Thomas Calculus Early Transcendentals 13Th Edition Thomas Solutions Manual Full Chapter PDF
Thomas Calculus Early Transcendentals 13Th Edition Thomas Solutions Manual Full Chapter PDF
Thomas Calculus Early Transcendentals 13Th Edition Thomas Solutions Manual Full Chapter PDF
x x
2
(diagonal)2 b 4 4
1. A( x) 2 x; a 0, b 4; V A( x) dx 2 x dx x 2 16
2 2 a 0 0
2 2
2 x 2 x2 2 1 x 2
2. A( x)
(diameter)2
4
4
4
1 2 x 2 x 4 ; a 1, b 1;
1
b 1
V A( x) dx 1 2 x 2 x 4 dx x 23 x3 x5 2 1 23 51 16
5
a 1 1 15
2 2
3.
A( x) (edge)2 1 x 2 1 x 2 2 1 x 2 4 1 x 2 ; a 1, b 1;
1
b 1
V A( x) dx 4 1 x 2 dx 4 x x3 8 1 13 16
3
a 1 1 3
2 1 x
2 2
1 x 2 1 x 2 2
4. A( x)
(diagonal)2
2
2
2 2
2 1 x 2 ; a 1, b 1;
1
1 x 2 dx 2 x x3 4 1 13 83
b 1 3 1
V A( x) dx 2
a 1
5. (a) STEP 1) A( x) 12 (side) (side) sin 3 12 2 sin x 2 sin x sin 3 3 sin x
STEP 2) a 0, b
b
STEP 3) V A( x) dx 3 sin x dx 3 cos x 3(1 1) 2 3
a 0 0
(b) STEP 1) A( x) (side)2 2 sin x 2 sin x 4 sin x
STEP 2) a 0, b
b
STEP 3) V A( x) dx 4 sin x dx 4 cos x 0 8
a 0
6. (a) STEP 1) A( x)
(diameter)2
4
4 (sec x tan x)2 4 sec2 x tan 2 x 2sec x tan x
4 sec2 x sec 2 x 1 2 sin2x
cos x
STEP 2) a 3 , b 3
/3
b
STEP 3) V A( x) dx
a
/3
/3 4
2 sec2 x 1 2 sin2 x dx 4 2 tan x x 2 cos1 x
cos x /3
4 2 3 3 2 11 2 3 3 2 11 4 4 3 23
2 2
STEP 2) a 0, b 2
0
6 x1/2 3x dx 4 x3/2 32 x 2 (32 24) 0 8
b 4 4
STEP 3) V A( x) dx
a 0
2
diameter
3/ 2 1 2
2 x 2x x x 4 x x x3/2 1 x 2
(b) STEP 1) A( x) 12 2
12 2 2 4 8 4
STEP 2) a 0, b 4
0
x x3/2 14 x 2 dx 12 x 2 25 x5/2 12
0 8 3 8
b 4 4
STEP 3) V A( x) dx 8 1 x3 8 64 16 (0)
a 5 15
2
9. A( y ) 4 (diameter)2 4 5 y2 0 54 y 4 ;
d
c 0, d 2; V A( y ) dy
c
2
2 5
0 4
y5
y 4 dy 54 5 4 25 0 8
0
2 2
10.
A( y ) 12 (leg)(leg) 12 1 y 2 1 y 2 12 2 1 y 2 2 1 y 2 ; c 1, d 1;
1
d
c
1
V A( y ) dy 2 1 y 2 dy 2 y
1
y3
3
1
4 1 13 83
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have b 4 h 34 b.
h 3
The equation of the line through (5, 0) and (0, 4) is y 54 x 4, thus the length of the base 45 x 4 and
the height 3
4 54 x 4 53 x 3. Thus A( x) 12 (base) (height) 12 54 x 4 53 x 3
b 5 6 2 5
6 x2 12 x 6 and V A( x) dx x 12 x 6 dx 25
2 x3 6 x 2 6 x (10 30 30) 0 10
25 5 a 0 25 5 5 0
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar
53 y
2 d 5 9 2
triangles we have b
h
35 b 35 h. Thus A( y ) (base)2 9 y2
25
V A( y ) dy y dy
c 0 25
5
25
3 y 3 15 0 15
0
13. (a) It follows from Cavalieri’s Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus,
STEP 1) A( x) (sidelength)2 s 2 ;
STEP 2) a 0, b h;
b h
STEP 3) V A( x) dx s 2 dx s 2 h
a 0
(b) From Cavalieri’s Principle we conclude that the volume of the column is the same as the volume of the
prism described above, regardless of the number of turns V s 2 h
14. 1) The solid and the cone have the same altitude
of 12.
2) The cross sections of the solid are disks of
diameter x 2x 2x . If we place the vertex of
the cone at the origin of the coordinate system
and make its axis or symmetry coincide with
the x-axis then the cone’s cross sections will
be circular disks of diameter 4x 4x 2x
(see accompanying figure).
3) The solid and the cone have equal altitudes and
identical parallel cross sections. From
Cavalier’s Principle we conclude that the solid
and the cone have the same volume.
2
2 2 2 2
R( x) y 1 2x V R ( x) dx 1 2x dx 1 x x4 dx x x2 12
x3
2 2 2
15.
0 0 0 0
8 2
2 42 12 3
dy
2 3y 2
2 29 2
R( y ) x 2 V R( y ) dy 2
3y 2
16. y 2 dy 43 y3 43 8 6
0 0 0 4 0
17. R( y ) tan 4 y ; u 4 y du 4 dy 4 du dy; y 0 u 0, y 1 u 4 ;
2
1
0
2 1
V R( y ) dy tan 4 y dy 4
0
0
/4
tan 2 u du 4
0
/4 /4
1 sec2 u du 4 u tan u 0
4 4 1 0 4
Copyright 2014 Pearson Education, Inc.
434 Chapter 6 Applications of Definite Integrals
2
x 0 u 0, x 2 u V 18 sin 2 u du 8 u2 14 sin 2u 8 2 0 0 16
0 0
2
19. R( x) x 2 V R ( x ) dx
2
0
x2
2 2 2 2
dx x 4 dx x5 325
5
0 0 0
2
20. R( x) x3 V R( x) dx
2
0
0
2 2 2 2
dx x 6 dx x7 128
7
x3
0 0 7
3
21. R( x) 9 x 2 V R ( x) dx
2
3
1
22. R( x) x x 2 V R ( x ) dx
2
0
0
1 2 1
x x 2 dx x 2 2 x3 x 4 dx
0
1
x3 24x x5 13 12 15
3 4 5
0
(10 15 6)
30 30
/2
R( x) dx
2
23. R( x) cos x V
0
/2 /2
cos x dx sin x 0 (1 0)
0
/4
R ( x ) dx
2
24. R( x) sec x V
/4
/4 /4
sec2 x dx tan x /4 [1 (1)] 2
/4
1 1
25. R( x) e x V [ R ( x)]2 dx (e x )2 dx
0 0
1 2 x 1
e dx 2 e 2 x 2 (e 2
1)
0 0
2 1 12
(e2 1)
e 2
2e
/6
/2 /2 2 /2 /2 cos x
26. R( x) cot x V [ R( x)]2 dx cot x dx cot x dx dx [ln(sin x)] /2
/6
/6 /6 /6 sin x
ln1 ln 12 ln 2
dx
2
27. R( x)
2 x
1 V
4
1/4
[ R( x)]2 dx
4
1/4 2 x
1 4 1
4 1/4 x
dx 4 [ln x]1/4
4
4 ln 4 ln 14 2 ln 4
3 3 3
28. R( x) e x 1 V [ R( x)]2 dx (e x 1 )2 dx e 2 x 2 dx 2 [e2 x 2 ]13 2 (e4 1) 84.19
1 1 1
/4
R( x) dx
2
29. R( x) 2 sec x tan x V
0
/4
2
2 sec x tan x dx
0
0
/4
2 2 2 sec x tan x sec2 x tan 2 x dx
/4 /4
2 dx 2 2 sec x tan x dx
0 0
/4
