dθ =∫ 2π
0
∫ 1
0
(sin r)r dr dθ = 2π∫ 1
0
r sin r dr = 2π(sin 1 − cos 1)
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2.∫∫Γ
(sin r)r dr dθ =∫ 2π
0
∫ 2
1
(sin r)r dr dθ = 2π∫ 2
1
r sin r dr = 2π[cos 1 − 2 cos 2 + sin 2 − sin 1]
7. (a) Γ : 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 1∫∫Γ
(r cos θ + r sin θ)r drdθ =∫ π/2
0
∫ 1
0
r2(cos θ + sin θ) dr dθ
=( ∫ π/2
0
(cos θ + sin θ) dθ)(∫ 1
0
r2 dr
)= 2(
13
)=
23
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
878 SECTION 17.4
(b) Γ : 0 ≤ θ ≤ π/2, 1 ≤ r ≤ 2∫∫Γ
(r cos θ + r sin θ)r drdθ =∫ π/2
0
∫ 2
1
r2(cos θ + sin θ) dr dθ
=(∫ π/2
0
(cos θ + sin θ) dθ)(∫ 2
1
r2 dr
)= 2(
73
)=
143
8. Γ : 0 ≤ θ ≤ π3 , 0 ≤ r ≤ 1
cos θ∫∫Γ
√x2 + y2 dx dy =
∫ π/3
0
∫ 1/ cos θ
0
r · r dr dθ =∫ π/3
0
dθ
3 cos3 θ=
13
∫ π/3
0
sec3 θ dθ
=13
[12
sec θ tan θ +12
ln | sec θ + tan θ|]π/30
=13
√3 +
16
ln(2 +√
3)
9.∫ π/2
−π/2
∫ 1
0
r2 dr dθ =13π 10.
∫ π/2
0
∫ 2
0
r2 dr dθ =π
2
∫ 2
0
r2 dr =43π
11.∫ 1
1/2
∫ √1−x2
0
dy dx =∫ π/3
0
∫ 1
12 sec θ
r dr dθ =∫ π/3
0
(12− 1
8sec2 θ
)dθ =
16π −
√3
8
12.∫ π/3
0
∫ 1/2 sec θ
0
r4 cos θ sin θ dr dθ +∫ π/2
π/3
∫ 1
0
r4 cos θ sin θ dr dθ
=∫ π/3
0
sin θ
5(32) cos4 θdθ +
∫ π/2
π/3
15
cos θ sin θ dθ =7
480+
140
=19480
13.∫ 1
0
∫ √1−x2
0
sin√x2 + y2 dy dx =
∫ π/2
0
∫ 1
0
sin(r) r dr dθ =∫ π/2
0
(sin 1 − cos 1) dθ =π
2(sin 1 − cos 1)
14.∫ 2π
0
∫ 1
0
e−r2r dr dθ = 2π
∫ 1
0
re−r2dr = π(1 − 1
e)
15.∫ 2
0
∫ √2x−x2
0
x dy dx =∫ π/2
0
∫ 2 cos θ
0
r cos θ r dr dθ =83
∫ π/2
0
cos4 θ dθ =83· 34· 12· π
2=
π
2
(See Exercise 62, Section 8.3)∧
16. The region is the inside of the circle (x− 1/2)2 + y2 = 1/4, which has polar equation r = cos θ.
So the integral becomes ∫ π/2
−π/2
∫ cos θ
0
r3 dr dθ =∫ π/2
−π/2
14
cos4 θ dθ =3π32
17. A =∫ π/3
0
∫ 3 sin 3θ
0
r dr dθ =92
∫ π/3
0
sin2 3θ dθ =94
∫ π/3
0
(1 − 6 cos θ) dθ =3π4
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.4 879
18. A =∫ 2π
0
∫ 2(1−cos θ)
0
r dr dθ =∫ 2π
0
2(1 − cos θ)2 dθ = 6π
19. First we find the points of intersection:
r = 4 cos θ = 2 =⇒ cos θ =12
=⇒ θ = ±π
3.
A =∫ π/3
−π/3
∫ 4 cos θ
2
r dr dθ =∫ π/3
−π/3
(8 cos2 θ − 2) dθ =∫ π/3
−π/3
(2 + 4 cos 2θ) dθ =4π3
+ 2√
3
20. A =∫ 2π
0
∫ 1+2cos θ
0
r dr dθ =∫ 2π
0
12(1 + 2 cos θ)2 dθ = 3π
21. A = 4∫ π/4
0
∫ 2√
cos 2θ
0
r dr dθ = 8∫ π/4
0
cos 2θ dθ = 4
22. A =∫ π/3
−π/3
∫ 3 cos θ
1+cos θ
r dr dθ =∫ π/3
−π/3
12[9 cos2 θ − (1 + cos θ)2
]dθ =
[3θ2
+ sin 2θ − sin θ
]π/3−π/3
= π
23.∫ 2π
0
∫ b
0
(r2 sin θ + br
)dr dθ =
∫ 2π
0
[13r3 sin θ +
b
2r2
]b0
dθ
= b3∫ 2π
0
(13
sin θ +12
)dθ = b3π
24. V =∫ 2π
0
∫ 1
0
(1 − r2)r dr dθ = 2π∫ 1
0
(r − r3) dr =π
2
25. 8∫ π/2
0
∫ 2
0
r
2
√12 − 3r2 dr dθ = 8
∫ π/2
0
[− 1
18(12 − 3r2
)3/2]20
dθ
= 8∫ π/2
0
43
√3 dθ =
163
√3π
26. V =∫ 2π
0
∫ √5
0
6√
5 − r2 r dr dθ = 2π∫ √
5
0
r6√
5 − r2 dr =307
(5)1/6π
27.∫ 2π
0
∫ 1
0
r√
4 − r2 dr dθ =∫ 2π
0
[−1
3(4 − r2
)3/2]10
dθ
=∫ 2π
0
(83−√
3)dθ =
23(8 − 3
√3 )π
28. V =∫ π/2
−π/2
∫ cos θ
0
(1 − r2)r dr dθ =∫ π/2
−π/2
(cos2 θ
2− cos4 θ
4
)dθ =
5π32
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
880 SECTION 17.5
29.∫ π/2
−π/2
∫ 2 cos θ
0
2r2 cos θ dr dθ =∫ π/2
−π/2
[23r3 cos θ
]2 cos θ
0
dθ
=∫ π/2
−π/2
163
cos4 θ dθ =323
∫ π/2
0
cos4 θ dθ =323
(316
π
)= 2π
Ex. 46, Sect. 8.3∧
30.∫ π/2
−π/2
∫ 2a cos θ
0
r2 dr dθ =∫ π/2
−π/2
8a3
3cos3 θdθ =
329a3
31.b
a
∫ π
0
∫ a sin θ
0
r√a2 − r2 dr dθ =
b
a
∫ π
0
[−1
3(a2 − r2
)3/2]a sin θ
0
dθ
=13a2b
∫ π
0
(1 − cos3 θ
)dθ =
13πa2b
32. (a) V = 2∫ 2π
0
∫ r
0
√R2 − s2 s ds dθ = 4π
∫ r
0
S√R2 − S2 dS =
4π3
[R3 − (R2 − r2)3/2
].
(b)43πR3 − V =
43π(R2 − r2)3/2
33. A = 2∫ π/4
0
∫ 2 cos 2θ
0
r dr dθ = 2∫ π/4
0
12 (2 cos 2θ)2 dθ = 2
∫ π/4
0
(1 + cos 4θ) dθ =π
2
34.∫∫
Ω
ex2+y2
dxdy =∫ 2π
0
∫ 4
2
rer2dr dθ = π
(e16 − e4
)
SECTION 17.5
1. M =∫ 1
−1
∫ 1
0
x2 dy dx =23
xM M =∫ 1
−1
∫ 1
0
x3 dy dx = 0 =⇒ xM = 0
yM M =∫ 1
−1
∫ 1
0
x2y dy dx =∫ 1
−1
12x2 dx =
13
=⇒ yM =1/32/3
=12
2. M =∫ 1
0
∫ √x
0
(x + y) dy dx =∫ 1
0
(x3/2 +
x
2
)dx =
1320
xMM =∫ 1
0
∫ √x
0
x(x + y) dy dx =∫ 1
0
(x5/2 +
x2
2
)dx =
1942
=⇒ xM =190273
yMM =∫ 1
0
∫ √x
0
y(x + y) dy dx =∫ 1
0
(x2
2+
x3/2
3
)dx =
310
=⇒ yM =613
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.5 881
3. M =∫ 1
0
∫ 1
x2xy dy dx =
12
∫ 1
0
(x− x5) dx =16
xM M =∫ 1
0
∫ 1
x2x2y dy dx =
12
∫ 1
0
(x2 − x6) dx =221
=⇒ xM =2/211/6
=47
yM M =∫ 1
0
∫ 1
x2xy2 dy dx =
13
∫ 1
0
(x− x7) dx =18
=⇒ yM =1/81/6
=34
4. M =∫ π
0
∫ sinx
0
y dy dx =∫ π
0
sin2 x
2dx =
π
4
xMM =∫ π
0
∫ sinx
0
xy dy dx =∫ π
0
xsin2 x
2dx =
π2
8=⇒ xM =
π
2
yMM =∫ π
0
∫ sinx
0
y2 dy dx =∫ π
0
sin3 x
3dx =
49
=⇒ yM =169π
5. M =∫ 8
0
∫ x1/3
0
y2 dy dx =13
∫ 8
0
x dx =323
xM M =∫ 8
0
∫ x1/3
0
xy2 dy dx =13
∫ 8
0
x2 dx =5129
=⇒ xM =512/932/3
=163
yM M =∫ 8
0
∫ x1/3
0
y3 dy dx =14
∫ 8
0
x4/3 dx =967
=⇒ yM =96/732/3
=97
6. M =∫ a
0
∫ √a2−x2
0
xy dy dx =∫ a
0
x
2(a2 − x2) dx =
a4
8
xMM =∫ a
0
∫ √a2−x2
0
x2y dy dx =∫ a
0
x2
2(a2 − x2) dx =
a5
15=⇒ xM =
815
a
yMM =∫ a
0
∫ √a2−x2
0
xy2 dy dx =∫ a
0
x
3(a2 − x2)3/2 dx =
a5
15=⇒ yM =
815
a
7. M =∫ 1
0
∫ 3x
2x
xy dy dx =52
∫ 1
0
x3 dx =58
xM M =∫ 1
0
∫ 3x
2x
x2y dy dx =52
∫ 1
0
x4 dx =12
=⇒ xM =1/25/8
=45
yM M =∫ 1
0
∫ 3x
2x
xy2 dy dx =193
∫ 1
0
x4 dx =1915
=⇒ yM =19/155/8
=15275
8. M =∫ 2
0
∫ 3− 32x
0
(x + y) dy dx =∫ 2
0
[x(3 − 3
2x) +
12(3 − 3
2x)2]dx = 5
xMM =∫ 2
0
∫ 3− 32x
0
x(x + y) dy dx =∫ 2
0
[x2(3 − 3
2x) +
x
2(3 − 3
2x)2]dx =
72
=⇒ xM =710
yMM =∫ 2
0
∫ 3− 32x
0
y(x + y) dy dx =∫ 2
0
[x
2(3 − 3
2x)2 +
(3 − 32x)3
3
]dx = 6 =⇒ yM =
65
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
882 SECTION 17.5
9. M =∫ 2π
0
∫ 1+cos θ
0
r2 dr dθ =13
∫ 2π
0
(1 + 3 cos θ + 3 cos2 θ + cos3 θ) dθ =5π3
xM M =∫ 2π
0
∫ 1+cos θ
0
r3 cos θ dr dθ =14
∫ 2π
0
(1 + cos θ)4 cos θ dθ
=14
∫ 2π
0
[cos θ + 4 cos2 θ + 6 cos3 θ + 4 cos4 θ + cos5 θ
]dθ
=7π4
Therefore, xM =7π/45π/3
=2120
.
yM M =∫ 2π
0
∫ 1+cos θ
0
r3 sin θ dr dθ =14
∫ 2π
0
(1 + cos θ)4 sin θ dθ =14
[15(1 + cos θ)5
]2π0
= 0
Therefore, yM = 0.
