CSM Chapters6

Chapter 6

Applications of the Integral

6.1 Rectilinear Motion Revisited

1. s(t) =

6 dt = 6t+ c; 5 = s(2) = 6(2) + c; c = 7; s(t) = 6t 7

2. s(t) =

(2t+ 1) dt = t2 + t+ c; 0 = s(1) = 12 + 1 + c = 2 + c; c = 2;

s(t) = t2 + t 2

3. s(t) =

(t2 4t) dt = 1

3t3 2t2 + c; 6 = s(3) = 9 + c; c = 15; s(t) = 1

3t3 2t2 + 15

4. s(t) =

4t+ 5 dt =

1

6(4t+ 5)3/2 + c; 2 = s(1) =

9

2+ c; c = 5

2;

s(t) =1

6(4t+ 5)2/3 5

2

5. s(t) =

10 cos

4t+

6

dt =

5

2sin

4t+

6

+ c;

5

4= s(0) = 5

2

1

2

+ c = 5

4+ c;

c =5

2; s(t) = 5

2sin

4t+

6

+

5

2

6. s(t) =

2 sin 3t dt = 2

3cos 3t+ c; 0 = s() =

2

3+ c; c = 2

3; s(t) = 2

3cos 3t 2

3

7. v(t) =

5 dt = 5t+ c; 4 = v(1) = 5 + c; c = 9; v(t) = 5t+ 9;

s(t) =

(5t+ 9) dt = 5

2t2 + 9t+ c; 2 = s(1) =

13

2+ c; c = 9

2;

s(t) = 52t2 + 9t 9

2

335

336 CHAPTER 6. APPLICATIONS OF THE INTEGRAL

8. v(t) =

6t dt = 3t2 + c; 0 = v(2) = 12 + c; c = 12; v(t) = 3t2 12;

s(t) =

(3t2 12) dt = t3 12t+ c; 5 = s(2) = 16 + c; c = 11;

s(t) = t3 12t+ 11

9. v(t) =

(3t2 4t+ 5) dt = t3 2t2 + 5t+ c; 3 = v(0) = c; v(t) = t3 2t2 + 5t 3;

s(t) =

(t3 2t2 + 5t 3) dt = 1

4t3 2

3t3 +

5

2t2 3t+ c; 10 = s(0) = c;

s(t) =1

4t4 2

3t3 +

5

2t2 3t+ 10

10. v(t) =

(t 1)2 dt = 1

3(t 1)3 + c; 4 = v(1) = c; c = 4; v(t) = 1

3(t 1)3 + 4;

s(t) =

1

3(t 3)3 + 4

dt =

1

12(t 1)4 + 4t+ c; 6 = s(1) = 4 + c; c = 2;

s(t) =1

12(t 1)4 + 4t+ 2

11. v(t) =

(7t1/3 1) dt = 21

4t4/3 t+ c; 50 = v(8) = 76 + c; c = 26;

v(t) =21

4t4/3 t+ 26;

s(t) =

21

4t4/3 t 26

dt =

9

4t7/3 1

2t2 26t+ c; 0 = s(8) = 48 + c; c = 48;

s(t) =9

4t7/3 1

2t2 26t 48

12. v(t) =

100 cos 5t dt = 20 sin 5t+ c; 20 = v(/2) = 20 + c; c = 40;

v(t) = 20 sin 5t 40;s(t) =

(20 sin 5t 40) dt = 4 cos 5t 40t+ c; 15 = s(/2) = 20 + c; c = 15 + 20;

s(t) = 4 cos 5t 40t+ 15 + 20

13. v(t) = 2t 2 = 2(t 1)

dist. =

50

|2(t 1)| dt = 2 10(t 1) dt+ 2

51(t 1) dt

= 2

12t2 + t

10

+ 2

1

2t2 t

51

= 2

1

2 0

+ 2

15

212

= 17 cm

6.1. RECTILINEAR MOTION REVISITED 337

14. v(t) = 2t+ 4 = 2(t 2)

dist. =

60

| 2(t 2)| dt = 2 20(t 2) dt+ 2

62(t 2) dt

= 2

12t2 + 2t

20

+ 2

1

2t2 2t

62

= 2(2 0) + 2[6 (2)] = 20 cm

15. v(t) = 3t2 6t 9 = 3(t+ 1)(t 3)

dist. =

40

|13t2 6t 9| dt = 3 30(t2 2t 3) dt+ 3

43(t2 2t 3) dt

= 3

13t3 + t2 + 3t

30

+ 3

1

3t3 t2 3t

43

= 3(9 0) + 320

3 (9)