(tan x)2 sec2 x dx
0
/4
/4 /4
2 x 0 2 2 sec x 0 tan3 x
3
0
2 0 2 2
2 1 13 (13 0)
2 2 2 11
3
/2
R ( x) dx
2
30. R( x) 2 2sin x 2(1 sin x) V
0
/2
0
4(1 sin x)2 dx 4
/2
0 1 sin 2 x 2sin x dx
/2
4 1 1 (1 cos 2 x) 2sin x dx
2
0
4
0
2
/2 3 cos 2 x
2sin x
2
/2
4 32 x sin42 x 2 cos x
0
4
3
4
0 0 (0 0 2) (3 8)
1 1
31. R( y ) 5 y 2 V R ( y ) dy
2 4
1 15 y dy
1
y5 [1 (1)] 2
1
2 2
32. R( y ) y 3/2 V R( y ) dy y 3 dy
2
0 0
2
y4
4 4
0
/2
R( y ) dy
2
33. R( y ) 2sin 2 y V
0
/2 /2
2sin 2 y dy cos 2 y 0
0
[1 (1)] 2
y 0
V R ( y ) dy
2
34. R( y ) cos 4 2
0
cos
2 dy 4 sin
y
4
y 0
4 2
4[0 (1)] 4
3 3 1
R( y ) dy 4
2 V 2
35. R( y ) y 1
dy
0 0 ( y 1) 2
3
4 y11 4 14 (1) 3
0
1
V R ( y ) dy
2y 2
36. R( y ) 2
y 1 0
1 2
2 y y2 1 dy; [u y 2 1 du 2 y dy;
0
y 0 u 1, y 1 u 2]
2 2
V u 2 du u1 12 (1) 2
1 1
39. r ( x) x and R( x) 1 V
1
0 R ( x) 2
r ( x)
2
dx
1
0
1
1 x 2 dx x
x3
3 0
1 13 0 23
40. r ( x) 2 x and R( x) 2 V
0
1
R( x) 2
r ( x)
2
dx
1
1
(4 4 x) dx 4 x x2 4 1 12 2
2
0 0
41. r ( x) x 2 1 and R( x) x 3
V
2
1 R( x) 2
r ( x)
2
dx
2
2
( x 3) 2 x 2 1 dx
1
2
x 2 6 x 9 x 4 2 x 2 1 dx
1
1
x 4 x 2 6 x 8 dx x5 x3 6 2x
2 2
8x
5 3 2
1
32
5 3 2
8 24 16 15 13 62 8
33
5
3 28 3 8 530533 1175
42. r ( x) 2 x and R( x) 4 x 2
V
2
1 R( x) 2
r ( x)
2
dx
2
(2 x)2 dx
2
4 x2
1
16 8 x 2 x 4 4 4 x x 2 dx
2
1
12 4 x 9 x 2 x 4 dx 12 x 2 x 2 3 x3 x5
2 2 5
1 1
24 8 24 32
5
12 2 3 15 15 33
5
1085
43. r ( x) sec x and R( x) 2
V
/4
/4
R ( x) r ( x)
2 2
dx
/4
/4 2 sec2 x dx 2x tan x/4/4
2 1 2 1 ( 2)
45. r ( y ) 1 and R( y ) 1 y V
1
0 R ( y ) 2
r ( y)
2
dy
1 1
(1 y )2 1 dy 1 2 y y 2 1 dy
0 0
1
1
2 y y 2 dy y 2
0
y3
3
0
1 13 43
46. R( y ) 1 and r ( y ) 1 y V
1
0 R( y ) 2
r ( y)
2
dy
1
0
1
0
1 (1 y ) 2 dy 1 1 2 y y 2 dy
1
1
2 y y 2 dy y 2
0
y3
3
0
1 13 23
47. R( y ) 2 and r ( y ) y V
0
4
R( y ) 2
r ( y )
2
dy
4
4 y2
(4 y ) dy 4 y 2
(16 8) 8
0 0
48. R( y ) 3 and r ( y ) 3 y 2
V
0
3
R( y ) 2
r ( y)
2
dy
3
3 3 y 2 dy 3 y 2 dy y 3
3 3
0 0 3 0
49. R( y ) 2 and r ( y ) 1 y V
1
0 R( y ) 2
r ( y)
2
dy
1
2
4 1 y dy 1 2 y y dy
0
1
1
3 2 y y dy 3 y 43 y 3/2
0
y2
2
0
3 43 12 18683 76
V
4
0 R ( x) 2
r ( x)
2
dx
4 4
(4 x) dx 4 x x2 (16 8) 8
2
0 0
2
2 2 y5
(b) r ( y ) 0 and R( y ) y 2 V R( y)2 r ( y )2 dy y 4 dy 5 325
0 0 0
(c) r ( x) 0 and R( x) 2 x V
0
4
R( x) 2
r ( x)
2
dx 2 x dx
4
0
2
4
4
4 4
x x dx 4 x 8 x3 x
3/ 2 2
16 64 16 83
0 2 0 3 2
R ( y ) dy 2
dy
2 2
r ( y)
2 2
(d) r ( y ) 4 y 2 and R ( y ) 4 V 16 4 y
2
0 0
2
2
0 2
16 16 8 y 2 y 4 dy 8 y 2 y 4 dy 83 y3
0
y5
5
0
643 325 22415
y
52. (a) r ( y ) 0 and R( y ) 1 2
2
dy
V R ( y ) r ( y )
0
2 2
1 dy 1 y dy
2 y 2
2 y2
0 2
0 4
2
y
y2
2
y3
12 2 42 12
0
8 2
3
y
(b) r ( y ) 1 and R ( y ) 2 2
V
2
0 R( y ) 2
r ( y )
2
dy 2
0
2 y 2
2
2
1 dy 4 2 y
0
y2
4
1 dy
2
2 y2 2 y
3
3 2y 8 2 8
0 4 dy 3 y y 12 6 4 12 2 3 3
0
(b) r ( x) 1 and R ( x) 2 x 2 V
1
1 R ( x) 2
r ( x)
2
dx 2 x
1
1
2 2
1 dx
4 4 x 2 x 4 1 dx 3 4 x 2 x 4 dx 3x 43 x3 x5 2 3 43 15
1 1 5 1
1 1 1
215 (45 20 3) 56
15
R( x) dx
dx
1 1 2
r ( x)
2 2
(c) r ( x) 1 x 2 and R( x) 2 V 4 1 x2
1 1
1
4 1 2 x 2 x 4 dx 3 2 x 2 x 4 dx 3x 23 x3 x5 2 3 23 15
1 1 5 1
1 1
215 (45 10 3) 64
15
V
b
0 R ( x) 2
r ( x)
2
dx
0
b
bh x h
2
dx
b h2
0 b2
2
x 2 2bh x h 2 dx
b
h2 x 2 x h2 b3 b b h3 b
3
x2 2
3b b 0
(b) r ( y ) 0 and R( y ) b 1 h V
y h
0 R( y ) 2
r ( y )
2
dy b 1
2 h
0
y 2
h
dy
h
h
0
2 y y2
h
b 2 1 h 2 dy b2 y
y2
h
y3
3h 2
0
2
h b2h
b h h 3 3
55. R( y ) b a 2 y 2 and r ( y ) b a 2 y 2
V
a
a R( y )
2
r ( y)
2
dy
2 2
b a 2 y 2 b a 2 y 2 dy
a
a
a a
4b a 2 y 2 dy 4b a 2 y 2 dy
a a
2
4b area of semicircle of radius a 4b 2a 2a 2 b 2
5 2 5
56. (a) A cross section has radius r 2 y and area r 2 2 y. The volume is 0 2 ydy y 0
25 .
ha
57. (a) R( y ) a 2 y 2 V
h a
a
a 2 y 2 dy a 2 y
y3
3
a
( h a )3 3
a 2 h a3 3 a3 a3
a 2 h 13 h3 3h 2 a 3ha 2 a3 a3 a 2 h h3 h 2 a ha 2
3
3
h2 (3a h)
3
h 2 (15 h ) 3
(b) Given dV
dt
0.2 m3 /sec and a 5 m, find dh
dt h 4
. From part (a), V (h) 3
5 h2 3h
dV
dh
10 h h 2 dV
dt
dV dh h(10 h) dh
dh dt dt
dh
dt h 4
0.2
4 (10 4)
1
(20 )(6)
1201 m/sec.
59. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a
disk of radius h has been removed. Thus its area is A1 R 2 h 2 ( R 2 h2 ). The cross section of the
2
hemisphere is a disk of radius R 2 h2 . Therefore its area is A2 R 2 h 2 R 2 h 2 .
We can see that A1 A2 . The altitudes of both solids are R. Applying Cavalieri’s Principle we find
Volume of Hemisphere (Volume of Cylinder) (Volume of Cone) R 2 R 13 R 2 R 32 R3 .
Copyright 2014 Pearson Education, Inc.
442 Chapter 6 Applications of Definite Integrals
60. x 36 x 2 V 6 R ( x ) 2 dx 6 x 2 36 x 2 dx 6 36 x 2 x 4 dx
R( x) 12 144
0 144 0 0
6
12 x3 x5 12 63 65 63 12 36
144
5 0 144 5 144 5 196
144 5
6036 36 cm3 .
5
The plumb bob will weigh about W (8.5) 365 192 gm, to the nearest gram.