10. M =∫∫Ω
y dx dy =∫ 5π/6
π/6
∫ 2 sin θ
1
r sin θ r dr dθ =∫ 5π/6
π/6
(83
sin4 θ − 13
sin θ
)dθ =
3√
3 + 8π12
xM = 0 by symmetry
yM M =∫ 5π/6
π/6
∫ 2 sin θ
1
r2 sin2 θ r dr dθ =∫ 5π/6
π/6
(4 sin6 θ − 1
4sin2 θ
)dθ =
11√
3 + 12π16
Therefore, yM =33√
3 + 36π12√
3 + 32π
11. Ω : −L/2 ≤ x ≤ L/2, −W/2 ≤ y ≤ W/2
Ix =∫∫Ω
M
LWy2 dxdy =
4MLW
∫ W/2
0
∫ L/2
0
y2dx dy =112
MW 2
symmetry∧
Iy =∫∫Ω
M
LWx2 dxdy =
112
ML2, Iz =∫∫Ω
M
LW
(x2 + y2
)dxdy =
112
M(L2 + W 2
)
Kx =√Ix/M =
W√
36
, Ky =√Iy/M =
L√
36
Kz =√Iz/M =
√3√L2 + W 2
6
12. λ(x, y) = k
(x +
L
2
)M =
∫ W/2
−W/2
∫ L/2
−L/2
k
(x +
L
2
)dx dy =
kWL2
2
Ix =∫ W/2
−W/2
∫ L/2
−L/2
k
(x +
L
2
)y2 dx dy =
(∫ W/2
−W/2
y2 dy
)(∫ L/2
−L/2
k
(x +
L
2
)dx
)
=(W 3
12
)(M
W
)=
112
MW 2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.5 883
Iy =∫ W/2
−W/2
∫ L/2
−L/2
k
(x +
L
2
)x2 dx dy =
kL4W
24=
112
ML2
Iz =∫ W/2
−W/2
∫ L/2
−L/2
k
(x +
L
2
)(x2 + y2) dx dy = Ix + Iy =
112
M(L2 + W 2).
13. M =∫∫Ω
k
(x +
L
2
)dxdy =
∫∫Ω
12kLdxdy =
12kL( area of Ω) =
12kL2W
symmetry∧
xMM =∫∫Ω
x
[k
(x +
L
2
)]dxdy =
∫∫Ω
(kx2 +
12Lx
)dxdy
=∫∫Ω
kx2 dxdy = 4k∫ W/2
0
∫ L/2
0
x2 dx dy =112
kWL3
symmetry∧
symmetry∧
= 16
(12kL
2W)L = 1
6 ML; xM = 16 L
yMM =∫∫Ω
y
[k
(x +
L
2
)]dxdy = 0; yM = 0
by symmetry∧
14. Iz =∫∫Ω
λ(x, y)[x2 + y2] dx dy =∫∫Ω
λ(x, y)x2 dx dy +∫∫Ω
λ(x, y)y2 dx dy = Ix + Iy.
Since Iz = Ix + Iy, we have MKz2 = MKx
2 + MKy2 therefore Kz
2 = Kx2 + Ky
2.
15. Ix =∫∫Ω
4MπR2
y2 dxdy =4MπR2
∫ π/2
0
∫ R
0
r3 sin2 θ dr dθ
=4MπR2
(∫ π/2
0
sin2 θ dθ
)(∫ R
0
r3 dr
)=
4MπR2
(π4
)(14R4
)=
14MR2
Iy = 14MR2, Iz = 1
2MR2
Kx = Ky = 12R, Kz = R/
√2
16. Iz = IM + d2M. Rotation doesn’t change d, doesn’t change M , and doesn’t change IM .
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
884 SECTION 17.5
17. IM , the moment of inertia about the vertical line through the center of mass, is∫∫Ω
M
πR2
(x2 + y2
)dxdy
where Ω is the disc of radius R centered at the origin. Therefore
IM =M
πR2
∫ 2π
0
∫ R
0
r3 dr dθ =12MR2.
We need I0 = 12 MR2 + d2M where d is the distance from the center of the disc to the origin. Solving
this equation for d, we have d =√I0 − 1
2 MR2
/√M.
18. Ix =∫ b
a
∫ f(x)
0
λy2 dy dx =λ
3
∫ b
a
[f(x)]3 dx
Iy =∫ b
a
∫ f(x)
0
λx2 dy dx = λ
∫ b
a
x2f(x) dx.
19. Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b
Ix =∫∫Ω
4Mπab
y2 dxdy =4Mπab
∫ a
0
∫ ba
√a2−x2
0
y2 dy dx =14Mb2
Iy =∫∫Ω
4Mπab
x2 dxdy =4Mπab
∫ a
0
∫ ba
√a2−x2
0
x2 dy dx =14Ma2
Iz = 14M(a2 + b2
)
20. Ix =∫ 1
0
∫ √x
0
(x + y)y2 dy dx =∫ 1
0
(x5/2
3+
x2
4
)dx =
528
Iy =∫ 1
0
∫ √x
0
(x + y)x2 dy dx =∫ 1
0
(x7/2 +
x3
2
)dx =
2572
; Iz = Ix + Iy.
21. Ix =∫ 1
0
∫ 1
x2xy3 dy dx =
14
∫ 1
0
(x− x9) dx =110
Iy =∫ 1
0
∫ 1
x2x3y dy dx =
12
∫ 1
0
(x3 − x7) dx =116
Iz =∫ 1
0
∫ 1
x2xy(x2 + y2) dy dx = Ix + Iy =
1380
22. Ix =∫ 8
0
∫ 3√x
0
y2 · y2 dy dx =∫ 8
0
x5/3
5dx =
965
Iy =∫ 8
0
∫ 3√x
0
y2 · x2 dy dx =∫ 8
0
x3
3dx =
10243
; Iz = Ix + Iy
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.5 885
23. Ix =∫ 2π
0
∫ 1+cos θ
0
r4 sin2 θ dr dθ =15
∫ 2π
0
(1 + cos θ)5 sin2 θ dθ =33π40
Iy =∫ 2π
0
∫ 1+cos θ
0
r4 cos2 θ dr dθ =15
∫ 2π
0
(1 + cos θ)5 cos2 θ dθ =93π40
Iz =∫ 2π
0
∫ 1+cos θ
0
r4 dr dθ = Ix + Iy =63π20
24. xM =x1M1 + x2M2
M1 + M2, yM =
y1M1 + y2M2
M1 + M2
25. Ω : r21 ≤ x2 + y2 ≤ r2
2, A = π(r22 − r2
1
)(a) Place the diameter on the x-axis.
Ix =∫∫Ω
M
Ay2 dxdy =
M
A
∫ 2π
0
∫ r2
r1
(r2 sin2 θ
)r dr dθ =
14M(r22 + r2
1
)(b) 1
4M(r22 + r2
1
)+ Mr2
1 = 14M(r22 + 5r2
1
)(parallel axis theorem)
(c) 14M(r22 + r2
1
)+ Mr2
2 = 14M(5r2
2 + r21
)26. Set r1 = r2 = r in the proceeding problem. Then the required moments of inertia are
(a) 12Mr2 (b) 3
2Mr2.
27. Ω : r21 ≤ x2 + y2 ≤ r2
2, A = π(r22 − r2
1
)I =∫∫Ω
M
A
(x2 + y2
)dxdy =
M
A
∫ 2π
0
∫ r2
r1
r3 dr dθ =12M(r2
2 + r21)
28. Let l be the x-axis and let the plane of the plate be the xy-plane. Then
I − IM =∫∫Ω
λ(x, y)y2 dx dy −∫∫Ω
λ(x, y)(y − yM )2 dx dy
=∫∫Ω
λ(x, y)[2yMy − y2M ] dx dy
= 2yM∫∫Ω
yλ(x, y) dx dy − y2M
∫∫Ω
λ(x, y) dx dy
= 2y2MM − y2
MM = y2MM = d2M.
29. M =∫∫Ω
k(R−√x2 + y2
)dxdy = k
∫ π
0
∫ R
0
(Rr − r2
)dr dθ =
16kπR3
xM = 0 by symmetry
yMM =∫∫Ω
y[k(R−√x2 + y2
)]dxdy = k
∫ π
0
∫ R
0
(Rr2 − r3
)sin θ dr dθ =
16kR4
yM = R/π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
886 SECTION 17.5
30. Ix =∫∫Ω
k(R−√x2 + y2)y2 dx dy = k
∫ π
0
∫ R
0
(R− r) r2 sin2 θ r dr dθ =kπR5
40=
3MR2
20.
Iy = k
∫ π
0
∫ R
0
(R− r)r2 cos2 θr dr dθ =kπR5
40=
3MR2
20
Iz = Ix + Iy =3MR2
10.
31. Place P at the origin.
M =∫∫Ω
k√x2 + y2 dxdy
= k
∫ π
0
∫ 2R sin θ
0
r2 dr dθ =329kR3
xM = 0 by symmetry
yMM =∫∫Ω
y(k√x2 + y2
)dxdy = k
∫ π
0
∫ 2R sin θ
0
r3 sin θ dr dθ =6415
kR4
yM = 6R/5
Answer: the center of mass lies on the diameter through P at a distance 6R/5 from P .
32. Putting the right angle at the origin, we have λ(x, y) = k(x2 + y2).
M =∫ b
0
∫ h−hb x
0
k(x2 + y2) dy dx =112
kbh(b2 + h2)
xMM =∫ b
0
∫ h−hb x
0
kx(x2 + y2) dy dx =kb2h(3b2 + h2)
60=⇒ xM =
b(3b2 + h2)5(b2 + h2)
yMM =∫ b
0
∫ h−hb x
0
ky(x2 + y2) dy dx =kbh2(b2 + 3h2)
60=⇒ yM =
h(b2 + 3h2)5(b2 + h2)
33. Suppose Ω, a basic region of area A, is broken up into n basic regions Ω1, · · · , Ωn with areas
A1, · · · , An. Then
xA =∫∫Ω
x dxdy =n∑
i=1
⎛⎝∫∫
Ωi
x dxdy
⎞⎠ =
n∑i=1
xi Ai = x1 A1 + · · · + xn An.
The second formula can be derived in a similar manner.
34. (a) M =∫ 2
0
∫ 1
x/2
(x + y) dy dx =43
xM M =∫ 2
0
∫ 1
x/2
x(x + y) dy dx =76; xM =
78
yM M =∫ 2
0
∫ 1
x/2
y(x + y) dy dx = 1; yM =34
(b) Ix =∫ 2
0
∫ 1
x/2
y2(x + y) dy dx =45
Iy =∫ 2
0
∫ 1
x/2
x2(x + y) dy dx =43
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.6 887
SECTION 17.6
1. They are equal; they both give the volume of T .
2. (a) Lf (P ) =m∑i=1
n∑j=1
q∑k=1
xi−1yj−1zk−1ΔxiΔyjΔzk, Uf (P ) =m∑i=1
n∑j=1
q∑k=1
xiyjzkΔxiΔyjΔzk
(b) xi−1yj−1zk−1 ≤(xi + xi−1
2
)(yj + yj−1
2
)(zk + zk−1
2
)≤ xiyjzk
xi−1yj−1zk−1ΔxiΔyjΔzk ≤ 18(xi
2 − xi−12) (
yj2 − yj−1
2) (
zk2 − zk−1
2)≤ xiyjzkΔxiΔyjΔzk
Lf (P ) ≤ 18
m∑i=1
n∑j=1
q∑k=1
(xi
2 − xi−12) (
yj2 − yj−1
2) (
zk2 − zk−1
2)≤ Uf (P ).