= 34 cm

16. v(t) = 4t3 64t = 4t(t+ 4)(t 4)

dist. =

51

|4t3 64t| dt = 4 41(t3 16t) dt+ 4

54(t3 16t) dt

= 4

14t4 + 8t2

41

+ 4

1

4t4 8t2

54

= 4

64 31

4

+ 4

175

4 (64)

= 306 cm

17. v(t) = 6 cost

dist. =

31

|6 cost| dt = 6 3/21

cost dt+ 6 5/23/2

cost dt+ 6

35/2 cost dt

= 6( sint)]3/21 + 6(sint)]5/23/2 + 6( sint)]35/2= 6[(1) 0] + 6[1 (1)] + 6[0 (1)] = 24 cm

18. v(t) = 2(t 3)

dist. =

72

|2(t 3)| dt = 2 32(t 3) dt+ 2

73(t 3) dt

= 2

12t2 + 3t

32

+ 2

1

2t2 3t

73

= 2

9

2 4

+ 2

7

292

= 17 cm

19. We first convert mi/h to mi/s: 60 mi/h = 60/3600 mi/s. Then the distance traveled is 20

60

3600dt =

60

3600t

20

=60

1800mi =

1

30mi 5280 ft/mi = 176 ft.

20. a(t) = 32; v(0) = 0; s(0) = 144; v(t) =32 dt = 32t+ c; 0 = v(0) = c; v(t) = 32t;

s(t) =

32t dt = 16t2 + c; 144 = s(0) = c; s(t) = 16t2 + 144

To find when the ball hits the ground, we solve s(t) = 16t2 + 144 = 0. This gives t = 3.The ball hits the ground in 3 seconds. Its speed at this time is |v(t)| = | 96| = 96 ft/s.


21. a(t) = 32; v(0) = 0; s(4) = 0; v(t) =32 dt = 32t+ c; 0 = v(0) = c; v(t) = 32t;

s(t) =

32t dt = 16t2 + c; 0 = s(4) = 256 + c; c = 256; s(t) = 16t2 + 256

The height of the building is s(0) = 256 ft.

22. Let the depth of the well be h.

a(t) = 32; v(0) = 0; s(0) = h; v(t) =32 dt = 32t+ c; 0 = v(0) = c; v(t) = 32t;

s(t) =

32t dt = 16t2 + c; h = s(0) = c; s(t) = 16t2 + h

If tr is the time for the rock to hit the water, then 0 = s(tr) = 16t2r + h, and h = 16t2r.Since the speed of sound is 1080 ft/s and the sound is heard after 2 seconds, h = 1080(2 tr).Then 16t2r = 1080(2 tr) or 2t2r + 135tr 270 = 0. Using the quadratic formula to find thepositive root, we obtain

tr =135 +18, 225 + 2, 160

4=13520, 385

4 1.9440 s.

Then the depth of the well is h = 1080(2 tr) 60.4669 ft.

23. a(t) = 9.8; v(0) = 24.5; s(0) = 0; v(t) =9.8 dt = 9.8t+ c; 24.5 = v(0) = c;

v(t) = 9.8t+ 24.5; s(t) =(9.8t+ 24.5) dt = 4.9t2 + 24.5t+ c; 0 = s(0) = c;

s(t) = 4.9t2 + 24.5tSolving v(t) = 9.8t + 24.5 = 0, we see that the maximum height is attained when t = 2.5seconds. The maximum height is s(2.5) = 30.625 m.