7
16
256 y 2 dy 256 y 3
7 7 y 3
R( y ) dy
2
61. R( y ) 256 y 2 V
16 16
(256)(7) 73 (256)(16) 163
3 3
73
3
256(16 7) 163
3
1053 cm 3
3308 cm3
63. Volume of the solid generated by rotating the region bounded by the x-axis and y f ( x ) from x a to
b
x b about the x-axis is V [ f ( x)]2 dx 4 , and the volume of the solid generated by rotating the same
a
b b b
a f ( x) 1 dx f ( x) dx 8 4
2 2
region about the line y 1 is V [ f ( x) 1]2 dx 8 . Thus
a a
b
a f ( x)
2
2 f ( x) 1 f ( x)
2
dx 4 (2 f ( x) 1) dx 4 2
a
b b
a
f ( x) dx dx 4
b
a
b b 4b a
f ( x) dx 12 (b a) 2 f ( x) dx 2
a a
64. Volume of the solid generated by rotating the region bounded by the x-axis and y f ( x ) from x a to x b
b
about the x-axis is V f ( x) dx 6 , and the volume of the solid generated by rotating the same
2
a
b
region about the line y 2 is V f ( x) 2 dx 10 . Thus
2
a
b
a f ( x) 2
2 b
dx f ( x) dx 10 6
a
2 b
a f ( x) 2
4 f ( x) 4 f ( x)
2
dx 4
b b b b b
(4 f ( x) 4) dx 4 4 f ( x) dx 4 dx 4 f ( x) dx (b a ) 1 f ( x) dx 1 b a
a a a a a
dx
2
b 2 2
2 x 1 x4 dx 2 x x4 dx 2 x2 16
x 4 2
24 1616
shell shell 2 3 2
V 2
a radius height 0 0 0
2 3 6
dx
b 2 2 2
2 x 2 x4 dx 2 2 x x4 dx 2 x 2 16
x 4 2 4 1 6
2 3
V 2
a
shell
radius
shell
height 0 0 0
V 2
b
a shell
radius shell
height dx 0
3
2 x x 2 1 dx;
u x 2 1 du 2 x dx; x 0 u 1, x 3 u 4
23 (8 1) 143
4 4
V u1/2 du 23 u 3/2 23 43/2 1
1 1
V 2
b
a shell
radius shell
height dx 2 x
0
3 9x
x3 9
dx;
[u x3 9 du 3x 2 dx 3 du 9 x 2 dx; x 0 u 9, x 3 u 36]
36 36
V 2 3u 1/2 du 6 2u1/2 12 36 9 36
9 9
7. a 0, b 2;
V 2
b
a shell
radius shell
height dx 0
2
2 x x 2x dx
2 2 2
2 x 2 32 dx 3 x 2 dx x3 8
0 0 0
8. a 0, b 1;
b
V 2
a shell
radius shell
height dx 2 x 2 x dx
0
1 x
2
2
1
0 dx 3x dx x
3 x2
2
1
0
2 3 1
0
9. a 0, b 1;
b
V 2
a shell
radius shell
height dx 2 x (2 x) x dx
0
1 2
1 1
2 2 x x 2 x3 dx 2 x 2 x3 x4
3 4
0 0
2 1 13 14 2 121243 1012 56
10. a 0, b 1;
b
V 2
a
dx 2 x 2 x x dx
shell
radius
shell
height
1
0
2 2
2 x 2 2 x dx 4 x x dx
1 2 1 3
0 0
1
4 x2 x4 4 12 14
2 4
0
11. a 0, b 1;
b
V 2
a shell
radius shell
height dx 2 x
0
1
x (2 x 1) dx
1 1
2 x3/2 2 x 2 x dx 2 25 x5/2 23 x3 12 x 2
0 0
2 25 23 12 2 12302015 715
12. a 1, b 4;
b
V 2
a shell
radius shell
height dx
32 x1/2 dx 1
4
2 x
x sin x , 0 x sin x, 0 x
13. (a) x f ( x) x x f ( x) ; since sin 0 0 we have
x, x0 0, x0
sin x, 0 x
x f ( x) x f ( x) sin x, 0 x
sin x, x0
(b) V 2
b
a shell
radius shell
height dx 2 x f ( x) dx and x f ( x) sin x, 0 x by part (a)
0
V 2 sin x dx 2 cos x 0 2 ( cos cos 0) 4
0
x tan 2 x , 0 x tan 2 x, 0 x /4
14. (a) x g ( x) x 4 x g ( x) ; since tan 0 0 we have
x 0, x0 0, x0
tan 2 x, 0 x /4
x g ( x) x g ( x) tan 2 x, 0 x /4
2
tan x, x0
(b) V 2
b
a shell
radius shell
height dx 0
/4
2 x g ( x) dx and x g ( x) tan 2 x, 0 x /4 by part (a)
V 2
/4
0
tan 2 x dx 2
/4
0 sec2 x 1 dx 2 tan x x0 /4 2 1 4 4 2 2
15. c 0, d 2;
V 2
c
d
dy
shell
radius
shell
height 0
2
2 y y ( y ) dy
2
2
0
2
y y dy 2
3/2 2 2 y 5/ 2
5
y3
3
0
2 52
2
5 3
23 2
8 2
5
83 16 5
2
13
16
3 2 5
15
16. c 0, d 2;
V 2
c
d
shell
radius dy
shell
height 0
2
2 y y 2 ( y ) dy
2
0 dy 2
2
2
y3 y 2
y4
4
y3
3
0
16 24 13
16 56 403
17. c 0, d 2;
V 2
c
d
dy 2 y 2 y y dy
shell
radius
shell
height 0
2 2
2
2
0
2
2 y y dy 2 2
2 3 2 y3
3
y4
4
0
16
3
16
4
18. c 0, d 1;
d
V 2
c shell
radius dy 2 y 2 y y
shell
height
1
0
2
y dy
1
2 y y y 2
0 dy 2 y y dy
1
0
2 3
1
y3
2 3
y4
4
0
2 13 14 6
19. c 0, d 1;
d
V 2
c shell
radius shell
height dy 2 y y ( y) dy
1
0
1 1
2 2 y 2 dy 4 y 3 43
0 3 0
20. c 0, d 2;
d
V 2
c shell
radius shell
height dy 2
0
2 y y 2 dy y
2 y2 2
2 dy 3 y 3 83
0 2 0
21. c 0, d 2;
d
V 2
c shell
radius shell
height dy 2
0
2 y (2 y ) y 2 dy
2
0
2 y y 2 y 3 dy 2 y 2 3 4
2 y y 3 4
2
0
2 4 83 16
4
6 (48 32 48) 163
22. c 0, d 1;
d
V 2
c shell
radius shell
height dy 2 y (2 y) y dy
0
1 2
1
1
0
2 2 y y 2 y 3 dy 2 y 2
y3
3
y4
4
0
2 1 13 14 6 (12 4 3) 56
dx
b 2 2 2
23. (a) V 2 shell shell
2 x (3x)dx 6 x 2 dx 2 x3 16
a radius height 0 0 0
2 dx 0
4 x x 2 dx 6 2 x 2 13 x3
b shell shell 2 2 2
(b) V 2 (4 x) (3x)dx 6
a radius height 0 0
6 8 83 32
dx x2 x dx 6 13 x3 12 x2 0
b shell shell 2 2 2
(c) V 2 radius height 2 ( x 1) (3 x)dx 6
a 0 0
6 83 2 28
dy 2 y 13 y 2 dy 2 y 2 19 y3 0
d shell shell 6 6 6
(d) V 2 radius height 2 y 2 13 y dy 2
c 0 0
2 (36 24) 24
(e) V 2
c
d
shell
radius shell
height dy 6
0
2 (7 y ) 2 13 y dy 2 14 13
3
y 13 y 2 dy
6
0
6
2 14 y 13 y2 1 y3 2 (84 78 24) 60
6 9 0
d
(f ) V 2
c shell
radius shell
height dy 6
0
2 ( y 2) 2 13 y dy 2 0
6
4 34 y 13 y 2 dy
6
2 4 y 23 y 2 19 y 3 2 (24 24 24) 48
0
dx
b 2 2 2
24. (a) V 2 shell shell
2 x 8 x3 dx 2 8 x x 4 dx 2 4 x 2 15 x5
a radius height 0 0 0
2 16 32
5
965
(b) V 2
a
b
shell
radius shell
height dx 0
2
2 (3 x) 8 x3 dx 2 0
2
24 8x 3x3 x4 dx
2
2 24 x 4 x 2 34 x 4 15 x5 2 48 16 12 32 264
0 5 5
(c) V 2
a
b
shell
radius shell
height dx 0
2
2 ( x 2) 8 x3 dx 2 16 8 x 2 x3 x 4 dx
2
0
2
2 16 x 4 x 2 12 x 4 15 x5 2 32 16 8 32 336
0 5 5
dy 2 y y dy 2 y dy
d 8 8 4/3 8
(d) V 2 shell shell 1/3 6 y 7/3 6 (128) 768
c radius height 0 0 7 0 7 7
2 dy 2 (8 y) y dy 2 8 y
d 8 8 8
(e) V shell shell 1/3 1/3
y 4/3 dy 2 6 y 4/3 73 y 7/3
c radius height 0 0 0
2 96 384
7 576
7
dy 2 ( y 1) y 0
y 4/3 y1/3 dy 2 73 y 7/3 34 y 4/3
d shell shell 8 1/3 8 8
(f ) V 2 dx 2
c radius height 0 0
2π 384
7
12 936π
7
25. (a) V 2
a
b
shell
radius shell
height dx 2
1
2 (2 x) x 2 x 2 dx 2 2
1 4 3x2 x3 dx
2
2 4 x x3 14 x 4 2 (8 8 4) 2 4 1 14 27
1 2
(b) V 2
a
b
shell
radius shell
height dx 2
1
2 ( x 1) x 2 x 2 dx 2 2
1 2 3x x3 dx
2
2 2 x 32 x 2 14 x 4 2 (4 6 4) 2 2 32 14 27
1 2
(c) V 2
c
d
shell
radius shell
height dy 2 y y y dy 2 y
1
0
4
1
y ( y 2) dy
y y 2 y dy y 2 y 5/2 1
1 4 4
4 y 3/2 dy 2 3/2 2 8 2 5/2 13 y 3 y 2
0 1 5 0 5 1
85 (1) 2 645 643 16 2 52 13 1 725
(d) V 2
c
d
shell
radius shell
height dy 2 (4 y) y y dy
1
0 1
4
2 (4 y ) y ( y 2) dy
1
4 4 y y 3/2
0 dy 2 y y 6 y 4 y 8 dy
4
1
2 3/2
1 4
4 83 y 3/2 52 y5/2 2 13 y 3 52 y 5/2 3 y 2 83 y 3/2 8 y
0 1
4 83 52 2 643 645 48 643 32 2 13 25 3 83 8 1085 .