The middle term can be written
18
(m∑i=1
xi2 − xi−1
2
)⎛⎝ n∑j=1
yj2 − yj−1
2
⎞⎠( q∑
k=1
zk2 − zk−1
2
)=
18(1)(1)(1) =
18.
Therefore I =18.
3.∫∫∫Π
αdx dy dz = α
∫∫∫Π
dx dy dz = α (volume of Π) = α(a2 − a1)(b2 − b1)(c2 − c1)
4. Since the volume is 1, the average value is∫∫∫Ω
xyz dx dy dz =18.
5. Let P1 = {x0, · · · , xm}, P2 = {y0, · · · , yn}, P3 = {z0, · · · , zq} be partitions of [ 0, a ], [ 0, b ], [ 0, c ]
respectively and let P = P1 × P2 × P3. Note that
xi−1yj−1 ≤(xi + xi−1
2
)(yj + yj−1
2
)≤ xiyj
and therefore
xi−1yj−1 Δxi Δyj Δzk ≤ 14
(xi
2 − x2i−1
) (yj
2 − y2j−1
)Δzk ≤ xiyj Δxi ΔyjΔzk.
It follows that
Lf (P ) ≤ 14
m∑i=1
n∑j=1
q∑k=1
(xi
2 − x2i−1
) (yj
2 − y2j−1
)Δzk ≤ Uf (P ).
The middle term can be written
14
(m∑i=1
xi2 − x2
i−1
)⎛⎝ n∑j=1
yj2 − y2
j−1
⎞⎠( q∑
k=1
Δzk
)=
14a2b2c.
6. Ix = Ixy + Ixz, Iy = Ixy + Iyz, Iz = Ixz + Iyz
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
888 SECTION 17.6
7. x1 = a, y1 = b, z1 = c; x0 = A, y0 = B, z0 = C
x1V1 + xV = x0V0 =⇒ a2bc + (ABC − abc)x = A2BC
=⇒ x =A2BC − a2bc
ABC − abc
similarly
y =AB2C − ab2c
ABC − abc, z =
ABC2 − abc2
ABC − abc
8. Encase T in a box Π. A partition P of Π breaks up II into little boxes Πijk. Since f is
nonnegative on Π, all the mijk are nonnegative. Therefore
0 ≤ Lf (P ) ≤∫∫∫T
f(x, y, z) dx dy dz.
9. M =∫∫∫Π
Kz dxdydz
Let P1 = {x0, · · · , xm}, P2 = {y0, · · · , yn}, P3 = {z0, · · · , zq} be partitions of [ 0, a ] and let
P = P1 × P2 × P3. Note that
zk−1 ≤ 12 (zk + zk−1) ≤ zk
and therefore
Kzk−1 Δxi Δyj Δzk ≤ 12K Δxi Δyj
(zk
2 − z2k−1
)≤ Kzk Δxi ΔyjΔzk.
It follows that
Lf (P ) ≤ 12K
m∑i=1
n∑j=1
q∑k=1
Δxi Δyj(zk
2 − z2k−1
)≤ Uf (P ).
The middle term can be written
12K
(m∑i=1
Δxi
)⎛⎝ n∑j=1
Δyj
⎞⎠ ( q∑
k=1
zk2 − z2
k−1
)=
12K(a) (a) (a2) =
12Ka4.
M = 12Ka4 where K is the constant of proportionality for the density function.
10. xMM =∫∫∫Π
Kzxdx dy dz =Ka5
4=⇒ xM =
12a
yMM =∫∫∫Π
Kzy dx dy dz =Ka5
4=⇒ yM =
12a
zMM =∫∫∫Π
Kz2 dx dy dz =Ka5
3=⇒ zM =
23a.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.7 889
11. Iz =∫∫∫Π
Kz(x2 + y2
)dxdydz
=∫∫∫Π
Kx2z dxdydz
︸ ︷︷ ︸I1
+∫∫∫Π
Ky2z dxdydz
︸ ︷︷ ︸I2
.
We will calculate I1 using the partitions we used in doing Exercise 9. Note that
x2i−1zk−1 ≤
(xi
2 + xixi−1 + x2i−1
3
)(zk + zk−1
2
)≤ xi
2zk
and therefore
Kx2i−1zk−1 Δxi Δyj Δzk ≤ 1
6 K(xi
3 − x3i−1
)Δyj(zk
2 − z2k−1
)≤ Kxi
2zk2 Δxi Δyj Δzk.
It follows that
Lf (P ) ≤ 16K
m∑i=1
n∑j=1
q∑k=1
(xi
3 − x3i−1
)Δyj(zk
2 − z2k−1
)≤ Uf (P ).
The middle term can be written
16K
(m∑i=1
xi3 − x3
i−1
)⎛⎝ n∑j=1
Δyj
⎞⎠( q∑
k=1
zk2 − z2
k−1
)=
16Ka3 (a)(a2) =
16Ka6.
Similarly I2 = 16 Ka6 and therefore Iz = 1
3 Ka6 = 23
(12 Ka4
)a2 = 2
3 Ma2.
by Exercise 9∧
12. (a) Lf (P ) ∼= 56.4803 Uf (P ) ∼= 57.5603 (c)∫∫R
(3y2 − 2x
)dxdy = 57
SECTION 17.7
1.∫ a
0
∫ b
0
∫ c
0
dx dy dz =∫ a
0
∫ b
0
c dy dz =∫ a
0
bc dz = abc
2.∫ 1
0
∫ x
0
∫ y
0
y dz dy dx =∫ 1
0
∫ x
0
y2 dy dx =∫ 1
0
x3
3dx =
112
.
3.∫ 1
0
∫ 2y
1
∫ x
0
(x + 2z) dz dx dy =∫ 1
0
∫ 2y
1
[xz + z2
]x0dx dy =
∫ 1
0
∫ 2y
1
2x2 dx dy
=∫ 1
0
[23x3
]2y1
dy =∫ 1
0
(163
y3 − 23
)dy =
23
4.∫ 1
0
∫ 1+x
1−x
∫ xy
0
4z dz dy dx =∫ 1
0
∫ 1+x
1−x
2x2y2 dy dx =∫ 1
0
2x2
3[(1 + x)3 − (1 − x)3
]dx =
119
5.∫ 2
0
∫ 1
−1
∫ 3
0
(z − xy) dz dy dx =∫ 2
0
∫ 1
−1
[12z2 − xyz
]31
dy dx
=∫ 2
0
∫ 1
−1
(4 − 2xy) dy dx =∫ 2
0
[2y − xy2
]1−1
dx =∫ 2
0
8 dy = 16
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
890 SECTION 17.7
6.∫ 2
0
∫ 1
−1
∫ 3
1
(z − xy) dy dx dz =∫ 2
0
∫ 1
−1
(2z − 4x) dx dz =∫ 2
0
4z dz = 8
7.∫ π/2
0
∫ 1
0
∫ √1−x2
0
x cos z dy dx dz =∫ π/2
0
∫ 1
0
[xy cos z]√
1−x2
0 dx dz
=∫ π/2
0
∫ 1
0
x√
1 − x2 cos z dx dz =∫ π/2
0
[−1
3(1 − x2)3/2 cos z
]10
dz =13
∫ π/2
0
cos z dz =13
8.∫ 2
−1
∫ y−2
1
∫ e2
e
x + y
zdz dx dy =
∫ 2
−1
∫ y−2
1
(x + y) dx dy =∫ 2
−1
[(y − 2)2 − 1
2+ y(y − 3)
]dy =
32
9.∫ 2
1
∫ y2
y
∫ lnx
0
yez dz dx dy =∫ 2
1
∫ y2
y
[yez]lnx0 dx dy
=∫ 2
1
∫ y2
y
y(x− 1) dx dy =∫ 2
1
[12x2y − xy
]y2
y
dy =∫ 2
1
(12y5 − 3
2y3 + y2
)dy =
4724
10.∫ π/2
0
∫ π/2
0
∫ 1
0
ez cosx sin y dz dy dx =∫ π/2
0
∫ π/2
0
(e− 1) cosx sin y dy dx
=∫ π/2
0
(e− 1) cosx dx = e− 1
11.∫∫∫Π
f(x)g(y)h(z) dxdydz =∫ c2
c1
[∫ b2
b1
(∫ a2
a1
f(x)g(y)h(z) dx)dy
]dz
=∫ c2
c1
[∫ b2
b1
g(y)h(z)(∫ a2
a1
f(x) dx)dy
]dz
=∫ c2
c1
[h(z)
(∫ a2
a1
f(x) dx)(∫ b2
b1
g(y) dy)dz
]
=(∫ a2
a1
f(x) dx) (∫ b2
b1
g(y) dy
)(∫ c2
c1
h(z) dz)
12.(∫ 1
0
x3 dx
)(∫ 2
0
y2 dy
)(∫ 3
0
z dz
)=(
14
)(83
)(92
)= 3
13.(∫ 1
0
x2 dx
)(∫ 2
0
y2 dy
)(∫ 3
0
z2 dz
)=(
13
)(83
)(273
)= 8
14. M =∫∫∫Π
kxyz dx dy dz = k
(∫ a
0
x dx
)(∫ b
0
y dy
)(∫ c
0
z dz
)=
18ka2b2c2
15. xMM =∫∫∫Π
kx2yz dxdydz = k
(∫ a
0
x2 dx
) (∫ b
0
y dy
) (∫ c
0
z dz
)
= k(
13 a
3) (
12 b
2) (
12 c
2)
= 112 ka
3 b2 c2.
By Exercise 14, M = 18 ka
2 b2 c2. Therefore x = 23 a. Similarly, y = 2
3 b and z = 23 c.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.7 891
16. (a)
I =∫∫∫Π
kxyz[(x− a)2 + (y − b)2
]dx dy dz
= k
(∫ a
0
x(x− a)2 dx)(∫ b
0
y dy
)(∫ c
0
z dz
)+ k
(∫ a
0
x dx
)(∫ b
0
y(y − b)2 dy)(∫ c
0
z dz
)
=148
ka2b2c2(a2 + b2).
Since M = 18ka
2b2c2, I = 16M(a2 + b2).
(b) IM − 118
(a2 + b2) by parallel axis theorem
17.