24. a(t) = 3.7; v(0) = 24.5; s(0) = 0; v(t) =3.7 dt = 3.7t+ c; 24.5 = v(0) = c;

v(t) = 3.7t+ 24.5; s(t) =(3.7t+ 24.5) dt = 1.85t2 + 24.5t+ c; 0 = s(0) = c;

s(t) = 1.85t2 + 24.5tSolving v(t) = 3.7t+24.5 = 0 we see that the maximum height is attained when t 6.6216seconds. The maximum height is s(6.6216) 81.1149 m.

25. a(t) = 32; v(0) = 32; s(0) = 384; v(t) =32 dt = 32t+ c; 32 = v(0) = c;

v(t) = 32t+ 32; s(t) =(32t+ 32) dt = 16t2 + 32t+ c; 384 = s(0) = c;

s(t) = 16t2 + 32t+ 384Solving v(t) = 32t + 32 = 0 we see that the maximum height is attained when t = 1second. The maximum height is s(1) = 400 ft. Setting s(t) = 16t2+32t+384 = 0, we havet2 2t 24 = (t 6)(t+ 4) = 0. Thus, the ball hits the ground at 6 seconds.

6.1. RECTILINEAR MOTION REVISITED 339

26. Setting s(t) = 16t2 + 32t + 384 = 256, we have t2 2t 8 = (t 4)(t + 2) = 0. Thus, theball passes the observer at 4 seconds. At this time v(4) = 96 ft/s.

27. a(t) = 32; v(0) = 16; s(0) = 102; v(t) =32 dt = 32t+ c; 16 = v(0) = c;

v(t) = 32t 16; s(t) =(32t 16) dt = 16t2 16t+ c; 102 = s(0) = c;

s(t) = 16t2 16t+ 102Solving s(t) = 16t2 16t+ 102 = 6, we see that the marshmallow hits the person at t = 2seconds. The impact velocity is v(2) = 80 ft/s.

28. a(t) = 32; v(0) = 96; s(0) = 22; v(t) =32 dt = 32t+ c; 96 = v(0) = c;

v(t) = 32t+ 96; s(t) =(32t+ 96) dt = 16t2 + 96t+ c; 22 = s(0) = c;

s(t) = 16t2 + 96t+ 22Solving s(t) = 16t2 + 96t+ 22 = 102, we see that the stone hits the culprit at t = 1 second(or t = 5 seconds if it misses on the way up and hits on its way back down). The impactvelocity is v(1) = 64 ft/s.

29. We measure upward from the top of the volcano, so that s(0) = 0. From a(t) = g = 1.8 weobtain v(t) = 1.8t+ v0 and s(t) = 0.9t2 + v0t. If the rock attains its maximum height attime t1, then v(t1) = 0 = 1.8t+ v0 and t1 = v0/1.8. Solving

200, 000 = 0.9t21 + v0t1 = 0.9 v01.8

2+ v0

v01.8

= 0.9

v01.8

2=

v203.6

gives v03.6(200, 000) 848.5 m/s.

30. (a) Taking a(t) = 32, v(0) = 2, and s(0) = 25, we have v(t) = 32t2and s(t) = 16t2 2t + 25. Using similar triangles, we obtain 25

s=

x

x 30 , 25(x 30) = sx and x =750

25 s . Then

dx

dt=

750

(25 s)2ds

dt=

750

(25 s)2 v(t) =750

(25 s)2 (32t 2)

= 1500(16t+ 1)(25 s)2 =

1500(16t+ 1)

(16t2 + 2t)2= 375(16t+ 1)

t2(8t+ 1)2.

s30

xx 30

(b) When t = 1/2,dx

dt= 375(8 + 1)1

4 (4 + 1)2= 540 ft/s.

31. From the hint, a =dv

dt=dv

dsv, and integrating with respect to s gives

a ds =

vdv

ds

ds.

Then as =1

2v2 + c, and solving for v we have v2 = 2as 2c. Since v = v0 when s = 0,

v20 = 2c and v2 = 2as+ v20 .


