Thomas Calculus 13th [Solutions]

Section 6.6 Moments and Centers of Mass 485
16. We use the vertical strip approach:
1 x x
2
M x y dm x x dx
0 2
1
1 2 4 1 3 5
x x 12x dx 6 x x dx
2 0 0
4 6
6 x x
1 6 1 1 6 1 1 ;
4 6
0
4 6 4 2
2
1 2 1 2 3 1 3 4
M 12 12 12 x x 1 1
y x dm x x x dx x x x dx x x dx 12
12 3
0 0 0 4 5
0
4 5 20 5 ;
3 4 1
4 5 1
1 2 1 2 3
M dm x x dx 1 1 12
0 12 x x dx
0 12 x x
3 4 12 0
3 4 12
1. So x 3 and
M 5
M
y x 1 3 , 1 is the center of mass.
M 2 5 2
b shell shell 4 4
17. (a) We use the shell method: V 2 4 4
radius height dx 2 x dx 16 x dx
a 1 x x 1 x
(b)
4 1/2 3/2
4
16 x dx 16 2 x 16 2 8 2 32 (8 1) 224
1 3 1 3 3 3 3
Since the plate is symmetric about the x -axis and its density ( x ) 1 is a function of x alone, the
x
distribution of its mass is symmetric about the x -axis. This means that y 0. We use the vertical strip
approach to find
8(2 2 2) 16;
x : M 4 4 4 1/2 1/2
4
4 4 8 1
y x dm 1 x dx 1 x x
dx 8 1 x dx 8 2x
x x x
1
M dm 4 4 4 3/2 1/2
4
4 4 1 1
1 dx 8 1 x
dx 8 1 x dx 8 2x
x x x
1
M y
(c)
M y
8 1 ( 2) 8. So x 16 2 ( x, y ) (2,0) is the center of mass.
M 8
18. (a) We use the disk method: V b 2 4
4
4 2
1
4
1
a
R( x) dx 1 dx 4 2
1 x dx 4 x
4 1 4
( 1)
x
[ 1 4] 3
(b)
4 4
We model the distribution of mass with vertical strips: M 2 2
x y dm dx x dx
1 2 x 1
2
x
4 4
3/2 2
4 4 3/2 4
2 x dx 2 2 1 ( 2) 2; M 2 2 1/2
2 2 x
1 x
y x dm x dx x dx
1
1 x
1 3 1
16 2 28
4 4
2 ;
2
4 4 1/2 1/2
3 3 3
M dm x
2 2 2 2 2(4 2) 4.
1 x
dx 1 x
dx 1 x dx x 1
So
28
3 7
x M y
M 4 3
and M
y x 2 1 ( x, y ) 7 , 1 is the center of mass.
M 4 2 3 2
2
x
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