Thomas Calculus 13th [Solutions]

Section 15.6 Moments and Centers of Mass 1121
1 1 4 1 1 6
1
4 2 2 2 64 4 64
1976 7904
2
0 0 4 4 0 0 4 3 4 y
Ix
y z dz dy dx y y dy dx
y
3 4 dx
0 105 105
;
1 1 4 1 1 6
1
4 2 2 2 64 2 4 64
8 2 128 4832
2
0 0 4 4 0 0 4 3 4 y
I y
x z dz dy dx x x y dy dx
y
3 4 x dx
0 3 7 63
;
1 1 4 2 2 1 1 2 2 2 2 4 1 2
I 2 2 256
z 4 2 x y dz dy dx
0 0 4 16 x x y y y dy dx
0 0 16 x dx
y
0 3 15 45
24. (a)
(b)
2 2
M 2 4 x 2 2 x 2 4 x 2 2 2
2 2 (2 ) (2 ) 4 4 ;
2 4 x 2 0 dz dy dx 2 4 x 2 x dy dx 2
x x dx
2 2
M 2 4 x 2 2 x 2 4 x 2 2 2
yz
2 2 (2 ) (2 ) 4 2 ;
2 4 x 2 0 x dz dy dx 2 4 x 2 x x dy dx 2
x x x dx
2 2
M 2 4 x 2 2 x 2 4 x 2 2
2 2
1
4 4
2 2
2 4 2 0 2 4 2 (2 ) 2 2
(2 ) x x
xz
x
y dz dy dx x
y x dy dx x 4 4
dx 0
x 1 and y 0
2
2 2
M 2 4 x 2 2 x
1
2 4 x 2 2 1
2 2 2
xy
2 2 2
2 4 x 2 0 z dz dy dx 2 4 x 2 (2 x) dy dx 2 2
(2 x) 4 x dx
5 z 5
4
25. (a)
(b)
2
2 4 x 4 /2 2 4 /2 2 3
/2
M 4 2 2 dz dy dx 4 2 r dz dr d 4 4r r dr d 4 4 d 8 ;
0 0 x y 0 0 r
0 0 0
2 2 4 2 2 4 2
M r
32 64 8
xy
2 zr dz dr d
0 0 r
0 0 2 16 r dr d d z
3 0 3 3
, and x y 0,
by symmetry
8 4 2 c c 2 c 3 2 2 2
2
M 2 r dz dr d cr r dr d c d c c
0 0 r
0 0 0 4 2
8 c 2 2,
since c 0
26. M 8; M xy
1 5 1 1 5
z
1 5 1 1 5
z dz dy dx dy dx 0; M
1 3 1 1 3 2
yz x dz dy dx 2 x dy dx
1
1 3 1 1 3
1
1 5 1 1 5 1
4 x dx 0; M
1
xz y dz dy dx 2 y dy dx 16 dx
1 3 1 1 3 1
32 x 0, y 4, z 0;
Ix
1 5 1 2 2 1 5 2 2 2
1
y z dz dy dx 400
1 3 1 1 3 y dy dx
3 3 1
dx
3
I y
1 5 1 2 2 1 5 2 2 4
1 2
x z dz dy dx 16
1 3 1 1 3 x dy dx
3 3 1
x 1 dx
3
Iz
1 5 1 2 2 1 5 2 2 1 2
x y dz dy dx 2 x y dy dx 2 2x 98 dx 400
1 3 1 1 3 1 3 3
2 1
2 4 (2 y)/2 2 2
27. The plane y 2z 2 is the top of the wedge IL
( y 6) z dz dy dx
2 2 1
2 4 2 3
( y 6) (4 y) (2 y) 1
4 3
2
dy dx ; let t 13 49
2 2 2 24 3
2 y I 4 t
L
2 24 5 t 16 t dt
3
1386;
M
1 (3)(6)(4) 36
2
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