32. Let a be the acceleration due to gravity, v(0) = v0, and s(0) = 0.

v(t) =

a dt = at+ c; v0 = v(0) = c; v(t) = at+ v0;

s(t) =

(at+ v0) dt =

1

2at2 + v0t+ c; 0 = s(0) = c; s(t) =

1

2at2 + v0t

Solving s(t) =1

2at2 + v0t = 0, we obtain t = 0 and t = 2v0

a. Then v(2v0/a) = v0, and

the speed at impact with the ground is the initial velocity v0.

33. Let a be the acceleration of gravity on the earth, v(0) = v0, and s(0) = 0.

v(t) =

a dt = at+ c; v0 = v(0) = c; v(t) = at+ v0;

s(t) =

(at+ v0) dt =

1

2at2 + v0t+ c 0 = s(0) = c; s(t) =

1

2at2 + v0t

To find the maximum height reached on earth, we solve v(t) = at + v0 = 0. The maximumheight is reached when t = v0/a and is s(v0/a) = v20/2a v20/a = v20/2a. On theplanet, the acceleration of gravity is a/2. Proceeding as on the earth, we obtain v(t) =1

2at + v0, and s(t) =

1

4at2 + v0t. To find the maximum height reached on the planet, we

solve v(t) =1

2at + v0 = 0. The maximum height is reached when t = 2v0/a and is

s(2v0/a) = v20/a 2v20/a = v20/a. Thus, the maximum height reached on the planet istwice that reached on earth.

34. Let a be the acceleration due to gravity on earth. Then, with initial velocity 2v0, we have

ve(t) = at + 2v0 and se(t) =1

2at2 + 2v0t. On the planet, with acceleration due to gravity

a/2 and initial velocity v0, we have vp(t) =1

2at + v0 and sp(t) =

1

4at2 + v0t. To find the

maximum height reached on earth, we solve ve(t) = at+ 2v0 = 0 and obtain t = 2v0a

. The

maximum height is se(2v0/a) = 2v20

a 4v

20

a= 2v

20

a. To find the maximum height reached

on the planet, we solve vp(t) =1

2at + v0 = 0 and obtain t = 2v0

a. The maximum height

is sp(2v0/a) = v20

a 2v

20

a= v

20

a. Thus, the maximum height reached on earth is twice

that reached on the planet. We want to find the initial velocity 0 on the earth so that the

maximum height reached on earth is v20

a, the maximum height reached on the planet. With

initial velocity 0, we have ve(t) = at+0 and se(t) =1

2at2+0t. Solving ve(t) = at+0 = 0

we obtain t = 0a. Then, we want s(0/a) =

20

2a

20

a=

20

2ato be equal to v

20

a. Solving

for 0 we see that the initial velocity on earth must be2v0.