18. V =∫∫∫T
dx dy dz =∫ 1
0
∫ 1
0
∫ 1−y
0
dz dy dx =12
19. center of mass is the centroid
x = 12 by symmetry
yV =∫∫∫T
y dxdydz =∫ 1
0
∫ 1
0
∫ 1−y
0
y dz dy dx =∫ 1
0
∫ 1
0
(y − y2
)dy dx
=∫ 1
0
[12y2 − 1
3y3
]10
dx =∫ 1
0
16dx =
16
zV =∫∫∫T
z dxdydz =∫ 1
0
∫ 1
0
∫ 1−y
0
z dz dy dx =∫ 1
0
∫ 1
0
12(1 − y)2 dy dx
=12
∫ 1
0
∫ 1
0
(1 − 2y + y2
)dy dx =
12
∫ 1
0
[y − y2 1
3y3
]10
dx =12
∫ 1
0
13dx =
16
V = 12 (by Exercise 18 ); y = 1
3 , z = 13
20. Ix =∫∫∫T
M
V(y2 + z2) dx dy dz =
13M
Iy =∫∫∫T
M
V(x2 + z2) dx dy dz =
12M Iz =
∫∫∫T
M
V(x2 + y2) dx dy dz =
12M
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
892 SECTION 17.7
21.∫ r
−r
∫ φ(x)
−φ(x)
∫ ψ(x,y)
−ψ(x,y)
k(r −√x2 + y2 + z2
)dz dy dx with φ(x) =
√r2 − x2,
ψ(x, y) =√r2 − (x2 + y2) , k the constant of proportionality
22.∫ 1
−1
∫ √1−x2
−√
1−x2
∫ 1
√x2+y2
k√x2 + y2 + z2 dz dy dx
23.∫ 1
0
∫ √x−x2
−√x−x2
∫ 1−y2
−2x−3y−10
dz dy dx
24.∫ √
2
−√
2
∫ √1−x2/2
−√
1−x2/2
∫ 4−x2−y2
2+y2dz dy dx
25.∫ 1
−1
∫ 2√
2−2x2
−2√
2−2x2
∫ 4−x2−y2/4
3x2+y2/4
k
(z − 3x2 − 1
4y2
)dz dy dx
26.∫ √
2
−√
2
∫ √2−y2
−√
2−y2
∫ 4−z2
z2+2y2k√x2 + y2 dx dz dy
27.∫∫T
∫ (x2z + y
)dx dy dz =
∫ 2
0
∫ 3
1
∫ 1
0
(x2z + y
)dx dy dz =
∫ 2
0
∫ 3
1
[13x3z + xy
]10
dy dz
=∫ 2
0
∫ 3
1
(13z + y
)dy dz =
∫ 2
0
[13zy +
12y2
]31
dz =∫ 2
0
(23z + 4
)dz =
283
28.∫ 1
0
∫ y
0
∫ x+y
0
2yex dz dx dy =∫ 1
0
∫ y
0
2y(x + y)ex dx dy =∫ 1
0
(4y2ey − 2yey + 2y − 2y2) dy = 4e− 293
29.∫∫T
∫x2y2z2 dx dy dz =
∫ 0
−1
∫ y+1
0
∫ 1
0
x2y2z2 dx dz dy +∫ 1
0
∫ 1−y
0
∫ 1
0
x2y2z2 dx dz dy
=∫ 0
−1
∫ y+1
0
[13x3y2z2
]10
dz dy +∫ 1
0
∫ 1−y
0
[13x3y2z2
]10
dz dy
=13
∫ 0
−1
∫ y+1
0
y2z2 dz dy +13
∫ 1
0
∫ 1−y
0
[y2z2]10dz dy
=13
∫ 0
−1
[13y2z3
]y+1
0
dy +13
∫ 1
0
[13y2z3
]1−y
0
dy
=19
∫ 0
−1
(y5 + 3y4 + 3y3 + y2
)dy +
19
∫ 1
0
(y2 − 3y3 + 3y4 − y5
)dy =
1270
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.7 893
30.∫ 2
0
∫ √4−x2
0
∫ √4−x2−y2
0
xy dz dy dx =∫ 2
0
∫ √4−x2
0
xy√
4 − x2 − y2 dy dx
=∫ π/2
0
∫ 2
0
r2 cos θ sin θ√
4 − r2r dr dθ
=12
∫ 2
0
r3√
4 − r2 dr =14
∫ 4
0
(4√u− u3/2) du =
3215
31.∫∫∫T
y2 dx dy dz =∫ 3
0
∫ 2−2x/3
0
∫ 6−2x−3y
0
y2 dz dy dx =∫ 3
0
∫ 2−2x/3
0
[y2z]6−2x−3y
0dy dx
=∫ 3
0
∫ 2−2x/3
0
(6y2 − 2xy2 − 3y3) dy dx
=∫ 3
0
[2y3 − 2
3xy3 − 3
4y4
]2−2x/3
0
dx
=14
∫ 3
0
(2 − 2
3x
)dx =
125
32.∫ 1
0
∫ 1−x2
0
∫ √1−y
0
y2 dz dy dx =∫ 1
0
∫ 1−x2
0
y2√
1 − y dy dx
=∫ 1
0
[−2
3x3 +
45x5 − 2
7x7 +
16105
]dx =
112
33. V =∫ 2
0
∫ x+2
x2
∫ x
0
dz dy dx =∫ 2
0
∫ x+2
x2x dy dx =
∫ 2
0
(x2 + 2x− x3
)dx =
83
xV =∫ 2
0
∫ x+2
x2
∫ x
0
x dz dy dx =∫ 2
0
∫ x+2
x2x2 dy dx =
∫ 2
0
(x3 + 2x2 − x4
)dx =
4415
yV =∫ 2
0
∫ x+2
x2
∫ x
0
y dz dy dx =∫ 2
0
∫ x+2
x2xy dy dx =
∫ 2
0
12(x3 + 4x2 + 4x− x5
)dx = 6
zV =∫ 2
0
∫ x+2
x2
∫ x
0
z dz dy dx =∫ 2
0
∫ x+2
x2
12x2 dy dx =
∫ 2
0
12(x3 + 2x2 − x4
)dx =
2215
x =1110
, y =94, z =
1120
34. (a) M =∫ 1
0
∫ 1
0
∫ 1
0
kz dx dy dz =12k
(b) M =∫ 1
0
∫ 1
0
∫ 1
0
k(x2 + y2 + z2) dz dy dx = k
35. V =∫ 2
−1
∫ 3
0
∫ 4−x2
2−x
dz dy dx = 272 ; (x, y, z) =
(12 ,
32 ,
125
)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
894 SECTION 17.7
36.∫∫∫T
(x− x) dx dy dz =∫∫∫T
x dx dy dz − x
∫∫∫T
dx dy dz = xV − xV = 0.
Similarly, the other two integrals are zero.
37. V =∫ a
0
∫ φ(x)
0
∫ ψ(x,y)
0
dz dy dx =16abc with φ(x) = b
(1 − x
a
), ψ(x, y) = c
(1 − x
a− y
b
)(x, y, z) =
(14 a,
14 b,
14 c)
38. Iz =∫∫∫T
M
V(x2 + y2) dx dy dz =
130
(M
V
)=
15M
39. Π: 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c
(a) Iz =∫ a
0
∫ b
0
∫ c
0
M
abc
(x2 + y2
)dz dy dx =
13M(a2 + b2
)(b) IM = Iz − d2M = 1
3 M(a2 + b2
)− 1
4
(a2 + b2
)M = 1
12M(a2 + b2
)∧
parallel axis theorem (17.5.7)
(c) I = IM + d2M = 112 M
(a2 + b2
)+ 1
4 a2M = 1
3 Ma2 + 112 Mb2
∧parallel axis theorem (17.5.7)
40. V =∫ 2
1
∫ 2
1
∫ 1+x+y
−2
dz dy dx =∫ 2
1
∫ 2
1
(3 + x + y) dy dx = 6
xV =∫ 2
1
∫ 2
1
∫ 1+x+y
−2
x dz dy dx =10912
=⇒ x =10972
= y by symmetry
zV =∫ 2
1
∫ 2
1
∫ 1+x+y
−2
z dz dy dx =7312
=⇒ z =7372
.
41. M =∫ 1
0
∫ 1
0
∫ y
0
k(x2 + y2 + z2
)dz dy dx =
∫ 1
0
∫ 1
0
k
(x2y + y3 +
13y3
)dy dx
=∫ 1
0
k
(12x2 +
13
)dx =
12k
(xM , yM , zM ) =(
712 ,
3445 ,
3790
)
42. T is symmetric (a) about the yz-plane, (b) about the xz-plane, (c) about the xy-plane,
(d) about the origin.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.7 895
43. (a) 0 by symmetry
(b)∫∫∫T
(a1 x + a2 y + a3z + a4) dxdydz =∫∫∫T
a4 dxdydz = a4 (volume of ball) = 43 πa4
by symmetry∧
44.∫ 2
0
∫ 2
0
∫ 4−y2
2−y
x2y2 dz dy dx =35245
45. V = 8∫ a
0
∫ √a2−x2
0
∫ √a2−x2−y2
0
dz dy dx = 8∫ a
0
∫ √a2−x2
0
√a2 − x2 − y2 dy dx
polar coordinates∧
= 8∫ π/2
0
∫ a
0
√a2 − r2 r dr dθ
= −4∫ π/2
0
[23(a2 − r2)3/2
]a0
dθ
=83
∫ π/2
0
dθ =43π a3
46. 8∫ a
0
∫ b√
1−x2/a2
0
∫ c√
1−x2/a2−y2/b2
0
dz dy dx =43πabc.