26. (a) V 2
b
a shell
radius shell
height dx 1
1
2 (1 x) 4 3 x 2 x 4 dx 2 1
1 x5 x4 3x3 3x2 4 x 4 dx
16 15 34 1 2 4 2 16 15 34 1 2 4 565
1
2 16 x 6 51 x5 34 x 4 x3 2 x 2 4 x 2
1
(b) V 2
c
d
shell
radius shell
height dy 2 y
1
0
4 y 4 y dy 14 2 y 4 y
3
4 y
3
dy
1 4
4 y 5/4 dy 4 y 4 ydy [u 4 y y 4 u du du; y 1 u 3, y 4 u 0]
0 3 1
0 1 3 3
169 y 9/4 4 (4 u ) u du 169 (1) 4 4 u u 3/2 du 169 4 83 u 3/2 52 u 5/2
0 3 3 3 0 3 0
169 4 8 3 18
5 3
3 169 885 872
45
1
27. (a) V 2
c
d
shell
radius shell
height 1
1
y 4 y5
dy 2 y 12 y 2 y 3 dy 24 y 3 y 4 dy 24 4 5
0 0 0
24 14 15 2420 65
(b) V 2
c
d
shell
radius shell
height dy 2 (1 y) 12 y
1
0
2
1
y 3 dy 24 (1 y ) y 2 y 3 dy
0
1
1
y3
24 y 2 2 y 3 y 4 dy 24 3
0
y4
2
y5
5
0
24 13 12 15 24 301 45
(c) V 2
c
d
shell
radius shell
height dy 2 y 12 y
1
0
8
5
2
y 3 dy 24
0
1
85 y y 2 y3 dy
1
24
1 8
0 5 y 2 13
5
y3 y dy 24 y y
4
8 3
15
13
20
4
y5
5
0
24 158 1320 51
24 (32 39 12) 24 2
60 12
(d) V 2
c
d
shell
radius shell
height dy 2 y 12 y y dy 24 y y
1
0
2
5
2 3 1
0
2
5
2
y 3 dy
1
1
0
24 y 3 y 4 25 y 2 y dy 24 y y y dy 24 y
2 3
5
1 2
0 5
2
3 3
5
4 2 3
15
3
20
y4
y5
5
0
24 152 203 51 2460 (8 9 12) 2412 2
28. (a) V 2
c
d
shell
radius shell
height dy 2
0
y2 y4 y2
2
0
y4
2
0
y5
2 y 2 4 2 dy 2 y y 2 4 dy 2 y 3 4 dy
2
y4 y6 24
14 244 32 14 61 32 242 83
6
2 4 24 2 2 32
24
0 4
(b) V 2
c
d
shell
radius shell
height dy 2
0
y2 y4 y2 2 y4
2 (2 y ) 2 4 2 dy 2 (2 y ) y 2 4 dy
0
2
2
2 2 y 2
0
y4
2
y3
y5
4
2 y3 y5
dy 2 3 10
y4
4
y6
24 2
0
163 1032 164 6424 85
(c) V 2
c
d
shell
radius shell
height dy 2
0
y2 y4 y2 2 y4
2 (5 y ) 2 4 2 dy 2 (5 y ) y 2 4 dy
0
2
2
2 5 y 2 54 y 4 y 3
0
y5
4
5 y3 5 y5
dy 2 3 20
y4
4
y6
24 2
0
403 160
20
16
4
24
64 8
(d) V 2
c
d
shell
radius shell
height dy 2 y 2
0
5
8
y2
2
y4
4
y2
2
2 5 2
dy 0 2 y 8 y
y4
4
dy
2
2 3 y5 5 y 4 dy 2 y y 5 y 5 y 2
164 6424 4024 160
160
4 6 3 5
2 y 58 y 2 32 4
0 4
4 24 24 160
0
About y -axis: V 2
a
b
shell
radius shell
height dx
1
2 x x x 2 dx 2 x 2 x3 dx
0 0
1
1
2 x3 x4 2 13 14 6
3 4
0
0
1
y2
2
y3
3
0
12 13 6
30. (a) V
a
b
R ( x) 2
r ( x)
2
dx 2 4
0
x
2
2
x 2 dx
0 4
3 x 2 2 x 4 dx x4 x 2 4 x
4 3 4
0
16 16 16 16
(b) V 2
b
a shell
radius shell
height dx 4
0
4
0
4 2
2 x 2x 2 x dx 2 x 2 2x dx 2 2 x x2 dx 0
4
2 x 2 x6 2 16 64
3
323
0 6
(c) V 2
b
a shell
radius shell
height dx 4
0
4
0
4 2
2 (4 x) 2x 2 x dx 2 (4 x) 2 2x dx 2 8 4 x x2 dx 0
4
2 8 x 2 x 2
x3
6 0
2 32 32 64
6
64
3
(d) V
b
a R( x) 2
r ( x)
2
dx 0
4 2 x
(8 x ) 6 2
2 4
2
dx 0 64 16 x x 36 6 x x2
4 dx
4 3 2 4
10 x 28 dx x4 5 x 2 28 x [16 (5) (16) (7) (16)] (3) (16) 48
3
x
0 4 0
31. (a) V 2
d
c shell
radius shell
height dy 1
2
2 y ( y 1) dy
2
2
1
2
y3 y 2
y 2 y dy 2 3 2
1
2
83 42 13 12
2 73 2 12 3 (14 12 3) 53
dx 2 x x2 dx 2 x2 x3 1
b 2 2 3 2
shell shell
(b) V 2 radius height 2 x(2 x) dx 2
a 1 1
2 4 83 1 13 2
1238 331 2 34 32 43
V 2 radius height dx 12 2 103 x (2 x) dx 2 12 203 163 x x2 dx
b shell shell
(c)
a
x 83 x 2 13 x3 2 40
3 3 3 3 3 3
32 8 20 8 1 2 33 2
2
2 20
3 1
2
(d) V 2
d
c shell
radius shell
height dy 1
2
1
2 ( y 1)3
2 ( y 1)( y 1) dy 2 ( y 1)2 2 3 23
1
32. (a) V 2
d
c shell
radius shell
height dy 0
2
2 y y 2 0 dy
8
2
2 y4 24
2 y 3 dy 2 4 2 4
0 0
(b) V 2
b
a shell
radius shell
height dx
4
2 x 2 x dx 2
0
4
0 2 x x3/2 dx
4 5
2 x 2 52 x5/2 2 16 252
0
2 16 64
5
25 (80 64) 32
5
(c) V 2
b
a shell
radius shell
height dx
0
4
2 (4 x) 2 x dx 2 8 4 x1/2 2 x x3/2 dx
0
4
4
2 8 x 83 x3/2 x 2 2 x 5/2 2 32 64 16 64 215 (240 320 192) 215 (112) 224
5 0 3 5 15
2
dy 0
2 y 2 y3 dy 2 23 y 3 4
y
d 2 2 4
shell shell
(d) V 2 radius height 2 (2 y ) y 2 dy 2
c 0 0
2 163 164 3212 (4 3) 83
33. (a) V 2
d
c
dy 2 y y y dy
shell
radius
shell
height
1
0
3
1
2 y y dy 2 2
1 2 4 y3 y5 1 1 4
0 3 5 3 5 15
0
V 2 dy 2 (1 y) y y dy
d shell shell 1 3
(b) radius height
c 0
1
2 y y y y dy 2
1 2 3 4 y2 y3 y4 y5
0 2 3 4 5
0
2 1
2
13 14 1
5 2
60
(30 20 15 12) 7
30
34. (a) V 2
d
c shell
radius shell
height dy 2 y 1 y y dy
1
0
3
1
1
y2
2 y y 2 y 4 dy 2 2
0
y3
3
y5
5
0
2 12 13 15 230 (15 10 6) 1115
(b) Use the washer method:
V
d
c R ( y ) 2
r ( y)
2
dy 1 y y dy 1 y
1
0
2 3 2 1
0
2
y 6 2 y 4 dy
1
y3 y7 2 y5
y 3 7 5 1 13 17 52 105
0
(105 35 15 42) 97
105
(c) Use the washer method:
1
dy
d 2 1 2
V R( y ) r ( y ) dy 1 y y 3 0 dy 1 2 y y 3 y y 3
2 2
c 0
0
1
1
0
1 y 2 y 6 2 y 2 y3 2 y 4 dy y
y3
3
y7
7
y2
y4
2
2 y5
5
0
(70 30 105 2 42) 121
1 13 17 1 12 52 210 210
(d) V 2
d
c shell
radius shell
height dy 2 (1 y) 1 y y dy 2 (1 y) 1 y y dy
1
0
3 1
0
3
0
y y y dy 2 1 2 y y y y dy 2 y y
1 1 y3 y4 y5
2 1 y y3 2 4 2 3 4 2
3 4
5
0 0
2 1 1 13 14 15 260 (20 15 12) 23
30
35. (a) V 2
d
c shell
radius shell
height dy 2 y 1
0
8 y y 2 dy
2
0
2 y4
2 2 2 y 3/2 y 3 dy 2 4 5 2 y 5/2 4
0
4 2
5
2
2
5
4
24 2
423
5
444
2 4 85 1 85 (8 5) 245
dx x
b 4 4 4
x8 dx 2 25 x5/2 32
x4
shell shell 2 3/2 3
(b) V 2 2 x x x8 dx 2
a radius height 0 0 0
2 22
5
5 4
4 2
32 2
5
6
32 160
8
160 5
7
2 2 (32 20) 2 3 2 3 9 4
48
5
36. (a) V 2
b
a
dx 2 x 2 x x x dx
shell
radius
shell
height
1
0
2
2 x x x dx 2 x x dx
1 2 1 2 3
0 0
1
2 x3 x4 2 13 14 6
3 4
0
(b) V 2
b
a shell
radius shell
height dx 2 1 x 2 x x x dx 2 1 x x x dx
1
0
2 1
0
2
1
1
2 x 2 x 2 x3 dx 2 x2 23 x3 x4 2 12 23 14 212 (6 8 3) 6
2 4
0 0
37. (a) V
b
a R( x) 2
r ( x)
2
dx x 1
1/16
1/2
1 dx
1
2 x1/2 x (2 1) 2 14 16
1
1/16
7 9
1 16 16
(b) V 2
d
c shell
radius shell
height dy 2 y 1
0
1
y4
1 dy
16
2
2 y2
y 3 16 dy 2 12 y 2 32
y
2
1 1
2 18 12 32
1 2
321
2 (8 1) 9
38. (a) V
d
c R ( y ) 2
r ( y)
2
dy 1
2 1
y4
1 dy
16
2
13 y 3 16 24
y
1
1 1 1 1
8 3 16
( 2 6 16 3) 11
48 48
2
23 12 23 18 321 34 1 61 161 48 (4 16 48 8 3) 1148
39. (a) Disk: V V1 V2
b b
V1 1 R1 ( x) dx and V2 2 R2 ( x) dx with R1 ( x)
2 2 x 2
3
and R2 ( x) x ,
a1 a2
a1 2, b1 1; a2 0, b2 1 two integrals are required
(b) Washer: V V1 V2
V1 1
a1
b
R ( x) r ( x) dx with R ( x)
1
2
1
2
1
x 2
3
and r1 ( x ) 0; a1 2 and b1 0;
R ( x) r ( x) dx with R ( x)
b2 2 2 x 2
V2 2 2 2 3
and r2 ( x) x ; a2 0 and b2 1
a2
two integrals are required
d shell
(c) Shell: V 2 radius
c shell
height dy c
d
shell
2 y height
dy where shell height y 2 3 y 2 2 2 2 y 2 ;
c 0 and d 1. Only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V .
