6
Applications 40
Chapter of Integration
Review Exercises for Chapter 6
5
1. A
1
y
1 dx x2
1 x
5 1
4 5
1
3. A
1
1 x2 1
1
dx
arctan x
1
1
(1, 1)
4
5, ,
2 3 4
4
y
2
1 25
x
(1, 0)
(5, 0)
2
1,
1 2
1
1,
1 (1, 0)
1 2
x
1 (1, 0)
1
2
5. A
2
0
x 12 x 2
x3 dx 14 x 4
1
7. A
0
e2 xe 2
ex dx
2
2 1 2
ex
0
0
e2
y y
1
(0, e 2)
6
(2, e 2)
1
(1, 1)
x
4
1
(0, 0) 1
1
(0, 1)
x
1
1
2
3
( 1,
1)
5
4
8
9. A
4
sin x cos x 1 2 4 2
cos x d x
5 4 4
11. A
0 8
3 16x
0
8x
x2
x2
8x
3 dx
sin x 1 2 1 2 1 2
2x 2 d x 23 x 3
8 0
8x 2
20
512 3
170.667
22
4
(8, 3) (0, 3)
10
16
Review Exercises for Chapter 6
2
2
41
13. y A
1
1
x 1
x dx 2x 1 4 32 x 3
2
2
(0, 1)
1
0 1
2
(1, 0)
1
0
x dx 12 x 2
1 0
1
x
1 6
0.1667
15. x A
y2
0
2y x 1
1
1 1 2y d y
y 1
1
2
y
1
0
x
1
3
y
x y2
x
2
1 dx
1
2x y2 13 y 3
1 dx
(0, 2)
2 0
2
A
0
0
2y
0
y2 dy
4 3
1
(0, 0)
2 1 1 x 2
2
17. A
0 2 0
1 x dx 2
1
3
x 2 3
2
3
y
dx
2
1
x
2 dx
3 2
x dx 2y
(0, 1)
(3, 1)
y y A
1 x
1
x x 2 2x y 2
2 y 2 32 y 2
1 0
(2, 0)
x
2, y
1
1
3
2y d y 3 2
0 1
3y d y
0
19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years. 21. (a) Disk
4
(b) Shell x2 dx
0
V
y 4 3
x3 3
4 0
64 3
4
V
y 4 3
2
0
x2 dx
23 x 3
4 0
128 3
2 1 x 1 2 3 4
2 1 x 1 2 3 4
CONTINUED
42
Chapter 6
Applications of Integration
21. CONTINUED (c) Shell
4
(d) Shell
4
V
2
0 4
4 4x
0
x x dx x2 dx x3 3
4 0
V
2
0 4
6 6x
0
x x dx x2 dx 13 x 3
4 0
2 2
y 4 3
2 64 3
y 5 4 3
2x 2
2
3x 2
160 3
2 1 x 1 2 3
2 1 x 1 1 2 3 4 5
23. (a) Shell
4
(b) Disk x
0
V
4 3
3 4
4
16 2 16 3
x2 dx
4
V 64
0
2
0
3 4
2
16 x3 3
4
x2 48
0
dx
1 2
y 4
x2
32
9 8
16x
y 4
2 1 3 2 1 2 4 x 1 2 3 3 2 1
2 1 x 1 2 3
2 4
25. Shell
1
y
V
2
0 1 0
x x4 2x x2
2
1
dx dx
1
1
1 2 0
x 1
arctan x 0
2
4
4
Review Exercises for Chapter 6 27. Shell u x dx V u2 2u du
6
43
y 3 2 1 x 1 1 2 3 4 5 6
x 2
2
2
2 2
x 1 u3 1 x 2u du u 12 u 2
0
2
2
dx 4
4
0 2
u2 1 u
2
2u du u 3 3 1 4 20 3 u du 9 ln 3 42.359
2 3
4
0
u2
0
4
13 u 3
3u
3 ln 1
u
0
29. Since y 0, A
1
xx
1 d x.