47. M =∫ 2
−2
∫ √4−x2/2
−√
4−x2/2
∫ 4−y2
x2+3y2k|x| dz dy dx = 4
∫ 2
0
∫ √4−x2/2
0
∫ 4−y2
x2+3y2kx dz dy dx
= 4 k∫ 2
0
∫ √4−x2/2
0
(4x− x3 − 4xy2
)dy dx =
43k
∫ 2
0
x(4 − x2
)3/2dx =
12815
k
48. using polar coordinates V = 2∫ 2π
0
∫ 1
0
(r − r3) dr dθ = π
49. M =∫ 2
−1
∫ 3
0
∫ 4−x2
2−x
k(1 + y) dz dy dx =1354
k; (xM , yM , zM ) =(
12,
95,
125
)
50. (a) V =∫ 2
0
∫ 2−z
0
∫ 9−x2
0
dy dx dz
(b) V =∫ 2
0
∫ 2−x
0
∫ 9−x2
0
dy dz dx
(c) V =∫ 5
0
∫ 2
0
∫ 2−x
0
dz dx dy +∫ 9
5
∫ √9−y
0
∫ 2−x
0
dz dx dy
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
896 SECTION 17.8
51. (a) V =∫ 6
0
∫ 3
z/2
∫ 6−x
x
dy dx dz
(b) V =∫ 3
0
∫ 2x
0
∫ 6−x
x
dy dz dx
(c) V =∫ 6
0
∫ 3
z/2
∫ y
z/2
dx dy dz +∫ 6
0
∫ (12−z)/2
3
∫ 6−y
z/2
dx dy dz
52. (a) V =∫∫Ωxy
2√
4 − y dx dy (b) V =∫∫Ωxy
(∫ √4−y
−√4−y
dz
)dx dy
(c) V =∫ 4
−4
∫ 4
|x|
∫ √4−y
−√4−y
dz dy dx (d) V =∫ 4
0
∫ y
−y
∫ √4−y
−√4−y
dz dx dy
53. (a) V =∫∫Ωxy
2y dydz (b) V =∫∫Ωxy
( ∫ y
−y
dx
)dydz
(c) V =∫ 4
0
∫ √4−y
−√4−y
∫ y
−y
dx dz dy (d) V =∫ 2
−2
∫ 4−z2
0
∫ y
−y
dx dy dz
54. (a) V =∫∫Ωxy
(4 − z2 − |x|) dx dz (b) V =∫∫Ωxy
(∫ 4−z2
|x|dy
)dx dz
(c) V =∫ 2
−2
∫ 4−z2
z2−4
∫ 4−z2
|x|dy dx dz
(d) V =∫ 0
−2
∫ √4+x
−√
4+x
∫ 4−z2
|x|dy dz dx +
∫ 2
0
∫ √4−x
−√
4−x
∫ 4−z2
|x|dy dz dx
55. (a)∫ 4
2
∫ 5
3
∫ 2
1
lnxy
zdz dy dx ∼= 6.80703 (b)
∫ 4
0
∫ 2
1
∫ 3
0
x√yz dz dy dx =
16√
33
(4√
2 − 2)
56. Let f(x, z) = 36 − 9x2 − 4z2 and g(x, z) = 1 − 12 x− 1
3 z. Then
V =∫ 2
0
∫ 3−(3/2)x
0
∫ f(x,z)
g(x,z)
1 dy dz dx = 71
SECTION 17.8
1. r2 + z2 = 9 2. r = 2 3. z = 2r
4. r cos θ = 4z 5. 4r2 = z2 6. r2 sin2 θ + z2 = 8
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.8 897
7.∫ π/2
0
∫ 2
0
∫ 4−r2
0
r dz dr dθ
=∫ π/2
0
∫ 2
0
(4r − r2
)dr dθ
=∫ π/2
0
4 dθ = 2π0
1
2
0 1 2
0
2
4
8.∫ π/4
0
∫ 1
0
∫ √1−r2
0
r dz dr dθ =π
12
9.∫ 2π
0
∫ 2
0
∫ r2
0
r dz dr dθ
=∫ 2π
0
∫ 2
0
r3 dr dθ
=∫ 2π
0
4 dθ = 8π-2
0
2
20
0
2
4
-
10.∫ 3
0
∫ 2π
0
∫ 3
r
r dz dθ dr = 9π
11.∫ 1
0
∫ √1−x2
0
∫ √4−(x2+y2)
0
dz dy dx =∫ π/2
0
∫ 1
0
∫ √4−r2
0
r dz dr dθ
=∫ π/2
0
∫ 1
0
r√
4 − r2 dr dθ
=∫ π/2
0
(83−√
3)
dθ =16
(8 − 3
√3)π
12.∫ π
0
∫ 1
0
∫ 1
r
z3r dz dr dθ =∫ π
0
∫ 1
0
14(1 − r4) r dr dθ =
π
12
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
898 SECTION 17.8
13.∫ 3
0
∫ √9−y2
0
∫ √9−x2−y2
0
1√x2 + y2
dz dx dy =∫ π/2
0
∫ 3
0
∫ √9−r2
0
1r· r dz dr dθ
=∫ π/2
0
∫ 3
0
√9 − r2 dr dθ
=∫ π/2
0
[r
2
√9 − r2 +
92
sin−1 r
3
]30
dθ
=9π4
∫ π/2
0
dθ =98π2
14.∫ π/2
0
∫ 1
0
∫ √1−r2
0
zr dz dr dθ =∫ π/2
0
∫ 1
0
r
2(1 − r2) dr dθ =
π
16
15.∫ 1
0
∫ √1−x2
0
∫ 2
0
sin(x2 + y2) dz dy dx =∫ π/2
0
∫ 1
0
∫ 2
0
sin(r2)r dz dr dθ
=∫ π/2
0
∫ 1
0
2r sin(r2) dr dθ = 12π(1 − cos 1) ∼= 0.7221
16.∫ 2π
0
∫ 1
0
∫ 2−r2
r2r2dz dr dθ = 4π
∫ 1
0
(1 − r2)r2 dr =8π15
17. (0, 1, 2) → (1, 12π, 2) 18. (0, 1,−2) → (1, 1
2π,−2)
19. (0,−1, 2) → (1, 32π, 2) 20. (0, 0, 0) → (0, arbitrary, 0)
21. V =∫ π/2
−π/2
∫ 2a cos θ
0
∫ r
0
r dz dr dθ =∫ π/2
−π/2
∫ 2a cos θ
0
r2 dr dθ
=∫ π/2
−π/2
83a3 cos3 θ dθ =
329
a3
22. V =∫ π/2
−π/2
∫ 2a cos θ
0
∫ r2/a
0
r dz dr dθ =32πa3
23. V =∫ π/2
−π/2
∫ a cos θ
0
∫ a−r
0
r dz dr dθ =∫ π/2
−π/2
∫ a cos θ
0
r(a− r) dr dθ
=∫ π/2
−π/2
a3
(12
cos2 θ − 13
cos3 θ)
dθ =136
a3(9π − 16)
24. V =∫ π/2
−π/2
∫ 2 cos θ
0
∫ 2+ 12 r cos θ
0
r dz dr dθ =52π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.8 899
25. V =∫ π/2
−π/2
∫ cos θ
0
∫ r cos θ
r2r dz dr dθ =
∫ π/2
−π/2
∫ cos θ
0
(r2 cos θ − r3
)dr dθ
=∫ π/2
−π/2
112
cos4 θ dθ =132
π
26. V =∫ 2π
0
∫ 3
0
∫ √25−r2
r+1
r dz dr dθ =413π
27. V =∫ 2π
0
∫ 1/2
0
∫ √1−r2
r√
3
r dz dr dθ =∫ 2π
0
∫ 1/2
0
(r√
1 − r2 − r2√
3)dr dθ =
13π(2 −
√3)
28. V =∫ 2π
0
∫ a
0
∫ √a2+r2
√2r
r dz dr dθ =2π3a3(
√2 − 1)
29. V =∫ 2π
0
∫ 3
1
∫ √9−r2
0
r dz dr dθ =∫ 2π
0
∫ 3
1
r√
9 − r2 dr dθ = 323 π
√2
30. V =∫ 2π
0
∫ 2
1
∫ 12
√36−r2
0
r dz dr dθ =13π(35
√35 − 128
√2).
31. Set the lower base of the cylinder on the xy-plane so that the axis of the cylinder coincides with the
z-axis. Assume that the density varies directly as the distance from the lower base.
M =∫ 2π
0
∫ R
0
∫ h
0
kzr dz dr dθ =12kπR2h2
32. xM = yM = 0 by symmetry
zMM =∫ 2π
0
∫ R
0
∫ h
0
kz2r dz dr dθ =13kπR2h3
M =12kπR2h2, zM =
23h
The center of mass lies on the axis of the cylinder at a distance 23h from the base of zero mass
density.
33. I = Iz = k
∫ 2π
0
∫ R
0
∫ h
0
zr3 dr dθ dz
= 14 kπR
4h2 = 12
(12 kπR
2h2)R2 = 1
2 MR2
∧from Exercise 31
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
900 SECTION 17.9
34. (a) I =M
πR2h
∫ 2π
0
∫ R
0
∫ h
0
r3 dz dr dθ =12MR2
(b) I =M
πR2h
∫ 2π
0
∫ R
0
∫ h
0
(r2 sin2 θ + z2)r dz dr dθ =14MR2 +
13Mh2
(c) I = 14MR2 + 1
3Mh2 −M( 12h)2 = 1
4MR2 + 112Mh2
35. Inverting the cone and placing the vertex at the origin, we have
V =∫ h
0
∫ 2π
0
∫ (R/h)z
0
r dr dθ dz =13πR2h.
36. xM = yM = 0 by symmetry
zMM =∫ h
0
∫ 2π
0
∫ (R/h)z
0
(M
V
)zr dr dθ dz =
(M
V
)πR2h2
4=⇒ zM =
πR2h2
4V=
34h
On the axis of the cone at a distance 34h from the vertex.
37. I =M
V
∫ h
0
∫ 2π
0
∫ (R/h)z
0
r3 dr dθ dz =310
MR2
38. I =M
V
∫ h
0
∫ 2π
0
∫ (R/h)z
0
z2 r dr dθ dz =35Mh2.
39. V =∫ 2π
0
∫ 1
0
∫ 1−r2
0
r dz dr dθ =12π 40. M =
∫ 2π
0
∫ 1
0
∫ 1−r2
0
kzr dz dr dθ =16πk
41. M =∫ 2π
0
∫ 1
0
∫ 1−r2
0
k(r2 + z2
)r dz dr dθ =
14kπ
SECTION 17.9
1.(√
3, 14 π, cos−1
[13
√3])
2.(
12
√6, 1
2
√2,
√2)
3. ( 34 ,
34
√3, 3
2
√3 ) 4. (2
√10, 2
3π, cos−1[ 310
√10])
5. ρ =√
22 + 22 + (2√
6/3)2 =4√
63
6.(
8, −π
4,
5π6
)
φ = cos−1
(2√
6/34√
6/3
)= cos−1(1/2) =
π
3
θ = tan−1(1) =π
4
(ρ, θ, φ) =
(4√
63
,π
4,π
3
)
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.9 901
7. x = ρ sinφ cos θ = 3 sin 0 cos(π/2) = 0
z = ρ cosφ = 3 cos 0 = 3
y = ρ sinφ sin θ = 3 sin 0 sin(π/2) = 0
(x, y, z) = (0, 0, 3)
8. (a) (5, 12π, arccos 4
5 )
(b) (5, 32π, arccos 4
5 )
9. The circular cylinder x2 + y2 = 1; the radius of the cylinder is 1 and the axis is the z-axis.
10. The xy-plane.
11. The lower nappe of the circular cone z2 = x2 + y2.
12. Vertical plane which bisects the first and third quadrants of the xy-plane.
13. Horizontal plane one unit above the xy-plane.
14. sphere x2 + y2 + (z − 12 )2 = 1
4 of radius 12 and center (0, 0, 1
2 )
15. Sphere of radius 2 centered at the origin:∫ 2π
0
∫ π
0
∫ 2
0
ρ2 sinφdρ dφ dθ =83
∫ 2π
0
∫ π
0
sinφdφ dθ =163
∫ 2π
0
dθ =32π3
16. That part of the sphere of radius 1 that lies in the first quadrant between the x, z-plane and the plane
y = x ∫ π/4
0
∫ π/2
0
∫ 1
0
ρ2 sinφdρ dφ dθ =π
12
17. The first quadrant portion of the sphere that lies between the x, y-plane and the plane z = 32
√3.
∫ π/2
π/6
∫ π/2
0
∫ 3
0
ρ2 sinφdρ dθ dφ = 9∫ π/2
π/6
∫ π/2
0
sinφdθ dφ
= 92 π
∫ π/2
π/6
sinφdφ
= 92 π [− cosφ]π/2π/6 = 9
4π√
3
18. A cone of radius 1 and height 1;∫ π/4
0
∫ 2π
0
∫ secφ
0
ρ2 sinφdρ dθ dφ =13π
19.∫ 1
0
∫ √1−x2
0
∫ √2−x2−y2
√x2+y2
dz dy dx =∫ π/4
0
∫ π/2
0
∫ √2
0
ρ2 sinφdρ dθ dφ
= 23
√2∫ π/4
0
∫ π/2
0
sinφdθ dφ
=√
23 π
∫ π/4
0
sinφdφ =√
26
π (2 −√
2)
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
902 SECTION 17.9
20.∫ π/4
0
∫ π/2
0
∫ 2
0
ρ4 sinφdρ dθ dφ =16π5
∫ π/4
0
sinφdφ =8π5
(2 −√
2)
21.∫ 3
0
∫ √9−y2
0
∫ √9−x2−y2
0
z√x2 + y2 + x2 dz dx dy
=∫ π/2
0
∫ π/2
0
∫ 3
0
ρ cosφ · ρ · ρ2 sinφdρ dθ dφ
=∫ π/2
0
12
sin 2φdφ
∫ π/2
0
dθ
∫ 3
0
ρ4 dρ =[− 1
4cos 2φ
]π/20
(π2
) [15ρ5
]30
=243π20
22.∫ π/2
0
∫ π/2
0
∫ 1
0
1ρ2
ρ2 sinφdρ dθ dφ =π
2
23. V =∫ 2π
0
∫ π
0
∫ R
0
ρ2 sinφdρ dφ dθ =43πR3
24. r = ρ sinφ, θ = θ, z = ρ cosφ
25. V =∫ α
0
∫ π
0
∫ R
0
ρ2 sinφdρ dφ dθ =23αR3
26. M =∫ 2π
0
∫ π
0
∫ R
0
k(R− ρ)ρ2 sinφdρ dφ dθ =13kπR4
27. M =∫ 2π
0
∫ tan−1(r/h)
0
∫ h secφ
0
kρ3 sinφdρ dφ dθ
=∫ 2π
0
∫ tan−1(r/h)
0
kh4
4tanφ sec3 φdφ dθ
=kh4
4
∫ 2π
0
13[sec3 φ
]tan−1(r/h)
0dθ =
kh4
4
∫ 2π
0
13
⎡⎣(√
r2 + h2
h
)3
− 1
⎤⎦ dθ
=16kπh(r2 + h2
)3/2 − h3
28. V =∫ 2π
0
∫ tan−1 r/h
0
∫ h secφ
0
ρ2 sinφdρ dφ dθ =2π3
∫ tan−1(r/h)
0
h3 tanφ sec2 φdφ
=π
3h3 tan2(tan−1
( rh
)=
13πr2h
29. center ball at origin; density =M
V=
3M4πR3
(a) I =3M
4πR3
∫ 2π
0
∫ π
0
∫ R
0
ρ4 sin3 φdρ dφ dθ =25MR2
(b) I = 25 MR2 + R2M = 7
5 MR2 (parallel axis theorem)
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.9 903
30. The center of mass is the centroid; V =23πR3
zV =∫ 2π
0
∫ π/2
0
∫ R
0
(ρ cosφ)ρ2 sinφdρ dφ dθ =14πR4
z =38R; (x, y, z) =
(0, 0,
38R
)
31. center balls at origin; density =M
V=
3M4π(R2
3 −R13)
(a) I =3M
4π(R2
3 −R13) ∫ 2π
0
∫ π
0
∫ R2
R1
ρ4 sin3 φdρ dφ dθ =25M
(R2
5 −R15
R23 −R1
3
)This result can be derived from Exercise 29 without further integration. View the solid as a ball
of mass M2 from which is cut out a core of mass M1.
