Thomas Calculus 13th [Solutions]

(a)
Chapter 6 Additional and Advanced Exercises 501
9
2 9
1 9 9 x 20
9 1/2
9
M x
dx ; M 3
1 2 3
x 2 2
1
9 y x dx x
1
3/2
x
1
3 2 12 ;
9
20
M 9
3 1/2
12
1 dx 6 x 4 x My
3/2
1
M 4
3 and M 9
y x
5
x
M 4 9
9 x 9 9 1
9
9
M x dx 4;
2
9
3
3/2
9 1/2
9
M 2 52;
1 x 2 x
3/2
1
y x dx x M x 3 dx 6 x
1
3/2
x
1
1 x
1
(b)
3
2
12
M y
M
13
3
M x
x and y 1
M
3
39.
40.
41.
b strip 2 2 2 2 y
F W depth L( y) dy F 2 (62.4)(2 y)(2 y) dy 249.6 2y y dy 249.6 y
a
0 0
3
(249.6) 4
8
(249.6)
4
332.8 lb
3 3
b strip 5/6
5
5/6
5 10 2
F W depth L( y) dy F 75 y (2y 4) dy 75 y 2y 4y dy
a
0 6 0 3 3
5/6 2 2 3
5/6
75
10 7 y 2y dy 75
10 y 7 y 2 y (75)
50 7 25 2 125
0 3 3 3 6 3 0
18 6 36 3 216
25 175 250 75
(75)(3075)
(75) (25 216 175 9 250 3) 118.63 lb.
9 216 3 216 9 216 9 216
b strip 4 y
4 1/2 3/2
F W depth L( y) dy F 62.4 (9 y) 2 dy 62.4 9y 3y dy
a
0 2
0
3 2
0
3/2
4
2 5/2 2 62.4
(62.4)(176)
y y
5 0
5 5 5
62.4 6 (62.4) 6 8 32 (48 5 64) 2196.48 lb
42. Place the origin at the bottom of the tank. Then
column, strip depth
2
2 h 849 2
2 2
849 h h . Now solve
2 2 3
s h 1 9.707 9.707 ft .
h
0
strip
depth
h y, L( y) 1 F
2 h
h h y
849( h y) 1 dy 849 ( h y) dy 849 h y
0 0
2
0
849
2
2
F W L( y) dy,
h the height of the mercury
h 40000 to get h 9.707 ft. The volume of the mercury is
CHAPTER 6
ADDITIONAL AND ADVANCED EXERCISES
1.
b
2 2 x 2 2
a
V f ( x ) dx b ab f ( t ) dt x ax for all
a
f ( x) 2x a
2
x a f ( x) 2x a
2.
a
0
2 2 x 2 2
a
V f ( x) dx a a f ( t)
dt x x for all
2 2x
1
x a f x x f x
( ) 2 1 ( )
3.
x
2 2 2
s x Cx f t dt Cx f x C f x C C
0
( ) 1 ( ) 1 ( ) ( ) 1 for 1
f ( x) x 2
C
0
1 dt k . Then
C
where 1.
x 2 2
f a a k f x C dt a f x x C a
0
(0) 0 ( ) 1 ( ) 1 ,
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