Thomas Calculus 13th [Solutions]

Section 6.6 Moments and Centers of Mass 487
24. From the symmetry about the line
follows that the line through the points
y
1 ( b 0) b
3 3
( x, y) a , b .
2 3
x a it follows that
2
a
2 , 0 and
x a
2 . It also
a
2 , b is a median
25.
1/2 1 1/2 2 2
y x dy x dx ds ( dx) ( dy)
1 1 ;
2
4x dx
2
1
2
M x x 1 dx x 1 dx 2 x 1
0 4x
0 4 3 4
3/2
2
0
3/2 3/2 3/2
2 2 9
3/2
2 1 1 1 2 27 1 13
3 4 4 3 4 4 3 8 8 6
26.
3 2 2 2
y x dy 3 x dx dx ( dx) 3x dx
2
4
1 9 x dx;
1
M 3
1 9 4
x x x dx;
0
4 3 1 3
[ u 1 9x du 36 x dx du x dx;
36
10 1/2 3/2
10
1 2
3/2
x 0 u 1, x 1 u 10] M x u du u
1 36 36 3 1 54
10 1
27. From Example 4 we have
2 2
a k sin 2 a k
2 2 0 2
;
2 2
M
0 ( sin )( sin ) 0 sin a k
x a a k d a k d
2 0
(1 cos 2 ) d
2 2
M
0 ( cos )( sin ) 0 sin cos a k
y a a k d a k d
2
sin 0; 0
M ak sin d ak cos 2 ak . Therefore, x 0 and
0
0
M
the center of mass.
M y
2
2
M 2
x a k a a
y 1 0,
M 2 2ak
4 4
is
28.
M x y dm
0 ( a sin ) a d
0
2
a sin 1 k cos d
2 /2
2
a (sin )(1 k cos ) d a (sin )(1 k cos ) d
0 /
2 /2 2 /2
2 2
a sin d a k sin cos d a /2 sin d a k sin cos d
0 0 /2
2 /2
2 /2 2 2 2
2
a cos a k sin a cos a k sin
0 2 /2
0
2
/2
2 2
2 2 1 2 2 1 2 2 2 2 2
a 0 ( 1) a k 0 a ( 1) 0 a k 0 a a k a a k 2 a a k a (2 k);
2 2 2 2
2
M y x dm ( a cos ) a d a cos 1 k cos d
0 0
2 /2
2
a (cos )(1 k cos ) d a (cos )(1 k cos ) d
0 /2
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