M2 =M
VV2 =
3M4π(R2
3 −R13) (4
3πR2
3
)=
MR23
R23 −R1
3 ; similarly M1 =MR1
3
R23 −R1
3 .
Then
I = I2 − I1 = 25 M2R2
2 − 25 M1R1
2 =25
(MR2
3
R23 −R1
3
)R2
2 − 25
(MR1
3
R23 −R1
3
)R1
2
=25M
(R2
5 −R15
R23 −R1
3
).
(b) Outer radius R and inner radius R1 gives
moment of inertia =25M
(R5 −R1
5
R3 −R13
). [part (a) ]
As R1 → R,
R5 −R15
R3 −R13 =
R4 + R3R1 + R2R12 + RR1
3 + R14
R2 + RR1 + R12 −→ 5R4
3R2=
53R2.
Thus the moment of inertia of spherical shell of radius R is
25M
(53R2
)=
23MR2.
(c) I = 23MR2 + R2M = 5
3 MR2 (parallel axis theorem)
32. (a) The center of mass is the centroid; using the result of Exercise 30,
x = y = 0
z =z2V2 − z1V1
V=
38R2
43πR2
3 − 38R1
43πR1
3
43π(R2
3 −R13)
=3(R2
2 + R12)(R2 + R1)
8(R22 + R2R1 + R1
2)
(b) Setting R1 = R2 = R in (a), we get x = y = 0, z = 12R
33. V =∫ 2π
0
∫ α
0
∫ a
0
ρ2 sinφdρ dφ dθ =23π (1 − cosα) a3
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
904 SECTION 17.9
34.∫ 2π
0
∫ π/4
0
∫ 1
0
eρ3ρ2 sinφdρ dφ dθ =
13π(e− 1)
(2 −
√2)
35. (a) Substituting x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ
into x2 + y2 + (z −R)2 = R2
we have ρ2 sin2 φ + (ρ cosφ−R)2 = R2,
which simplifies to ρ = 2R cosφ.
(b) 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ
36. (a) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 sinφdρ dφ dθ =85kπR4
(b) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 sin2 φdρ dφ dθ =14kπ2R4
(c) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 cos2 θ sin2 φdρ dφ dθ =18kπ2R4
37. V =∫ 2π
0
∫ π/4
0
∫ 2
0
ρ2 sinφdρ dφ dθ +∫ 2π
0
∫ π/2
π/4
∫ 2√
2 cosφ
0
ρ2 sinφdρ dφ dθ
=13
(16 − 6
√2)π
38. V =∫ 2π
0
∫ π
0
∫ 1−cosφ
0
ρ2 sinφdρ dφ dθ =83π
39. Encase T in a spherical wedge W . W has spherical coordinates in a box Π that contains S. Define
f to be zero outside of T . Then
F (ρ, θ, φ) = f (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)
is zero outside of S and
∫∫T
∫f(x, y, z) dxdydz =
∫∫W
∫f(x, y, z) dxdydz
=∫∫
Π
∫F (ρ, θ, φ) ρ2 sinφdρdθdφ
=∫∫S
∫F (ρ, θ, φ) ρ2 sinφdρdθdφ.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.9 905
40. Break up T into little basic solids T1, . . . , TN . Choose a point (x∗i , y
∗i , z
∗i ) from each Ti and view
all the mass as concentrated there. Now Ti attracts m with a force
Fi∼= −Gmλ(x∗
i , y∗i , z
∗i )(Volume of Ti)ri3
ri
where ri is the vector from (x∗i , y
∗i , z
∗i ) to (a, b, c). We therefore have
Fi∼= Gmλ(x∗
i , y∗i , z
∗i )[(x
∗i − a)i + (y∗i − b)j + (z∗i − c)k]
[(x∗i − a)2 + (y∗i − b)2 + (z∗i − c)2]3/2
(Volume of Ti).
The sum of these approximations is a Riemann sum for the triple integral given and tends to thattriple integral as the maximum diameter of the Ti tends to zero.
41. T is the set of all (x, y, z) with spherical coordinates (ρ, θ, φ) in the set
S : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ.
T has volume V = 23 πR
3. By symmetry the i, j components of force are zero and
F =
⎧⎨⎩3GmM
2πR3
∫∫T
∫z
(x2 + y2 + z2)3/2dxdydz
⎫⎬⎭k
=
⎧⎨⎩3GmM
2πR3
∫∫S
∫ (ρ cosφρ3
)ρ2 sinφdρdθdφ
⎫⎬⎭k
=
{3GmM
2πR3
∫ 2π
0
∫ π/4
0
∫ 2R cosφ
R secφ
cosφ sinφdρ dφ dθ
}k
=GmM
R2
(√2 − 1
)k.
42. With the coordinate system shown in the figure, T
is the set of all points (x, y, z) with cylindrical coor-dinates (r, θ, z) in the set
S : 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, α ≤ z ≤ α + h.
The gravitational force is
F =
⎡⎣∫∫
T
∫ (GmM
V
)z
(x2 + y2 + z2)3/2dxdydz
⎤⎦k
=
⎡⎣GmM
πR2h
∫∫S
zr
(r2 + z2)3/2dr dθ dz
⎤⎦k
=[GmM
πR2h
∫ 2π
0
∫ R
0
∫ α+h
α
zr
(r2 + z2)3/2dzdrdθ
]k
=2GmM
R2h
(√R2 + α2 −
√R2 + (α + h)2 + h
)k
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
906 SECTION 17.10
SECTION 17.10
1. ad− bc 2. 1 3. 2(v2 − u2
)4. u ln v − u 5. −3u2v2 6. 1 +
1uv
7. abc 8. 2 9. ρ2 sinφ
10. |J(r, θ, z)| =
∣∣∣∣∣∣∣∣cos θ sin θ 0
−r sin θ r cos θ 0
0 0 1
∣∣∣∣∣∣∣∣ = r.
11. J(ρ, θ, φ) =
∣∣∣∣∣∣∣∣sin φ cos θ sin φ sin θ cos θ
−ρ sin φ sin θ ρ sin φ cos θ 0
ρ cos φ cos θ ρ cos φ sin θ −ρ sin φ
∣∣∣∣∣∣∣∣ = −ρ2 sin φ; |J(ρ, θ, φ)| = ρ2 sin φ.
12. (a) dx− by = (ad− bc)u0 (b) cx− ay = (bc− ad)v0
13. Set u = x + y, v = x− y. Then
x =u + v
2, y =
u− v
2and J (u, v) = −1
2.
Ω is the set of all (x, y) with uv-coordinates in
Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2.
Then∫ ∫Ω
(x2 − y2
)dxdy =
∫ ∫Γ
12uv dudv =
12
∫ 1
0
∫ 2
0
uv dv du
=12
(∫ 1
0
u du
)(∫ 2
0
v dv
)=
12
(12
)(2) =
12.
14. Using the changes of variables from Exercise 13,∫∫Ω
4xy dx dy =∫ 1
0
∫ 2
0
4(u2 − v2
4
)12dv du =
12
∫ 1
0
∫ 2
0
(u2 − v2
)dv du = −1
15.12
∫ 1
0
∫ 2
0
u cos (πv) dv du =12
(∫ 1
0
u du
)(∫ 2
0
cos (πv) dv)
=12
(12
)(0) = 0
16. Set u = x− y, v = x + 2y. Then
x =2u + v
3, y =
v − u
3, and J(u, v) =
13
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.
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SECTION 17.10 907
Therefore∫∫Ω
(x + y) dxdy =∫∫Γ
19(u + 2v) du dv =
19
∫ π
0
∫ π/2
0
(u + 2v) dv du =118
π3.
17. Set u = x− y, v = x + 2y. Then
x =2u + v
3, y =
v − u
3, and J(u, v) =
13.
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.
Therefore∫∫Ω
sin (x− y) cos (x + 2y) dxdy =∫∫Γ
13
sinu cos v dudv =13
∫ π
0
∫ π/2
0
sinu cos v dv du
=13
(∫ π
0
sinu du
)(∫ π/2
0
cos v dv)
=13(2)(1) =
23.
18. Using the change of variables from Exercise 16,∫∫Ω
sin 3x dx dy =∫ π
0
∫ π/2
0
sin(2u + v)13du dv = 0.
19. Set u = xy, v = y. Thenx = u/v, y = v and J (u, v) = 1/v.
xy = 1, xy = 4 =⇒ u = 1, u = 4
y = x, y = 4x =⇒ u/v = v, 4u/v = v =⇒ v2 = u, v2 = 4u
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 1 ≤ u ≤ 4,√u ≤ v ≤ 2
√u.
(a) A =∫∫Γ
1vdudv =
∫ 4
1
∫ 2√u
√u
1vdv du =
∫1
4
ln 2 du = 3 ln 2
(b) xA =∫ 4
1
∫ 2√u
√u
u
v2dv du =
73
; x =7
9 ln 2
yA =∫ 4
1
∫ 2√u
√u
dv du =143
; y =14
9 ln 2
20. J(r, θ) = abr, Γ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
A =∫∫
Γ
abr dr dθ = ab
∫ 2π
0
∫ 1
0
r dr dθ = πab
21. Set u = x + y, v = 3x− 2y. Then
x =2u + v
5, y =
3u− v
5and J (u, v) = −1
5.
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908 SECTION 17.10
With Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2
M =∫ 1
0
∫ 2
0
15λ dv du =
25λ where λ is the density.
Then
Ix =∫ 1
0
∫ 2
0
(3u− v
5
)2 15λ dv du =
8λ375
=475
(25λ
)=
475
M,
Iy =∫ 1
0
∫ 2
0
(2u + v
5
)2 15λ dv du =
28λ375
=1475
(25λ
)=
1475
M,
Iz = Ix + Iy =1875
M.
22. x =u + v
2, y =
v − u
2, J(u, v) =
12
Γ : −2 ≤ u ≤ 2, −4 ≤ v ≤ −u2
A =∫∫Γ
12du dv =
12
∫ 2
−2
∫ −u2
−4
dv du =163
23. Set u = x− 2y, v = 2x + y. Then
x =u + 2v
5, y =
v − 2u5
and J (u, v) =15.