31. From Exercise 23(a) we have: V 1 V 4
y0
64 ft 3 16 16 1 9
u x dx A
x u du
1 1 Disk:
3 1 1
16 9 9
y0
y2 dy y2 dy 13 y 3 27
y0 3
u
0
1
u du
0
u3 2 4 15
u1 2 du
1 9
9
3
2 52 u 5
y
2 32 u 3
1 0
9y 9y0 13 y 30 y03
9 27
9 0
1
x
27y0
By Newtons Method, y0
1
1.042 and the depth of the 1.958 ft.
gasoline is 3
1.042
33.
fx fx 1 fx
2
4 54 x 5 x1 4 1 1 u 2u
4
35. y x x 1
2
y
300 cosh
x 2000
280,
2000 x 2000
3 x sinh 20 2000
2000
u x dx s
s
2000
1 1 20
2000
3 x sinh 20 2000 400 9 sinh2
2
dx
1 du
3
2000
x dx 2000
4018.2 ft (by Simpsons Rule or graphing utility) uu 1 du
1
0 3
x dx
2
1
2
1
u3 2 2 52 u 5
u1 2 du 2 32 u 3
3 1
2
4 32 u 3u 15
3
5
1
8 1 15
63
6.076
44
Chapter 6 3 x 4 y 1 y
2
Applications of Integration
37. y
39. F 3 4 25 16
4
kx k1 4x
5 5
4 F W
4x d x
0
2x 2
0
S
y 4
2
0
3 x 4
25 dx 16
15 8
x2 4 2
0
15
50 in
lb
4.167 ft
lb
(4, 3)
3
2 1 x 1 2 3 4
41. Volume of disk: Weight of disk: 62.4 Distance: 175 W 62.4 9
150
1 3
2
y 1 3
2
43. Weight of section of chain: 5 x Distance moved: 10 y W 62.4 9 ton y2 2
150 0 10
x x dx 5 10 2 250 ft lb
10
5
0
10
x
2 0
y 175 y dy 163.4 ft 175y
0
104,000 ft
b
lb
45. W
a 4
F x dx a x 2dx
0
80 a
ax3 3
4 0
64 a 3
3 80 64
a
15 4
3.75
a
47. A
0
a 1 A x y 6 a2 6 a2
a
x
2
dx
0
a
2 a x1 2
x dx
ax
4 3
a x3 2
12 x 2
a 0
a2 6
x
0 a
a a
0
x
2
dx
4
6 a2 dx 6ax
a
ax
0
2 a x3 2
x2 dx
a
y
61 a2 2 3 a2
a
x
( x, y )
a2
0
4a 2x 3 1 2 8 32 32 ax 3
4a 1 2x 3 2 8 12 52 ax 5
x2 dx 13 x 3
a 0
a
x
32 ax a2 x, y aa , 55
3ax 2
a 5
R eview Exercises for Chapter 6 49. By symmetry, x
1
45
0. x2 dx 2 a 2x x3 3
a 0
y
A 1 A y
2
0
a2
4a 3 3
a2
3 4a 3 31 4a 3 2 6 8a 3
a a
( x, y )
a2
a
x2 2 dx x4 dx 15 x 5 15 a 5
a 0
a
a
x
a4
0
2a 2x 2 2a 2 3 x 3 25 a 3
6 a 4x 8a 3 6 a5 8a 3 x, y 0, 2a 2 5
2a 2 5
51. y
0 by symmetry
4 3 2
y 1 y= 6x+1
For the trapezoid: m My
0 6 0
(6, 2)
46
6
16 1 x 6 12 x 3 1 2x d x
18 1 x 6 x3 9 1 x2
0
1 x 1 2 3 4 5 7
x
dx
6
1 2 3 4
(6, 2)
60
For the semicircle: m My
6
1 2
8
2 x
2
2
8
4
x u 6 2 4 3 4
6
2
4
x
6
2
dx 6, u u2 du 16 3
2
6
x4
x 8, u
6 2 dx 2.