Vi i
ci
d
R ( y )
i
2
ri ( y )
2
dy, i 1, 2 with R ( y) 1, r ( y) 1 1 y , c1 0 and d1 1;
R( x) dx
2
41. (a) V
b
a
2
r ( x)
2 4
4
4
4
4
4
25 x 2 (3) 2 dx 25 x 2 9 dx 16 x 2 dx
4
16 x 13 x3 64 64 64 64 256
4 3 3 3
(b) Volume of sphere 43 (5)3 500 Volume of portion removed 500 256 244
3 3 3 3
b
shell
42. V 2 radius
a shell
height dx 1
1
2 x sin x 2 1 dx; [u x 2 1 du 2 x dx;
x 1 u 0, x 1 u ] sin u du cos u 0 (1 1) 2
0
dx
b r r r
shell
43. V 2 radius shell
2 x hr x h dx 2 hr x 2 h x dx 2 3hr x3 h2 x 2
a height 0 0 0
2
2 r 3h r 2h 13 r 2 h
2
d shell
44. V 2 radius
c shell
height dy r
0
2 y r 2 y 2 r 2 y 2 dy 4 y r 2 y 2 dy
r
0
0 r 2 1/2 r2
2 2
[u r y du 2 y dy; y 0 u r , y r u 0] 2 2
u du 2 u du 4 u 3 2
r2 0 3 0
43 r 3
f (a) a
45. W (a) [( f 1 ( y ))2 a 2 ]dy 0 2 x[ f (a) f ( x)]dx S (a);
f ( a) a
46. V
0
/3
[22 (sec y )2 ]dy [4 y tan y ]0 /3 43 3
shell shell 1
b 1
2 2
47. V 2 dx 2 xe x dx e x (e1 e0 ) 1 1e
a radius height 0 0
x2 2 x
dy 1/2
1. dx
13 32 x 2 2 2x
L
3
0
1 x 2 2 x 2 dx
0
3
1 2 x 2 x 4 dx
3 27
3
12
dy 4
2. dx
3
2
xL 1 94 x dx;
0
u 1 9 x du 9 dx 4 du dx;
4 4 9
x 0 u 1; x 4 u 10]
2
3. dx
dy
y2 1 dx
dy
y 4 12 1
4 y2 16 y 4
3 3
L 1 y 4 12 1 4 dy y 4 12 1 4 dy
1 16 y 1 16 y
2
3 2 3
y 1 2 dy y 2 1 2 dy
1 4y 1 4y
3
y 3 y 1
3 4 27
1
1 13 14 9 12
3 12
1 11
3 4
( 1 4 3) ( 2)
9 12
9 12 53
6
y2
2
4. dx
dy
12 y1/2 12 y 1/2 dy
dx 1
4
1
y
L
1
9
1 14 y 2 dy y 2 dy
1
y 1
9 1
4
1
y
2
9 9
12 y 1 dy 12 y1/2 y 1/2 dy
1 y 1
9 y 3/ 2 9
12 23 y 3/2 2 y1/2 3 y1/2
1 1
3
33 3 13 1 11 13 32
3
2
5. dx
dy
y 3 1 3 dx
dy
y 6 12 1 6
4y 16 y
2 2
L 1 y 6 12 1 6 dy y 6 12 1 6 dy
1 16 y 1 16 y
2 2
2 3 y 3 2 3 y 3 y 4 y 2
y 4 dy 1 y 4 dy 4 8
1 1
16 1
4 (16)(2)
14 81 4 32
1 11
4 8
12832
18 4 123
32
y2 2
6. dx
dy
2 1 2 dx
dy
14 y 4 2 y 4
2y
L
3
2
1 14 y 4 2 y 4 dy
3
2
1
4 y 4 2 y 4 dy
y 2 y2 dy 12 23 y2 y 2 dy
3 2
12
2
3
y3
12 3 y 1 12 27
2 3 3
1 83 12
12 26
3
83 12 12 6 12 13
4
7.
dy
dx
x1/3 14 x 1/3 dx dy 2 2/3
x 2/3 12 x16
8 2/3
L 1 x 2/3 12 x16 dx
1
x1/3 14 x1/3 dx
8 2/3 8 2
x 2/3 12 x16 dx
1 1
1
x1/3 14 x 1/3 dx 34 x 4/3 83 x 2/3
8 8
1
dy
8. dx
x2 2 x 1 4 x 2 2 x 1 14 1
(4 x 4) 2 (1 x )2
(1 x)2 14 1
(1 x ) 2
dx
dy 2
(1 x)4 12 1
16(1 x )4
2 (1 x )4
L 1 (1 x) 4 12 16
dx
0
2 (1 x )4
(1 x)4 12 16
dx
0
2
2 (1 x )2
(1 x)2 4 dx
0
2
0
(1 x )2
1
3
(1 x) 2 4 dx; [u 1 x du dx; x 0 u 1, x 2 u 3] L u 2 14 u 2 du
3
u3 14 u 1 9 12
3
1 1 1 1081 4 3 106 53
1 3 4 12 12 6
2 y
dx 2 2
9. dy
1x 4x dx
dy
1
x
4x 12 1 x
x 2 16 0.5
2 2 x2
L 1 1 1 x2 dx 1 1 2
x dx y ln x
1 x2 2 16 1 x2 2 16 8
2 x
1 1x 4x
2 2 2 1
dx 4x dx ln x x2 0 1 2
1 x 8
1
ln 2 48 ln1 18 ln 2 83 0.5
y
dy 2
dy 2
10. dx
x 41x dx x 41x x 2 21 1 2
16 x 5
3 2 x 2 ln x
L 1 x 1
2
12 dx 4 y
2
4
1 16 x 3
x
3 3 1 2 2
x 2 12 1 2 dx 4x
dx
1 16 x 1 1
3 x
1
3
x 41x dx x2 14 ln x
2 0 1 2 3
1
92 14 ln 3 12 14 ln1 4 14 ln 3
x
2 y
dy dy 2
11. dx
x2 1 dx
2
1 x 4 12 1
4 x2 4 x2 16 x 4 10
3 4
L 1 x 1
2
14 dx 8
y
x3 1
1 16 x 6 3 4x
x
3 3 2 4
1 x 4 12 1 dx 2
1 dx 2
16 x 4 1 4 x2
x
1
3
9 121 13 13 536
3 0 1 2 3
dx x3 41x
3
x2 1
4 x2 1
x
2
dy dy 2
12. dx
x4 1 dx
4
1 x8 12 1
4 x4 4 x4 16 x8 y
1
L 1 x8 12 1 dx 1
1/2 16 x8 x5 1
y
x
1 1 2 5 12 x 3
1/2 x8 12 1 dx 4
1 dx 0.5
16 x8 1/2 4 x4
1/2 x dx
1 1
4
1 x5 1
3
4 x4 5 12 x 1/2 x
0 0.5 1 1.5
15 121 1601 23 373
480
2
13. dx
dy
sec 4 y 1 dx
dy
sec4 y 1
L
/4
/4
1 sec4 y 1 dy /4
/4
sec 2 y dy
/4
tan y /4 1 (1) 2
14.
dy
dx
3x4 1
dy 2
dx
3x4 1
L
1
2
1 3x 4 1 dx 1
2
3 x 2 dx
1
3 x3
3 3 1 (2)3 3
(1 8) 7 3
2 3 3 3
4x
dy dy 2 2
(b)
15. (a) dx
2x dx
1 dx
2 dy 2
L dx
1
2
1 4x 2 dx
1
(c) L 6.13
16. (a)
dy
dx
sec2 x
dy 2
dx
sec4 x (b)
0
L 1 sec4 x dx
/3
(c) L 2.06
2 (b)
17. (a) dx
dy
cos y dx
dy
cos 2 y
L 1 cos 2 y dy
0
(c) L 3.82
y 2 y2 (b)
18. (a) dx dx
dy 1 y 2 dy 1 y 2
1/2 2 1/2
y
L 1 dy 1 dy
1/2 1 y 2 1/2 1 y 2
1 y 2
1/2 1/2
dy
1/2
(c) L 1.05
2 (b)
19. (a) 2 y 2 2 dx
dy
dx
dy
( y 1)2
3
L 1 ( y 1)2 dy
1
(c) L 9.29
dy dy 2 (b)
20. (a) dx
cos x cos x x sin x dx
x 2 sin 2 x
L 1 x 2 sin 2 x dx
0
(c) L 4.70
dy dy 2 (b)
21. (a) dx
tan x dx
tan 2 x
/6 /6 sin 2 x cos 2 x
L 1 tan 2 x dx dx
0 0 cos 2 x
/6 dx /6
cos x 0
sec x dx
0
(c) L 0.55
2
22. (a) dx sec2 y 1 dx sec2 y 1 (b)
dy dy
L
/4
/3
1 sec 2 y 1 dy /4
/3
| sec y | dy
/4
sec y dy
/3
(c) L 2.20
23. (a)
dy 2
dx
corresponds to 1
4x
here, so take
dy
dx
as 1 .
2 x
Then y x C and since (1, 1) lies on the curve,
2 dy
24. (a) dx corresponds to 14 here, so take dx as 12 . Then x 1y C and, since (0, 1) lies on the curve,
dy y y
C 1. So y 1 .