Γ is the region between the parabola v = u2 − 1 and the line v = 2u + 2. A sketch of the curves showsthat
Γ : −1 ≤ u ≤ 3, u2 − 1 ≤ v ≤ 2u + 2.Then
A =15
(area of Γ) =15
∫ 3
−1
[(2u + 2) −
(u2 − 1
) ]du =
3215
.
24. xA =12
∫ 2
−2
∫ −u2
−4
u + v
2dv du = −32
5yA =
12
∫ 2
−2
∫ −u2
−4
v − u
2dv du = −32
5
A =163
=⇒ x = y = −65
25. The choice θ = π/6 reduces the equation to 13u2 + 5v2 = 1. This is an ellipse in the uv-planewith area πab = π/
√65. Since J(u, v) = 1, the area of Ω is also π/
√65.
26.∫∫Sa
e−(x−y)2
1 + (x + y)2dx dy =
12
∫∫Γ
e−u2
1 + v2du dv
where Γ is the square in the uv-plane with vertices (−2a, 0), (0, −2a), (2a, 0), (0, 2a).Γ contains the square −a ≤ u ≤ a, −a ≤ v ≤ a and is contained in the square−2a ≤ u ≤ 2a, −2a ≤ v ≤ 2a. Therefore
12
∫ a
−a
∫ a
−a
e−u2
1 + v2du dv ≤ 1
2
∫∫Γ
e−u2
1 + v2dudv ≤ 1
2
∫ 2a
−2a
∫ 2a
−2a
e−u2
1 + v2du dv.
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SECTION 17.10 909
The two extremes can be written
12
(∫ a
−a
e−u2du
) (∫ a
−a
11 + v2
dv
)and
12
(∫ 2a
−2a
e−u2du
) (∫ 2a
−2a
11 + v2
dv
).
As a → ∞ both expressions tend to 12 (
√π) (π) = 1
2π3/2. It follows that∫ ∞
−∞
∫ ∞
−∞
e−(x−y)2
1 + (x + y)2dx dy =
12π3/2.
27. J = abcρ2 sinφ; V =∫ 2π
0
∫ π
0
∫ 1
0
abcρ2 sinφdρ dφ dθ =43πabc
28. x = y = 0
zV =∫ 2π
0
∫ π/2
0
∫ 1
0
(cρ cosφ)abcρ2 sinφdρ dφ dθ =πabc2
4=⇒ z =
38c .
29. V =23πabc, λ =
M
V=
3M2πabc
Ix =3M
2πabc
∫ 2π
0
∫ π/2
0
∫ 1
0
(b2ρ2 sin2 φ sin2 θ + c2ρ2 cos2 φ
)abcρ2 sinφdρ dφ dθ
= 15 M(b2 + c2
)Iy = 1
5 M(a2 + c2
), Iz = 1
5 M(a2 + b2
)30. I =
∫ 2π
0
∫ 1
0
∫ π
0
ρ2(abcρ2 sinφ) dφ dρ dθ =45πabc
PROJECT 17.10
1. (a) θ = tan−1
[(aybx
)1/α], r =
[(xa
)2/α+(yb
)2/α]α/2(b) ar1(cos θ1)α = ar2(cos θ2)α
br1(sin θ1)α = br2(sin θ2)α
r1 > 0, 0 < θ < 12π
⎫⎪⎪⎬⎪⎪⎭ =⇒ r1 = r2, θ1 = θ2
2. J = abαr cosα−1 θ sinα−1 θ
3. (a)
x
y
a
a
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910 REVIEW EXERCISES
(b) x = ar cos3 θ, y = ar sin3 θ; x23 + y
23 = a
23 =⇒ r = 1 and x = a cos3 θ, y = a sin3 θ
A =∫ 0
π2
y(θ)x′(θ) dθ =∫ 0
π2
a sin3 θ(3a cos2 θ[− sin θ]) dθ
= 3a2
∫ π2
0
sin4 θ cos2 θ dθ = 3a2
∫ π2
0
(sin4 θ − sin6 θ) dθ
= 3a2
[3 · 14 · 2
π
2− 5 · 3 · 1
6 · 4 · 2π
2
](See Exercise 62(b) in 8.3)
=3a2π
32
(c) Entire area enclosed: 4 · 3a2π
32=
3a2π
8
4. (a)
a = 3, b = 2 a = 2, b = 3
(b) From Problem 2, Jacobian J = 8abr cos7 θ sin7 θ
A =∫ π
2
0
∫ 1
0
8abr cos7 θ sin7 θ dθ = 4ab∫ π
2
0
cos7 θ sin7 θ dθ =ab
70
REVIEW EXERCISES
1.∫ 1
0
∫ √y
y
xy2 dx dy =∫ 1
0
[12x2y2]√y
ydy =
∫ 1
0
(12y3 − 1
2y4
)dy =
[18y4 − 1
10y5]10
=140
2.∫ 1
0
∫ y
−y
ex+y dx dy =∫ 1
0
[ex+y]y−y
dy =∫ 1
0
(e2y − 1) dy =[12e2y − y
]10
=e2
2− 3
2
3.∫ 1
0
∫ 3x
x
2yex3dy dx =
∫ 1
0
[y2ex
3]3xx
dx =∫ 1
0
(9x2ex3 − x2ex
3) dx =
[3ex
3 − 13ex
3]10
=83e− 8
3
4.∫ 2
1
∫ lnx
0
xey dy dx =∫ 2
1
[xey]lnx
0dx =
∫ 2
1
x(x− 1) dx =[13x3 − 1
2x2]21
=56
5.∫ π/4
0
∫ 2 sin θ
0
r cos θ dr dθ =∫ π/4
0
[12r2 cos θ
]2 sin θ
0dθ =
∫ π/4
0
2 sin2 θ cos θ dθ =[23
sin3 θ]π/40
=√
26
6.∫ 2
−1
∫ 4
0
∫ 1
0
xyz dx dy dz =∫ 2
−1
∫ 4
0
[12x2yz]10dy dz =
∫ 2
−1
∫ 4
0
12yz dy dz =
∫ 2
−1
4z dz = 6
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REVIEW EXERCISES 911
7.∫ 2
0
∫ 2−3x
0
∫ x+y
0
x dz dy dx =∫ 2
0
∫ 2−3x
0
[xz]x+y
0dy dx =
∫ 2
0
∫ 2−3x
0
(x2 + xy) dy dx
=∫ 2
0
[x2y +
12xy2
]2−3x
0
dx =∫ 2
0
(32x3 − 4x2 + 2x
)dx
=[38x4 − 4
3x3 + x2
]20
= −23
8. ∫ π2
0
∫ π2
z
∫ sin z
0
3x2 sin y dx dy dz =∫ π
2
0
∫ π2
z
[x3 sin y
]sin z
0dy dz =
∫ π2
0
∫ π2
z
sin3 z sin y dy dz
=∫ π
2
0
sin3 z cos z dz =14
sin4 z]π
2
0=
14
9.∫ 0
−π2
∫ 2 sin θ
0
∫ r2
0
r2 cos θ dz dr dθ =∫ 0
−π2
∫ 2 sin θ
0
r4 cos θ dr dθ =∫ 0
−π2
325
sin5 θ cos θ dθ = −1615
10.∫ π
2
−π6
∫ π2
0
∫ 1
0
ρ3 sinϕ cosϕdρ dθ dϕ =∫ π
2
−π6
∫ π2
0
14
sinϕ cosϕdθ dϕ
=∫ π
2
−π6
π
16sin 2ϕdϕ = −
[ π32
cos 2ϕ]π
2
−π6
=3π64
11.∫ 1
0
∫ 1
y
ex2dx dy =
∫ 1
0
∫ x
0
ex2dy dx =
∫ 1
0
ex2y∣∣∣x0dx =
∫ 1
0
xex2dx =
12ex
2∣∣∣10
=e− 1
2
12.∫ 2
0
∫ 1
x2
cos y2 dy dx =∫ 1
0
∫ 2y
0
cos y2 dx dy =∫ 1
0
x cos y2∣∣∣2y0
dy =∫ 1
0
2y cos y2dy = sin y2∣∣∣10
= sin 1
13.∫ 1
0
∫ √1−y2
0
1√1 − y2
dxdy =∫ 1
0
dy = 1
14.∫ 1
0
∫ 1−x
0
y cos(x + y) dy dx =∫ 1
0
∫ 1−y
0
y cos(x + y) dx dy
=∫ 1
0
y sin(x + y)]1−y
0dy =
∫ 1
0
y(sin 1 − sin y) dy
=[12y2 sin 1 + y cos y − sin y
]10
= cos 1 − 12
sin 1
15.∫ 1
0
∫ √1−x2
0
xy dy dx =∫ 1
0
[12xy2]√1−x2
0dx =
∫ 1
0
(12x− 1
2x3
)dx =
18
16.∫ √
3
−√
3
∫ 4−y2
y2/3
(x− y) dx dy =∫ √
3
−√
3
[x2
2− xy)
]4−y2
y2/3dy =
∫ √3
−√
3
(49y4 +
43y3 − 4y2 − 4y + 8
)dy =
48√
35
17.∫ 2
0
∫ 3x−x2
x
(x2 − xy) dy dx =∫ 2
0
[x2y − 1
2xy2]3x−x2
xdx =
∫ 2
0
(2x4 − 2x3 − 1
2x5
)dx = − 8
15
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912 REVIEW EXERCISES
18.∫ 2
0
∫ 2−x
0
x(x− 1)exy dy dx =∫ 2
0
[(x− 1)exy
]2−x
0dx =
∫ 2
0
(x− 1)e2x−x2dx−
∫ 2
0
(x− 1) dx
= −[12e2x−x2
]20−[12x2 − x
]20
= 0
19.∫ 2
0
∫ y
0
∫ √4−y2
0
2xyz dz dx dy =∫ 2
0
∫ 2
x
xyz2∣∣∣√4−y2
0dx dy =
∫ 2
0
∫ 2
0
xy(4 − y2) dx dy
=∫ 2
0
[12y3(4 − y2)
]dy =
83
20.∫∫∫T
z dxdydz = 2∫ 1
0
∫ x
0
∫ 1−x
0
z dz dy dx =∫ 1
0
∫ x
0
(1 − x)2 dy dx =∫ 1
0
(1 − x)2x dx =112
21.∫ 2
0
∫ √4−x2
0
∫ √4−x2−y2
0
xy dz dy dx =∫ 2
0
∫ √4−x2
0
[xyz]√4−x2−y2
0dy dx
=∫ 2
0
[− 1
3x√
4 − x2 − y2]√4−x2
0dx
=∫ 2
0
13x(4 − x2)
32 dx =
3215
22.∫∫∫T
(x2 + 2z) dx dy dz =∫ 2
−2
∫ 4
x2
∫ 4−y
0
(x2 + 2z) dz dy dx
=∫ 2
−2
∫ 4
x2(4x2 − x2y + 16 − 8y + y2) dy dx
=∫ 2
−2
(16x6 − 8x2 +
643
)dx =
210
21
23.∫ 2
0
∫ √4−y2
0
e√
x2+y2dx dy =
∫ π/2
0
∫ 2
0
err dr dθ =π
2
∫ 2
0
rerdr =π
2
[rer − er
]20
=π
2(e2 + 1)
24.∫ 1
−1
∫ √1−x2
0
arctan (y/x) dy dx =∫ π/2
0
∫ 1
0
rθ dr dθ +∫ π
π/2
∫ 1
0
(θ − π)r dr dθ
=∫ π/2
0
θ
2dθ +
∫ π
π/2
12(θ − π) dθ = 0
25. V =∫ 3
0
∫ 2π
0
(9 − r2)r dr dθ = 2π∫ 3
0
(9 − r2)r dr =81π2
26.∫ 1
0
∫ √x
x2(2 − x2 − y2) dy dx = −
∫ 1
0
(2x1/2 − x5/2 − 1
3x3/2 − 2x2 + x4 + 1
3x6)dx =
52105
27. V =∫ 1
0
∫ 1−x
0
(x2 + y2) dy dx =∫ 1
0
[x2 − x3 +
13(1 − x)3
]dx =
[13x3 − 1
4x4 − 1
12(1 − x)4
]10
=16
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REVIEW EXERCISES 913
28. V =∫ 3
0
∫ π/2
0
r2 sin θ dr dθ =∫ π/2
0
9 sin θ dθ = 9
29. M =∫ π/2
−π/2
∫ cosx
0
y dy dx =∫ π/2
−π/2
12
cos2 x dx =π
4
xMM =∫ π/2
−π/2
∫ cosx
0
xy dy dx = 0 by symmetry
yMM =∫ π/2
−π/2
∫ cosx
0
y2 dy dx =∫ π/2
−π/2
13
cos3 x dx =49
The center of mass is: (0, 169π )
30. M =∫ 1
0
∫ y
y22x dx dy =
∫ 1
0
(y2 − y4) dy =215
xMM =∫ 1
0
∫ y
y22x2 dx dy =
∫ 1
0
(23y3 − 2
3y6
)dy =
114
yMM =∫ 1
0
∫ y
y22xy dx dy =
∫ 1
0
(y3 − y5) dy =112
The center of mass is: (15/28, 5/8)
31. M =∫ π/2
0
∫ R
r
u3 du dθ =π
8(R4 − r4); (polar coordinates [u, θ])
By symmetry, x = y.
