Thomas Calculus 13th [Solutions]

1132 Chapter 15 Multiple Integrals
73. We orient the cone with its vertex at the origin and axis along the z-axis
is through the vertex and parallel to the base of the cone
2 1 1 2 2 2
0 0
. We use the -axis
4 x which
Ix = r sin z dz r dr d
2 1 3 2 4 2 2
r sin r sin
r r
dr d
sin 1
d
sin 2
0 0 3 3 0 20 10 40 80 10 0 20 5 4
4 2
2
r
74.
75.
z
2 2
2 a a r 3 2 a 3 2 2 2 2 2
2 2 2
2 2
2 2
r a
0 0 a r
0 0 0 5 15
I r dz dr d r a r dr d a r d
2
2
2 5
a d
8 a
0 15 15
z
5
2 a h 2 2 2 a h 3 2 a 3
h
hr
0 0 r
0 0 0 0
I x y dz r dr d r dz dr d hr dr d
a
h a
r r
a a ha ha
0 4 5a
d h 0 0 4 5a
d 20 d 0 10
2 4 5
2 4 5 4 2
4
a
4
hr
a
3/2
a
0
76. (a)
(b)
77. (a)
(b)
78. (a)
2
2 1 r
2 1
1 5 1
2
0 0 0 0 0 2 12 0 6
M z dz r dr d r dr d d
2
2 1 r 2 1
2 1 7
1
2
1
0 0 0 3 0 0 24 0 12 2
M z dz r dr d r dr d d z , and x y 0, by
xy
symmetry;
z
2
2 1 r 3 1
2 1 7 1
2
0 0 0 2 0 0 16 0 8
I zr dz dr d r dr d d
2
2 1 r 2 2 1 4 1
2
2
0 0 0 0 0 5 0 5
M r dz dr d r dr d d
2
2 1 r 2 1
2 1 6 1
2
5
0 0 0 2 0 0 14 0 7 14
M zr dz dr d r dr d d z , and x y 0, by
xy
symmetry;
z
2
2 1 r 4 2 1 6 1
2
2
0 0 0 0 0 7 0 7
I r dz dr d r dr d d
2 1 1
1
2 1 3
1
2
;
0 0 r
2 0 0 8 0 4
xy
2 1 1 2 1
2 1 4
1
2
4
,
0 0 r
3 0 0 10 0 5 5
x y by
Iz
2 1 1 3 1
2 1 3 5
1
2
zr dz dr d r r dr d d
0 0 r
2 0 0 24 0 12
M xy
2 1 1 3 1
2 1
z dz r dr d r
0 0 r
4 0 0
5
r dr d
1
2
5
2 1 1
d z , and x y 0, by symmetry; I 2 3
12 0 6 6
z z r dz dr d
0 0 r
1
2 1 3 6
1
2
r r dr d d
3 0 0 28 0 14
M z dz r dr d r r dr d d
M z dz r dr d r r dr d d z and 0,
symmetry;
2 1 1
M z 2
dz r dr d from part (a);
0 0 r
5
2 a 4 a
2
2a 2
4 a
0 0 0 5 0 0 5 0 5
5 5 5
M sin d d d sin d d d ;
;
;
z
2 a
7 2 7 2 3
cos
6 3 2
sin
a
1 cos sin
a
cos
0 0 0 7 0 0 7 0
3
I d d d d d d
7 2
7
4a
d
8a
21 0 21
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