Let u My
x 2
6, then x
2
6 and dx u2 du
2
du. When x
2
0. When x
2
u
0
2
0
u4 2 4
2
12
0
4 44 3
u2 du 9
2 Thus, we have: x 18 2
1 2
u2
32 0
12
12
60 x 180
44 3 44 3 29 3
9 9 29 49 ,0 . 9 1 29 3 49 9
The centroid of the blade is
46
Chapter 6
Applications of Integration weight per cubic volume. g y dy
d
y
53. Let D F
c
surface of liquid;
d
D
d
y fy
D
Dfy
c
g y dy
c
yfy
d
g y dy
d g c f
( x, y )
d
yfy fy g y dy D
c d
g y dy
x
c c
fy y
g y dy
Area D
Area depth of centroid
Problem Solving for Chapter 6
1. T R
0
12 cc 2
c
13 c 2 x2 dx cx2 2 x3 3
c 0
cx
c3 2
c3 3
c3 6
T lim c0 R
13 c 2 lim c0 1 c3 6
1
3
3. (a)
1 V 2
2
0 1
1 41
y2
2
2 1 y2
1 4
y2
2
dy y2 1 y2 dy
4
0 1
y2
41
8
0
1 2
y2 dy
2
Integral represents 1 4 area of circle 4 R
2 2
8 (b) x 1 V 2 R
2 r
4 y2 R
V
r2 x r2 y2
r2 R
y2 r2 y2
2
dy
0 r
4R r2
0
y2 dy
22
4R V 2
r 22
12 r 4
rR
rR
5. V
22
r2 h2 4
x r2 22 r 3 h3 8
r
x2 dx
r
r2 h2 4
2 r2 h 4 r
x2 + y2 = r2
2 4 3
x2
32
h 2
h3 which does not depend on r! 6
Problem Solving for Chapter 6 7. (a) Tangent at A: y y x3, y 1 y To find point B: x3 x Tangent at B: y y 1
2
47
3x2 3x 3x 1 2 x3 3x 0 0B 2, 8 2
(b) Tangent at A a, a3 : y
a3
3a2 x y 3a2x 2a3 2a 2a, 12a2 x 12a2x 16a3 x 4a
a 2a 3
To find point B:
x3 x
3a2x a
2
0 0 8a3 2a 16a3 0 0
x B
3x x
2 2 3x2
Tangent at B:
y
8a3 y
x3, y 8 y
12 x 12x x3
2 16 12x 0 0C 27 4 108 16 4, 64 16
To find point C:
x3 x
12a2x 2a
2
To find point C: x3 x
1
C
a
4a, 64a3 2a3 dx
3
12x 2
2
16 4 2 dx x3 dx area S area R
Area of R
2a 4a
x3 12a x
2a 2
3a2x 16a
x
27 4 a 4 108a
4
Area of R
2 4
x3 12x
2
3x 16
Area of S Area of S
x dx
3
Area of S Area of S
16 area of R
16 area of R
x
9. s x (a) s x (b) ds ds (c) s x
2
1 ds dx 1 1
x
f t 2 dt 1 fx
2
f x 2 dx fx 1
2
dx 31 t 2
2
1
2
dy dx
x
2
dx 1
2
dx
2
dy
2
2
dt
1
1 2
9 t dt 4 22 27 1 to x 22 2.
b
(d) s 2
1
1
9 t dt 4
8 1 27
9 t 4
32 2 1
13 27
13
2.0858
This is the length of the curve y
x3 2 from x
11. (a) y My
0 by symmetry
6
(b) m 1 x3 1 x2
6
2
1 6
1 dx x3
b2 b2 2b b
1 1 2b b 1 2 2b b 2, 0 1
x
1 6
1 x3 1 dx x3 12 7
dx
1 6 1
2 dx x2
2
1 x
6 1
5 3
My x
2
1
1 dx x2 1b 1 b2
m x
y
3 2 1
2
1
35 36 12 ,0 7
2b b2
b
x, y x, y
,0
53 35 36
x, y
(c) lim x
b
lim
2b b 1
x
1 2 3 2 3 4 5 6
48
Chapter 6 area area
Applications of Integration 2 3 4 1 6 1 12 2 1 2 7 1 2 15. Point of equilibrium: 50 x P0, x0 10, 80
80
13. (a) W (b) W
0.5x 80, p
0.125x 10
Consumer surplus
0 80
50 10
0
0.5x 0.125x dx
10 dx 400
1600
Producer surplus
17. (a) Wall at shallow end From Exercise 22: F (b) Wall at deep end From Exercise 22: F (c) Side wall From Exercise 22: F1
4
62.4 2 4 20
9984 lb
62.4 4 8 20
39,936 lb
20 15
y
62.4 2 4 40 y 10y d y y2 dy 624 4y 2
19,968 lb
y=8
10
x = 40
F2
62.4
0 4
8 8y
0
5
624
y3 3
4
5 10 15 20 25 40 45 1 y = 10 x
x
0
26,624 lb Total force: F1 F2 46,592 lb