xMM =∫ π/2
0
∫ R
r
u4 cos θ du dθ =15(R5 − r5); xM =
8(R5 − r5)5π(R4 − r4)
32. M =∫ π
0
∫ 2(1+cos θ)
0
r2 dr dθ =∫ π
0
83(1 + cos θ)3 dθ =
203π
xMM =∫ π
0
∫ 2(1+cos θ)
0
r3 cos θ dr dθ =∫ π
0
4(1 + cos θ)4 cos θ dθ = 14π
yMM =∫ π
0
∫ 2(1+cos θ)
0
r3 sin θ dr dθ =∫ π
0
4(1 + cos θ)4 sin θ dθ = −45(1 + cos θ)5|π0 =
1285
The center of mass is: ( 2110 ,
9625π )
33. Introduce a coordinate system as shown in the figure.
(a) A = 12bh; by symmetry, x = 0
y A =∫ 0
−b/2
∫ 2hb (x+ b
2 )
0
y dy dx +∫ b
2
0
∫ − 2hb (x− b
2 )
0
y dy dx
=bh2
6=⇒ y =
h
3 x
y
−b 2 b 2
h
(b) I =∫ 0
−b/2
∫ 2hb (x+ b
2 )
0
λy2 dy dx +∫ b/2
0
∫ − 2hb (x− b
2 )
0
λy2 dy dx =λbh3
12= 1
6Mh2
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914 REVIEW EXERCISES
(c) I = 2∫ b/2
0
∫ − 2hb (x− b
2 )
0
λx2 dx dy =148
λhb3 = 124Mb2
34. Let λ =k√
x2 + y2
(a) M =∫ π
0
∫ R
r
k dr dθ = kπ(R− r)
By symmetry, xM = 0; yMM =∫ π
0
∫ R
r
kr sin θ dr dθ = k(R2 − r2)
The center of mass is: (0, R+rπ )
(b) Ix =∫ π
0
∫ R
r
kr2 sin2 θ dr dθ =k
3(R3 − r3)
∫ π
0
sin2 θ dθ =kπ
6(R3 − r3)
(c) Iy =∫ π
0
∫ R
r
kr2 cos2 θ dr dθ =k
3(R3 − r3)
∫ π
0
cos2 θ dθ =kπ
6(R3 − r3)
35. V =∫ 2
0
∫ x
0
∫ 2x+2y+1
0
dz dy dx =∫ 2
0
∫ x
0
(2x + 2y + 1) dy dx = 10
36. V =∫ 1
0
∫ x
x2
∫ 4(x2+y2)
−1
dz dy dx =∫ 1
0
∫ x
x2(4x2 + 4y2 + 1) dy dx
=∫ 1
0
(x +
163x3 − x2 − 4x4 − 4
3x6
)dx =
107210
37. The curve of intersection of the two surfaces is the circle: x2 + y2 = 4, x = 3
V =∫ 2
−2
∫ √4−x2
−√
4−x2
∫ 12−x2−2y2
2x2+y2dz dy dx =
∫ 2
−2
∫ √4−x2
−√
4−x23(4 − x2 − y2
)dy dx
= 3∫ 2π
0
∫ 2
0
(4 − r2
)r dr dθ
= 3∫ 2π
0
[2r2 − 1
4 r4]20dθ = 12
∫ 2π
0
= 24π
38. V =∫ 1
0
∫ √1−y2
0
∫ (2−y−z)/2
0
dx dz dy =∫ 1
0
∫ π/2
0
2 − r cos θ − r sin θ
2r dθ dr
=∫ 1
0
12(π − 2r)r dr =
3π − 412
39. V =∫ 2π
0
∫ π/3
0
∫ 2
secφ
ρ2 sinφdρ dφ dθ =∫ 2π
0
∫ π/3
0
[13 ρ
3]2secφ
dφ dθ
=13
∫ 2π
0
∫ π/3
0
(8 − sec3φ
)sinφdφ dθ
=13
∫ 2π
0
[− 8 cosφ− 1
2 sec2 φ]π/30
dθ
=13
(52
)(2π) =
5π3
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REVIEW EXERCISES 915
40. V =∫ 4
0
∫ (12−3x)/4
0
∫ 16−x2
0
dz dy dx=∫ 4
0
∫ 12−3x4
0
(16 − x2) dy dx=34
∫ 4
0
(64 − 16x− 4x2 + x3) dx = 80
41. V =∫ 2π
0
∫ π/2
π/4
∫ 1
0
ρ2 sinφdρ dφ dθ =∫ 2π
0
∫ π/2
π/4
13
sinφdφ dθ =√
2π3
42. V =∫ 2π
0
∫ π/4
0
∫ 1
0
ρ2 sinφdρ dφ dθ =2 −
√2
3π
43. (a) V =∫ 1
0
∫ x
0
∫ √1−x2
0
dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
dz dx dy
= 2∫ 1
0
∫ x
0
√1 − x2 dy dx = 2
∫ 1
0
x√
1 − x2dx =23
By symmetry, x = y.
xV =∫ 1
0
∫ x
0
∫ √1−x2
0
x dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
x dz dx dy
For the first integral:
∫ 1
0
∫ x
0
∫ √1−x2
0
x dz dy dx =∫ 1
0
∫ x
0
x√
1 − x2 dy dx
=∫ 1
0
x2√
1 − x2 dx =∫ π/2
0
sin2 u cos2 u du =π
16
x = sinu∧
For the second integral:∫ 1
0
∫ y
0
∫ √1−y2
0
x dz dx dy =∫ 1
0
∫ y
0
x√
1 − y2dx dy =∫ 1
0
12y2√
1 − y2 dy =π
32
Thus, xV =3π32
=⇒ x = y =9π64
Now calculate z:
z V =∫ 1
0
∫ x
0
∫ √1−x2
0
z dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
z dz dx dy;
∫ 1
0
∫ x
0
∫ √1−x2
0
z dz dy dx =∫ 1
0
∫ x
0
12(1 − x2)dy dx =
12
∫ 1
0
(x− x3
)dx =
18
and similarly,∫ 1
0
∫ y
0
∫ √1−y2
0
z dz dy dx =18.
Therefore, zV =14
=⇒ z =38
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916 REVIEW EXERCISES
(b) Iz =∫ 1
0
∫ x
0
∫ √1−x2
0
λ(√
x2 + y2)2
dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
λ(√
x2 + y2)2
dz dx dy;
∫ 1
0
∫ x
0
∫ √1−x2
0
λ(√
x2 + y2)2
dz dy dx =∫ π/4
0
∫ sec θ
0
∫ r sin θ
0
λr3dz dr dθ =320
λ
and∫ 1
0
∫ y
0
∫ √1−y2
0
λ(√x2 + y2)2dz dx dy =
320
λ =⇒ Iz =310
λ
44. (a) V =∫ 2π
0
∫ 1
0
(r − r2)r dr dθ = 2π∫ 1
0
(r2 − r3) dr =π
6By symmetry, x = y = 0
zV =∫ 2π
0
∫ 1
0
∫ r
r2zr dz dr dθ =
π
12and hence z = 1
2
(b) Iz = K
∫ 2π
0
∫ 1
0
∫ r
r2r2 dz dr dθ = 2πK
∫ 1
0
r2(r − r2)dr =πK
10
Here, K is the density of the mass.
45. Denote polar coordinates by [u, θ].
(a) M =∫ 2π
0
∫ r
0
∫ h
0
u3 dz du d θ = 2πh∫ r
0
u3 du =πhr4
2
(b) By symmetry, xM = yM = 0
(c) zMM =∫ 2π
0
∫ r
0
∫ h
0
u3z dz du dθ =πh2r4
4=⇒ zM = h/2
46. λ =√x2 + y2
(a) M =∫ 2π
0
∫ π/2
0
∫ r
0
ρ3 sin2 φdρ dφ dθ =r4π
2
∫ π2
0
sin2 φdφ =r4π2
8
(b) By symmetry, xM = yM = 0
zMM =∫ 2π
0
∫ π/2
0
∫ r
0
ρ4 sin2 φ cosφdρ dφ dθ =2r5π
15=⇒ zM =
16r15π
47. (a) M =∫ 1
0
∫ 2π
0
∫ 1
r
r2 dz dθ dr = 2π∫ 1
0
∫ 1
r
r2 dz dr = 2π∫ 1
0
r2(1 − r) dr =π
6
(b) By symmetry, xM = yM = 0
zMM =∫ 1
0
∫ 2π
0
∫ 1
r
r2z dz dθ dr = π
∫ 1
0
r2(1 − r2) dr =2π15
=⇒ zM =45
(c) Iz =∫ 1
0
∫ 2π
0
∫ 1
r
r4 dz dθ dr =π
15
48. J(u, v) =
∣∣∣∣∣ 2u 2v
−2v 2u
∣∣∣∣∣ = 4u2 + 4v2
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REVIEW EXERCISES 917
49. J(u, v) =
∣∣∣∣∣ eu cos v eu sin v
−eu sin v eu cos v
∣∣∣∣∣ = e2u
50. Ju, v, w) =
∣∣∣∣∣∣∣∣2u 2w vw
2w 2v uw
2v 2u uv
∣∣∣∣∣∣∣∣ = (vw − u2)(4uw − 4v2)
51. Set x =v − u
2, y =
v + u
2=⇒ u = y − x, v = y + x, 1 ≤ v ≤ 2, J = −1
2at x = 0, y = u, y = v =⇒ u = v
at y = 0, −x = u, x = v =⇒ u = −v∫∫Ω
cos(y − x
y + x
)dx dy =
∫ 2
1
∫ v
−v
12
cos(uv
)du dv =
∫ 2
i
v sin 1 dv =32
sin 1
52. By the hint, J =
∣∣∣∣∣∣∣∣∣∣∣
cos θu
sin θ
u2r
−r cos θu2
−r sin θ
u20
−r sin θ
u
r cos θu
0
∣∣∣∣∣∣∣∣∣∣∣= −2r3
u3
∫∫∫T
dx dy dz =∫ 2π
0
∫ 2
1
∫ 2
1
2r3
u3dr du dθ =
45π8