Thomas Calculus 13th [Solutions]

496 Chapter 6 Applications of Definite Integrals
14. disk method:
/4 2 /4 2
/4
V x dx x dx x x
0 0
0
2 4 tan 8 sec 1 8 tan 2 (4 )
15. The material removed from the sphere consists of a cylinder
and two caps. From the diagram, the height of the cylinder
2
2
2
is 2h, where h 3 2 , i.e. h 1. Thus
cy1
2
3
V (2 h ) 3 6 ft . To get the volume of a cap,
use the disk method and
2 2 2
x y
3 2
2 :
V
cap
2 2
x dy
1
2 2 y
4 y dy 4y
8
8
4
1
1
3 3 3
1
3
Vremoved Vcy1 2Vcap 6
3
53 ft . Therefore, 10
28
3
3 ft .
16. We rotate the region enclosed by the curve
b
2
4
121
y 12 1
x
and the x -axis around the x -axis. To find the
2
2 11/2 2 11/2
2
4x
4x
11/2 121 11/2 121
volume we use the disk method: V R( x) dx 12 1 dx 12 1 dx
a
17.
11/2
2 3 11/2 3
2
4x
dx x 4x
11 4 11 4 11 1
11/2 121 363
11/2
2 363 2 363 4 3
12 1 12 24 132 1 132 1
264
3
88 276 in
3
3/2 2
1/2 dy 1 1/2 1 1/2 dy 1 1
4
y x x
x x 2 x L 1
1 1
2 x dx
3 dx 2 2 dx 4 x 1 4 x
4 4 1/2 1/2
2 4 1/2 1/2 1/2 3/2
4
L 1 1
2 x dx 1 x x dx 1 x x dx 1
2x 2 x
1 4 x
1 4 1 2 2 3 1
1
4
2
8 2
2 1
2
14 10
2 3 3 2 3 3
18.
2 2/3
2
2/3 2 1/3
4 y
8 8
4
8 9 y 4
x y dx y dx L 1
dx dy 1 dy dy
dy 3 dy 9 1 dy 1
2/3
1
1/3
9 y 3y
2/3
1
8 2/3 1/3
2/3 1/3
9y 4 y dy;
[ u 9y 4 du 6 y dy; y 1 u 13, y 8 u 40]
3 1
40 1/2 3/2
40
L 1 u du 1 2 u 1
40 3/2
13 3/2
7.634
18 13 18 3 13 27
19.
2
y x ln x y 2x
1
8
8x
4 2 2 2
2
2 2
1 256 32 1 (16 x 1)
y x x x 16x
1 x 1
8x 2 2
64 x
(8 x)
8x 8x
1 ( ) 1 2 2
2 2
2
2 1 2
y dx x dx x 1 x 1 1 1
1 1 8x
8
1
8 8 8
Length 1 ( ) 2 ln 4 ln 2 1 ln1 3 ln 2
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