1 x
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
x dy /4 2 /4 /4
25. y cos 2t dt dx
cos 2 x L 1 cos 2 x dx 1 cos 2 x dx 2 cos 2 x dx
0 0 0 0
/4
0
2 cos x dx 2 sin x 0
/4
2 sin 4 2 sin(0) 1
2
1 x 2/3 1/ 2 1 x 2/3 1/ 2
26.
y 1 x
2/3 3/2
, 4
2
x 1
dy
dx
3
2 1 x
2/3 1/2
23 x 1/3
x1/3
L
1
2 /4
1
x1/3
dx
1 1 1
2/3
1 1 x2/3 dx 1 1 dx 1 1 1
x 1/3 dx 32 x 2/3
2 /4 x 2 /4
1 2/3
x 2 /4 x
1
2/3
dx 1
2 /4 x1/3
dx
2 /4 2 /4
2/3
32 (1)2/3 32 42
32 32 12 43 total length 8 34 6
dy 2 2 2
27. y 3 2 x, 0 x 2 dx 2 L 1 (2)2 dx 5 dx 5 x 2 5.
0 0 0
d (2 0)2 (3 (1))2 2 5
28. Consider the circle x 2 y 2 r 2 , we will find the length of the portion in the first quadrant, and multiply our
result by 4.
2
dy r r 2 r r 2 dx
y r 2 x 2 , 0 x r dx 2 x 2 L 4 1 2 x 2 dx 4 1 2x 2 dx 4
r x 0 r x 0 r x 0 r x2
2
r r
4 r dx 4r dx
0 r 2 x2 0 r 2 x2
30. 4 x 2 y 2 64 d
dx
4 x 2 y 2
d
dx 64 8x 2 y dy
dx
dy
0 dx 4yx dy 4yx dx;
2
2 y 2 16 x 2
ds 2 dx 2 dy 2 dx 2 4yx dx dx 2 16 x2 dx 2 1 16 x2 dx 2
2
dx 2 4 x 2 6416 x 2 dx 2
y y y2 y2
20 x 2 64 dx 2 4 (5 x 2 16) dx 2
y2 y2
31. 2x
0
x
1
dy 2
dt
dt , x 0 2 1 dy 2
dx
dy
dx
1 y f ( x) x C where C is any real
number.
xk 2 (dy)2 xk 2 f ( xk 1 ) xk 2 .
n n
(length of kth tangent fin) nlim xk f ( xk 1 ) xk
2 2
(b) Length of curve lim
n k 1 k 1
n b
1 f ( xk 1 ) xk 1 f ( x) dx
2 2
lim
n k 1 a
4
33. x 2 y 2 1 y 1 x 2 ; P 0, 14 , 12 , 43 , 1 L xi xi 1 yi yi 1
2 2
k 1
2 2 2 2
14 0 12 14 34 12 1 34
2 15 2 3 15 2 7 3 2 7
4
1 2
4
4
2
0 4
1.55225
y2 y1 dy
34. Let ( x1 , y1 ) and ( x2 , y2 ), with x2 x1 , lie on y mx b, where m x2 x1
, then dx
m
L
x2
x1
1 m 2 dx 1 m2 x x2 1 m 2 x2 x1 1
x
1
y2 y1 2
x2 x1 x2 x1
x2 x1 2 y2 y1 2 x2 x1 2 y2 y1 2
x2 x1
x2 x1
x2 x1 x2 x1 2 y2 y1 2 .
x2 x1 2
dt 0x
dy x 2
35. y 2 x3/2 dx
3x1/2 ; L( x) 1 3t1/2 1 9t dt ;
0
19 x 1 9 x
[u 1 9t du 9dt ; t 0 u 1, t x u 1 9 x] 19 u du 2 u 3/2 2 (1 9 x)3/2 2 ;
27
1 27 1 27
36. y x3 x 2 x 4 x1 4
dy
x2 2 x 1 1 ( x 1)2 1 ;
3 dx 4( x 1)2 4( x 1)2
2 2
x x 4(t 1)4 1 x [4(t 1)4 1]2
L( x) 1 (t 1) 2 1 2 dt 1 dt 0 1 dt
4(t 1)
2
0 4(t 1) 0 16(t 1) 4
x 16(t 1) 4 16(t 1)8 8(t 1)4 1 x 16(t 1)8 8(t 1)4 1 x [4(t 1)4 1]2 x 4(t 1) 4 1
4
dt 4
dt 4
dt dt
0 16(t 1) 0 16(t 1) 0 16(t 1) 0 4(t 1)2
x x 1
(t 1) 2 1 2 dt ; [u t 1 du dt ; t 0 u 1, t x u x 1] u 2 14 u 2 du
0 4(t 1) 1
x 1
13 u 3 14 u 1 1 ( x 1)3 4( x11) 1 1 1 ( x 1)3 4( x11) 12
1 ; L(1) 8 1 1 59
1 3 3 4 3 3 8 12 24
1. (a)
dy
dx
sec2 x dx
dy 2
sec4 x (b)
/4
S 2 (tan x) 1 sec4 x dx
0
(c) S 3.84
2. (a)
dy
dx
2 x dx dy 2
4x2 (b)
2
S 2 x 2 1 4 x 2 dx
0
(c) S 53.23
2
3. (a) xy 1 x 1 dx 1 dx
dy 14 (b)
y y2 dy y
2
S 2 1y 1 y 4 dy
1
(c) S 5.02
2
4. (a) dx cos y dx cos 2 y (b)
dy dy
S 2 (sin y ) 1 cos 2 y dy
0
(c) S 14.42
2 (b)
5. (a) x1/2 y1/2 3 y 3 x1/2
dy
dx 2 3 x1/2 12 x1/2
dx 1 3x 1/2
dy 2 2
S 2 3 x1/2 1 1 3x 1/2 dx
4 2 2
1
(c) S 63.37
1 y
2
1/2 2
(b)
6. (a) dx
dy
1 y 1/2 dx
dy
y2 y dx
2 2
S 2 1 1 y 1/2
1
(c) S 51.33
2 (b)
7. (a) dx
dy
tan y dx
dy
tan 2 y y
/3 y
2
S 2 0 tan t dt 1 tan y dy 1
0 y
x tan t dt
/3 y
2 tan t dt sec y dy
0
0.5
0 0
(c) S 2.08 x
0 0.2 0.4 0.6 0.8
dy dy 2 (b)
8. (a) dx
x 2 1 dx x2 1 y
S 2
1
5 x
1
t 2 1 dt 1 x 2 1 dx 3
y
x
t 2 1 dt
1
2
5 x
2 t 2 1 dt x dx
1 1
1
(c) S 8.55 x
0 1 2 3
dx S 2
4
b dy 2 4 4
1 14 dx 2 5 x dx 2 5 x2 4 5;
dy 2
9. y x dx 12 ; S 2 y 1 dx x
2 a 0 2 0 0
Geometry formula: base circumference 2 (2), slant height 42 22 2 5
Lateral surface area 12 (4 ) 2 5 4 5 in agreement with the integral value
dy
d 2 2 2 2
10. y x x 2y dx 2; S 2 x 1 dx 2 2 y 1 22 dy 4 5 y dy 2 5 y 2
2 dy c dy 0 0 0
2 5 4 8 5; Geometry formula: base circumference 2 (4), slant height 42 22 2 5
Lateral surface area 12 (8 ) 2 5 8 5 in agreement with the integral value
dx 2
3
dy 2
b 3 2 3
1 12 dx 2 5 ( x 1) dx 2 5 x2 x
( x 1) 2
11. dx 12 ; S 2 y 1
dy a dx 1 2 1
1
25
92 3 12 1 2 5 (4 2) 3 5; Geometry formula: r1 12 12 1, r2 32 12 2, slant height
dy
d 2 2 2
12. y x
2
12 x 2 y 1 dx
dy
2; S 2 x 1 dx
dy
2 (2 y 1) 1 4 dy 2 5 (2 y 1) dy
c 1 1
2
2 5 y 2 y 2 5 (4 2) (1 1) 4 5; Geometry formula: r1 1, r2 3,
1
slant height (2 1)2 (3 1)2 5 Frustum surface area (1 3) 5 4 5 in agreement with
the integral value
13.
dy
dx
x2
3
dy 2
dx
x4
9
S
0
2 2 x3
9
4
1 x9 dx;
u 1 x4 du 94 x3 dx 1 du x3 dx;
9 4 9
25 25/9 1/2 1
x 0 u 1, x 2 u 9
S 2 u 4 du
1
125 1 3 12527 27 98
25/9
2 23 u 3/2 3
1 27 81
14.
dy
dx
12 x 1/2 dy 2
dx
1
4x
15/4 15/4
S 2 x 1 41x dx 2 x 14 dx
3/4 3/4
3/2 15/4
154 14 34 14
3/2 3/2
2 23 x 14 43
3/4
24
3
43 1 43 (8 1) 28
3
15.
dy
dx
1 (2 2 x )
2 2 x x2
1 x
2xx 2
dy 2
dx
(1 x )2
2 x x2
1.5 (1 x )2
S 2 2 x x 2 1 dx
0.5 2 x x2
1.5 2 2
2 2 x x 2 2 x x 122 x x dx
0.5 2xx
1.5
dx 2 x 0.5 2
1.5
2
0.5
16.
