Thomas Calculus 13th [Solutions]

460 Chapter 6 Applications of Definite Integrals
25.
26.
x
dy
/4 2 /4 /4 2
y cos 2t dt cos 2x L 1 cos 2x dx 1 cos 2x dx 2cos x dx
0 dx
0 0 0
/4 /4
2 cos x dx 2 sin x 2 sin 2 sin(0) 1
0
0 4
2/3
1/2
2/3
1/2 2
2/3
3/2 1/2 1 1
2 dy 2/3 1/3
x
3 2
1
x
y 1 x , x 1 1 x x L 1
dx
4 dx 2 3 1/3
2 /4
1/3
x x
1 2/3 1 1 1 1 1/3 2/3
1 1 x dx 1 1 1 dx 1 dx 1 dx x dx 3 x
2/3 2/3 2/3 1/3
2 /4 x 2/4 x 2/4 x 2 /4 x
2/4 2 2 /4
2/3
3 2/3 3 2
(1) 3 3 1 3 total length 8 3 6
2 2 4 2 2 2 4
4
1
27.
3 2 ,0 2 dy
2 2 2
2
y x x 2 L
0 1 ( 2) dx
dx
0 5 dx 5 x 2 5. 0
d
2 2
(2 0) (3 ( 1)) 2 5
28. Consider the circle
result by 4.
2 2
x y r 2 , we will find the length of the portion in the first quadrant, and multiply our
2
y r 2 x 2
2 2
, 0 x r dy x
r
4 1
x
r
4 1
x
r
4
r
dx L 2 2 0 dx 2 2
0 dx 2 2
0
dx
2 2
r x r x
r x r x
r
4
r
r
dx 4r
dx
0 2 2 0 2 2
r x r x
29.
2 2 d 2 d 2 dx
2
x y y x y y x y y y y y
dy dy dy
9 ( 3) 9 ( 3) 18 2 ( 3) ( 3) 3( 3)( 1)
dx ( y 3)( y 1) ( y 3)( y 1)
dx
dy
dy 6x 6x
2 2 2 2 2
;
ds 2
2 2
2 dx 2 dy 2 ( y 3)( y 1) 2 ( y 3) ( y 1) 2 2
6x
dy dy dy dy
2
36x
( y 3) ( y 1) 2 2 ( y 1) 2 y 2y 1 4 y 2 ( y 1) 2
dy dy 1 dy dy dy
2
4 y( y 3)
4 y 4 y 4 y
30.
2 2 2 2 dy dy
x y d x y d x y 4x dy 4x
dx
dx dx dx dx y y
4 64 4 64 8 2 0 ;
2
2 2 2 2
2 2 2 2 16
2 2
4 2 16 2 16 2 y x 2 2
ds dx dy dx x dx dx x dx 1
x dx dx 4x 64 16x
dx
y 2 2 2 2
y y y y
2
x
y
dx 2 x 2 dx
2
y
20 64 4 (5 16)
2 2
31.
x dy
2
dy
2
dy
0 dt dx dx
2 x 1 dt , x 0 2 1 1 y f ( x ) x C where C is any real
number.
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