Thomas Calculus 13th [Solutions]

482 Chapter 6 Applications of Definite Integrals
7. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
cos x
of mass: ( x, y) x , , length: cos x, width: dx,
2
area: dA cos x dx , mass: dm dA cos x dx . The moment
of the strip about the x -axis is y dm cos x cos x dx
2
2
1 cos 2x
cos x dx dx (1 cos 2 x) dx ; thus,
2 2 2 4
/2 sin 2
/2
6
/2 4 (1 cos 2 ) x
M x y dm x dx x
4 2
/2
4 2 0 2 4
; M dm
/2 /2
/2 cos x dx sin x /2
2 . Therefore, M
y x
4 2 8 ( x, y ) 0,
M
8 is the center of mass.
8. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
sec x
of mass: ( x, y) x , , length:
2
area: dA
2
sec x dx , mass:
moment about the x -axis is
2
2
sec x , width: dx,
2
dm dA sec x dx . The
y dm
sec 2 x 2
( sec x)
dx
2
4
2 sec x dx .
/4 /4 4 /4 2 2
M x y dm sec x dx tan x 1 sec x dx
/4 2 /4 2 /4
3 /4
/4 2 2 /4 2
(tan x) /4
tan x sec x dx sec x dx tan x
2 /4 2 /4
2 3 2 /4
/4
1 1
4
/4 2
/4
1 ( 1) ; M dm x dx x
2 3 3 2 3 3
/4 /4
M x
Therefore, y
4 1 2
( x, y ) 0,
2
is the center of mass.
M
3 2 3 3
sec tan 1 ( 1) 2 .
9.
2
1
y
1,
1 x
2 2
1 1 1
2
1 1 1
x ,
1 2x x 2 1
2
x 2x
1 4
2
1
2
M
ln ln 2
y
1 x
1
M ln 2
M x dx
M dx dx
M dx x X
1
1.44 and
y
M x
M
1
4
ln 2
0.36
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