dy
dx
1
2 x 1
dy 2
dx
1
4( x 1)
5 5
S 2 x 1 1 4( x11) dx 2 ( x 1) 14 dx
1 1
3/2 5
2
1
5
x 54 dx 2 32 x 54
1
254 94
3/2 3/2 3/2 3/2
43 5 54 1 54
4
3
43 53
23
3
33 6 (125 27) 986
2
49
3
2 1 2 y 3
17. dx
dy
y2 dx
dy
y4 S 1 y 4 dy;
0 3
u 1 y 4 du 4 y3 dy 1 du y3 dy;
4
y 0 u 1, y 1 u 2 S 2
1
2
13 u1/2 14 du
2 2
6 u1 2 du 6 32 u 3/2 9 ( 8 1)
1 1
S 2 13 y 3/2 y1/2 1 14 y 2 y 1 dy
3
1
y
2
1/ 2
y 1/ 2
2
1 3
3 1 3/2
y y 1/2
2
3
3
dy y1/2 13 y 1 y1/2 1/1 2 dy 13 y 1 ( y 1) dy
1 y 1
3
3 1 2
1 3
y y3 y3
32 y 1 dy 9 3 y 27
1 9 3
9 3 19 13 1 3 19 13 1
9 (18 1 3) 169
2 15/4 15/4
19. dx
dy
1 dy
dx 4 1 y S 2 2 4 y 1 41 y dy 4 (4 y ) 1 dy
4 y 0 0
15/4 15/4 3/2 3/2 3/2
4 5 y dy 4 23 (5 y )3/2 83 5 15 53/2 83 54 5
0 0 4
83 5 5 5 8 5 83 40 585 5 353 5
2 1 1 1
20. dx
dy
1 dx
dy 2 y11 S 2 2 y 1 1 2 y11 dy 2 (2 y 1) 1 dy 2 2 y1/2 dy
2 y 1 5/8 5/8 5/8
1
2 2 23 y 3/2 43 2 13/2 85
5/8
3/2 4 2 5 5 4 2 82 2 5 5
3 1 8 8 3 82 2
16 2 5 5
12
1 dy 2 1 (e 2 e
ln 2 e y e y 2 ln 2 e y e y
21. S 2 e y e y 1 2y 2 y
) dy
0 2 2 0 2 4
dy 2 dy (e 2 e
ln 2 e y e y 2 ln 2 e y e y 2 ln 2
2 e y e y 2y 2 y
)dy
0 2 2 0 2 2 0
ln 2
2 12 e2 y 2 y 12 e 2 y 2 12 e 2 ln 2 2 ln 2 12 e2 ln 2 12 0 12
0
2 12 4 2 ln 2 12 14 2 2 18 2 ln 2 16
15 ln 2
3/2 2
22. y 13 x 2 2 dy x x 2 2 dx ds 1 2 x 2 x 4 dx S 2 x 1 2 x 2 x 4 dx
0
x2 1 dx 2 0 2 x x2 1 dx 2 0 2 x3 x dx 2 x4 x2 0
2
2 2 4 2
2 x 2 44 22 4
0
2
23. ds dx 2 dy 2 y3 1 3 1 dy y 6 12 1 6 1 dy y 6 12 1 6 dy
4y 16 y 16 y
2
4 y
4 y
1
2
1
2
4 y
1
2
y 3 1 3 dy y 3 1 3 dy; S 2 y ds 2 y y 3 1 3 dy 2π y 4 14 y 2 dy
2
y5
2 5 14 y 1 2 32
1 5 8
1 15 14 2 31
5 8
1 240 (8 31 5) 253
20
24.
dy
y cos x dx sin x dx
dy 2
sin 2 x S 2
/2
/2
(cos x) 1 sin 2 x dx
1/2 dy 2
25.
dy
y a 2 x 2 dx 12 a 2 x 2 ( 2 x ) x dx x2
a2 x2 a2 x2
S 2
a
a
a 2 x 2 1 2x
2
a x 2
dx 2
a
a a2 x2 x2 dx 2 aa a dx 2 a xa a
2
2 a [a (a)] (2 a)(2a) 4 a
26.
dy dy 2 2
h 2 h 2 2
y hr x dx hr dx r 2 S 2 hr x 1 r 2 dx 2 hr x h 2r dx 2h r
h 0 h 0 h
h2 r 2 h x
h2
0
dx
h
22r h2 r 2 x2 22r h2 r 2 h2 r h2 r 2
2 2
h
0 h
dy. Now, x
d 2
2
27. The area of the surface of one work is S 2 x 1 dx
dy
y 2 162 x 162 y 2
c
162 y 2 y 2 dy
y 2 y2 7 y2 7
dx
dy
dx
dy
2 2
; S 2 162 y 2 1 2 2
dy 2
2
16 y 2 16 y 16 16 y 16
7
2 16 dy 32 9 288 904.78 cm 2 . The enamel needed to cover one surface of one wok is
16
V S 0.5 mm S 0.05 cm (904.78)(0.05) cm3 45.24 cm3 . For 5000 woks, we need
5000 V 5000 45.24 cm3 (5)(45.24) L 226.2 L 226.2 liters of each color are needed.
dy 2 2 a h 2
28. y r 2 x 2 dx 12 22 x 2 2 x 2 dy
dx 2x 2 ; S 2 r 2 x 2 1 2x dx
r x r x r x a r x2
2
a
ah
r 2 x2 x2 dx 2 r aah dx 2 rh, which is independent of a.
2 a h
29.
dy
y R 2 x 2 dx 12 22 x 2 2 x 2 dy
dx x2 ; S 2 R2 x2 1 x2 dx
R x R x R x2
2 a R x2
2
2
a
ah
R2 x2 x2 dx 2 R aah dx 2 Rh
y 2 y2
30. (a) x 2 y 2 452 x 452 y 2 dx
dy
dx
dy
;
452 y 2 45 y 2
2
S
45
22.5
2 452 y 2 1
y2
452 y 2
dy 2
45
22.5 452 y 2 y 2 dy 2 454522.5 dy
(2 )(45)(67.5) 6075 square feet
(b) 19,085 square feet
x
f (mk ) f (mk ) 2k f (mk ) f (mk ) 2k ;
x
when x xk we have
r2 f (mk ) f (mk ) xk mk
x
f (mk ) f (mk ) 2k ;
2
(b) L2k xk r2 r1 xk f (mk ) 2k f (mk ) 2k xk f (mk )xk
2 2 2 x x 2 2
parts (a) and (b) above. Thus, Sk 2 f (mk ) 1 f (mk ) xk .
2
n n b
Sk nlim 2 f (mk ) 1 f (mk ) xk 2 f ( x) 1 f ( x) dx
2 2
(d) S lim
n a
k 1 k 1
1 x
1/ 2
1 x
2/3
2/3 3/2 dy 2/3 1/2 1/3 dy 2 2/3
32. y 1 x dx
3
2
23 x 1/3
dx 1 x2/3 2/3
1 1
x x x
2/3 3/2
2/3 3/2
1 1 1 3/2 1/3
S 2 2 1 x 1 1 1 dx 4 1 x x 2/3 dx 4 1 x 2/3 x dx;
0 x 2/3 0 0
0 0
S 4 u 3/2 32 du 6 25 u 5/2 6 0 25 125
1 1
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F ( x ) kx.
3 3 3
The work done by F is W F ( x) dx k x dx k2 x 2 92k . This work is equal to 1800 J
0 0 0
92 k 1800 k 400 N/m
0
(c) We substitute F 1600 into the equation F 200 x to find 1600 200 x x 8 in.
3. We find the force constant from Hooke’s law: F kx. A force of 2 N stretches the spring to 0.02 m
N . The force of 4 N will stretch the rubber band y m, where F ky y F
2 k (0.02) k 100 m k
4N 0.04
y N
y 0.04 m 4 cm. The work done to stretch the rubber band 0.04 m is W kx dx
100 m 0
0.04
0.04 (100)(0.04)2
x dx 100 x2
2
100 0.08 J
0 0 2
21,714 21,714 lb
5. (a) We find the spring’s constant from Hooke’s law: F kx k Fx 85 3
k 7238 in
0.5 0.5
(b) The work done to compress the assembly the first half inch is W kx dx 7238 x dx
0 0
0.5
(0.5)2
7238 x2 (7238) 2
2 (7238)(0.25)
905 in-lb. The work done to compress the assembly the
0 2
second half inch is:
1.0 1.0 2 1.0
W kx dx 7238 x dx 7238 x2 7238 1 (0.5)2 (7238)(0.75) 2714 in-lb
0.5 0.5
0.5 2 2
6. First, we find the force constant from Hooke’s law: F kx k Fx 150 lb . If someone
16 150 2, 400 in
1 16
7. The force required to haul up the rope is equal to the rope’s weight, which varies steadily and is proportional
50 50
to x, the length of the rope still hanging: F ( x ) 0.624 x. The work done is: W F ( x) dx 0.624x dx
0 0
50
0.624 x2 780 J
2
0
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the
b 18
bag is x ft off the ground is F ( x) 144 4 x. The work done is: W F ( x)dx (144 4 x) dx
a 0
18
144 x 2 x 2 1944 ft-lb
0
9. The force required to lift the cable is equal to the weight of the cable paid out: F ( x ) (4.5)(180 x )
180 180
where x is the position of the car off the first floor. The work done is: W F ( x) dx 4.5 (180 x ) dx
0 0
180
4.5 180 x x2
2 2
4.5 1802 180 2
4.5180 72,900 ft-lb
0 2 2
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F ( x) k2 .
x
b b b k ( a b )
The work done is W k2 dx k 12 dx k 1x k 1b 1a ab
a k a x a
11. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (20 x), the distance the bucket is being
raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is
20 20
0.8 (20 x )dx 0.8 20 x x2 160 ft-lb.
2
F 0.8(20 x). So: W
0
0
12. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (20 x), the distance the bucket is being
raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F 2(20 x ).
20 20
2(20 x ) dx 2 20 x x2 400 ft-lb.
2
So: W
0 0
Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is
also 2.5 times as great.
c) W (62.26)(240) 125
2 933,900
933,900 ft-lb; t 275 3396 sec 0.94 hours 56.6 min
c) W (62.59)(240) 125
2 938,850
938,850 ft-lb; t 275 3414 sec 0.95 hours 56.9 min
, thickness y, and height below the top of the tank (10 y). So the
15. The slab is a disk of area x 2 2
y 2
work to pump the oil in this slab, W , is 57 (10 y ) π . The work to pump all the oil to top of the tank is
y 2
2
10
10 57
10 y 2 y3 dy 57π4 103y 4 11,875 ft lb 37,306 ft-lb
3
y4
W
0 4 0
16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 y )( ) 2 y 2
and since
the tank is half full and the volume of the original cone is V 13 r 2 h 13 52 (10) 250
3
ft 3 , half the
2
volume 250π ft 3 , and with half the volume the cone is filled to a height y, 250 1 y y y 3 500 ft.
6 6 3 4
3
500
3
500 57 57π 14 y y4
3
2 3
So W 4
14 y y dy 4 3
4
60,042 ft-lb.
0 0
17. The typical slab between the planes at y and y y has a volume of V (radius)2 (thickness)
2
20
2
y 100 y ft 3 . The force F required to lift the slab is equal to its weight:
F 51.2V 51.2 100 y lb F 5120 y lb The distance through which F must act is about
30 30
(30 y ) ft. The work it takes to lift all the kerosene is approximately W W 5120 (30 y )y ft-lb
0 0
which is a Riemann sum. The work to pump the tank dry is the limit of these sums:
30
W 5120 (30 y ) dy 5120 30 y 2 5120 900
30 y2
2
(5120)(450 ) 7,238,229.48 ft-lb
0 0
18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 lb3 for 57 lb3 . Then,
ft ft
8
10 y y 64.5 108 8 64.5
83 103 2 64.53 8
8 64.5 3 4 3 4 3
W (10 y ) y 2 dy 64.5
4 3 4
0 4 4 3 4 4
0
21.5 83 34,582.65 ft-lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance (13 y ) ft.
8
8 13 y3 y 4
83 133 2 57384 7
3 4 3
Then W 574 (13 y ) y 2 dy 574 3 4 574 1338 84 574
0 0
(19 )(82 )(7)(2) 53,482.5 ft-lb
19. The typical slab between the planes at y and y y has a volume of about V (radius) 2 (thickness)
y y ft3. The force F ( y) required to lift this slab is equal to its weight: F ( y ) 73 V
2
73 y y 73 y y lb. The distance through which F ( y ) must act to lift the slab to the top of the
2
reservoir is about (4 y ) ft, so the work done is approximately W 73 y (4 y ) y ft-lb. The work done
n
lifting all the slabs from y 0 ft to y 4 ft is approximately W 73 yk 4 yk y ft-lb. Taking the limit
k 0
0 4 y y dy
4 4 2
of these Riemann sums as n , we get W 73 y (4 y ) dy 73
0
4
73 2 y 2 13 y 3 73 32 64 2336 ft-lb 2446.25 ft-lb.
0 3 3
20. The typical slab between the planes at y and y y has volume of about V (length)(width)(thickness)
2 25 y 2 (10)y ft 3 . The force F ( y ) required to lift this slab is equal to its weight:
F ( y ) 53 V 53 2 25 y 2 (10) y 1060 25 y 2 y lb. The distance through which F ( y ) must act to
lift the slab to the level of 15 m above the top of the reservoir is about (20 y ) ft, so the work done is
approximately W 1060 25 y 2 (20 y ) y ft-lb. The work done lifting all the slabs from y 5 ft to
n
y 5 ft is approximately W 1060 25 yk2 20 yk y ft-lb. Taking the limit of these Riemann sums as
k 0
5 5
n , we get W 1060 25 y 2 (20 y )dy 1060 (20 y ) 25 y 2 dy
5 5
21. The typical slab between the planes at y and y y has a volume of about V (radius) 2 (thickness)
2
25 y 2 y m3 . The force F ( y ) required to lift this slab is equal to its weight:
2
F ( y ) 9800 V 9800 25 y 2 y 9800 25 y 2 y N. The distance through which F ( y ) must
act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y ) m, so the work done is
approximately W 9800 25 y 2 (4 y )y N m. The work done lifting all the slabs from y 5 m to
0
y 0 m is approximately W 9800 25 y 2 (4 y )y N m. Taking the limit of these Riemann sums,
5
0
we get W 9800 25 y 2 (4 y ) dy 9800
5
0
5 100 25 y 4 y2 y3 dy
0
9800 100 y 25
2
y4
5
y 2 43 y 3 4 9800 500 25225 34 125 625
4
15, 073, 099.75 J
22. The typical slab between the planes at y and y y has a volume of about V (radius) 2 (thickness)
2
100 y 2 y 100 y 2 y ft 3 . The force is F ( y ) 3 V 56 100 y 2 y lb. The
56 lb
ft
Copyright 2014 Pearson Education, Inc.
474 Chapter 6 Applications of Definite Integrals
distance through which F ( y ) must act to lift the slab to the level of 2 ft above the top of the tank is about
(12 y ) ft, so the work done is W 56 100 y 2 (12 y ) y lb ft. The work done lifting all the slabs from
10
y 0 ft to y 10 ft is approximately W 56 100 y 2 (12 y )y lb ft. Taking the limit of these
0
10
Riemann sums, we get W 56 100 y 2 (12 y ) dy 56
0 10
0 100 y2 (12 y) dy
10
0
1200 100 y 12 y 2 y 3 dy 56 1200 y 2 3 4
10 100 y 12 y y 2 3 4
56
0
56 12,000 2 4 1000 4 (56 ) 12 5 4 2 (1000) 967,611 ft-lb. It would cost
10,000 10,000 5
(0.5)(967, 611) 483,805¢= $4838.05. Yes, you can afford to hire the firm.
v dvdx dx m 12 v2 ( x) x
x2 x2 x2
23. F m dv
dt
mv dv
dx
by the chain rule W mv dv
dx
dx m
x1 x1 1
12 m v ( x2 ) v ( x1 )
2 2 1 mv 2
2 1 mv 2 ,
1 as claimed.
2 2
W 12 320.3125
2
lb
ft/sec
2
(132ft/sec) 85.1 ft-lb
29. We imagine the milkshake divided into thin slabs by planes perpendicular to the y -axis at the points of a
partition of the interval [0, 7]. The typical slab between the planes at y and y y has a volume of about
weight: F ( y ) 4
9
V 49 y 17.5 2
14 y oz. The distance through which F ( y ) must act to lift this slab
to the level of 1 inch above the top is about (8 y ) in. The work done lifting the slab is about
( y 1417.5)
2
W 49 2
(8 y )y in oz. The work done lifting all the slabs from y 0 to y 7 is approximately
7
W 4 2 ( y 17.5)2 (8 y )y in oz which is a Riemann sum. The work is the limit of these sums as the
9.14
0
norm of the partition goes to zero:
W
7 4
0 9142
( y 17.5)2 (8 y ) dy 4π 7
9142 0 2450 26.25 y 27 y2 y3 dy
7
4 2 4 9 y 3 26.25
y4
y 2 2450 y 4 2 74 9 73 26.25
4
72 2450 7 91.32 in-oz
914 2 0 914 2
30. We fill the pipe and the tank. To find the work required to fill the tank note that radius = 10 ft, then
V 100y ft 3 . The force required will be F = 62.4 V = 62.4 100 y = 6240 y lb. The distance
through which F must act is y so the work done lifting the slab is about W1 6240 y y lb ft. The work it
385 385
takes to lift all the water into the tank is: W1 W1 6240 y y lb ft. Taking the limit we end up
360 360
385
6240 y dy 6240 2
385 y2 6240
with W1 2
[3852 3602 ] 182,557,949 ft-lb
360 360
To find the work required to fill the pipe, do as above, but take the radius to be 4 in. 1 ft. Then
2 6
1 y ft 3 and F 62.4 V 62.4 y. Also take different limits of summation and integration:
V 36 36
(1000) 5.975 10 4
6.672 10 11 1
6,370,000
1
35,780,000 5.144 1010
J
(b) W W1 W2 where W1 is the work done against the field of the first electron and W2 is the work done
against the field of the second electron. Let be the x-coordinate of the third electron. Then r12 ( 1) 2
and r22 ( 1)2
5
5 5
29 29
W1 23102 d 2310 2 d 23 1029 11 (23 1029 ) 14 12 23 1029 , and
3 r1 3 ( 1) 3 4
61 14 231210
5 231029 5 231029 5 29
W2 d d 23 1029 11 (23 1029 ) (3 2)
3 r22 3 ( 1) 2
3
23 1029.
12
It was proposed by Leconte and Horn to separate this series from all
the other Coleoptera as a primary division, and they looked on it as
of lower or more imperfect structure. Packard has very properly
protested against this interpretation; and there seems to be no
reason whatever for considering the Rhynchophora as "lower" than
other beetles; indeed we should be inclined to place such forms as
Calandrides amongst the most perfect of Insects; their external
structure (as shown by Eugnoristus monachus, Fig. 147) being truly
admirable.
The Platypides bore into the wood of trees and stumps; they are
chiefly exotic, and little is known about them. They are the most
aberrant of all Rhynchophora, the head being remarkably short, flat
in front, with the mouth placed on the under surface of the head,
there being no trace of a rostrum: the tarsi are elongate and slender,
the third joint not being at all lobed, while the true fourth joint is
visible. Hence they have not the appearance of Rhynchophora.
Some authorities treat the Platypides as a distinct family.
Some of the members of the group Tomicides also bore into the
wood. Recent observations have shown that there is an important
feature in the economy of certain of these wood-borers, inasmuch as
they live gregariously in the burrow, and feed on peculiar fungi that
develop there, and are called ambrosia. According to Hubbard[154],
some species cultivate these fungi, making elaborate preparations to
start their growth. The fungi, however, sometimes increase to such
an extent as to seal up the burrows, and kill the Insects by
suffocation.