6.2 Area Revisited

6.2. AREA REVISITED 341

-2 -1 1 2

-2

-1

1

2

1. A =

11(x2 1) dx =

13x3 + x

11

=2

323

=

4

3

-1 1 2 3

-1

1

2

3

2. A =

10(x2 1) dx+

21(x2 1) dx =

13x3 + x

10

+

1

3x3 x

21

=

2

3 0

+

2

323

= 2

-3 -2 -1 1 2

-30

-20

-10

10

3. A =

03x3 dx = 1

4x403

= 081

4

=

81

4

-1 1 2

-5

4. A =

10(1 x3) dx+

21(1 x3) dx =

x 1

4x41

0

+

x+ 1

4x42

1

=

3

4 0

+

2

34

=

7

2

-3

3

3

5. A =

30(x2 3x) dx =

13x3 3

2x23

0

=9

2 0 = 9

2

-3

-2 -1 1

6. A =

01

(x+ 1)2 dx =1

3(x+ 1)3

01

=1

3 0 = 1

3

-6

-3

3

6

-2 -1 1 2

7. A =

01

(x3 6x) dx+ 10(x3 6x) dx

=

1

4x4 3x2

01

+

14x4 3x2

10

=

0

11

4

+

11

4 0

=

11

2

-1 1 2

-5

8. A =

10(x3 3x2 + 2) dx+

21(x3 3x2 + 2) dx

=

1

4x4 x3 + 2x

10

+

14x4 + x3 2x

21

=

5

4 0

+

0

54

=

5

2


-6

-4

-2

2 49. A =

10(x3 6x2 + 11x 6) dx+

21(x3 6x2 + 11x 6) dx

+

32(x3 6x2 + 11x 6) dx

=

14x4 + 2x3 11

2x2 + 6x

10

+

1

4x4 2x3 + 11

2x2 6x

21

+

14x4 + 2x3 11

2x2 + 6x

32

=

9

4 0

+

(2)

94

+

9

4 2

=

11

4

-1

1

-1 1

10. A =

01

(x3 x) dx+ 10(x3 x) dx

=

1

4x4 1

2x20

1+

14x4 +

1

2x21

0

=

0

14

+

1

4 0

=

1

2

-2

211. A =

11/2(1 x2) dx+

31(1 x2) dx =

x 1

x

11/2

+

x+

1

x

31

=

2

52

+

10

3 2

=

11

6

-2

212. A =

21(1 x2) dx =

x+

1

x

21

=5

2 2 = 1

2

-2

2

2 4

13. A =

10(x1/2 1) dx+

41(x1/2 1) dx

=

23x3/2 + x

10

+

2

3x3/2 x

41

=

1

3 0

+

4

313

= 2

-2

2

4 8

14. A =

40(2 x1/2) dx+

94(2 x1/2) dx

=

2x 2

3x3/2

40

+

2x+ 2

3x3/2

94

=

8

3 0

+

0

83

=

16

3


-2

2

3

15. A =

02x1/3 dx+

30x1/3 dx = 3

4x4/3

02

+3

4x4/3

30

=

0 +

3

4(24/3)

+

3

4(34/3) 0

=

3

4(24/3 + 34/3)

3

4 8

16. A =

81

(2 x1/3) dx =2x 3

4x4/3

81

= 411

4

=

27

4

-1

1

-! !

17. A =

0 sinx dx+

0

sinx dx = cosx]0 cosx]0= [1 (1)] (1 1) = 4

-1

1

2

3

! 2! 3!

18. A =

30

(1 + cosx) dx = (x+ sinx)]30 = 3 0 = 3

-2

2

19. A =

/23/2

(1 + sinx) dx = (x+ cosx)]/23/2 =

23

2

= 2

2

4

-! !

20. A =

/30

sec2 x dx = tanx]/30 =3 0 = 3

-2

2

-2 2

21. A =

02x dx+

10x2 dx = 1

2x202

+1

3x310

= (0 2) +1

3 0

=

7

3

-2

2

-2 2

22. A =

23

(x+ 2) dx+ 02

(x+ 2) dx+

20

(2 x2) dx+ 22(2 x2) dx

= 1

2x2 + 2x

23

+

1

2x2 + 2x

02

+

2x 1

3x32

0

2x 1

3x32

2

= 2 + 3

2

+ (0 + 2) +

4

3

2 0

4

3 4

3

2

=

7

6+

8

3

2


-6

-3

3

-4 4

23. A =

30[x (2x)] dx =

303x dx =

3

2x230

=27

2

3

6

9

-4 4

24. A =

20(4x x) dx =

203x dx =

3

2x220

= 6

3

6

-4 4

25. A =

22

(4 x2) dx =4x 1

3x22

2=

16

316

3

=

32

3

1

2

-1 1

26. A =

10(x x2) dx =

1

2x2 1

3x31

0

=1

6

5

-2 2

27. A =

21

(8 x3) dx =8x 1

4x42

1= 12

33

4

=

81

4

-1

1

-1 1

28. A =

10(x1/3 x3) dx =

3

4x4/3 1

4x41

0

=1

2

2

4

-2 2

29. A =

11

[4(1 x2) (1 x2)] dx = 11

(3 3x2) dx = (3x x3)11= 2 (2) = 4

3

-2 2

30. A =

11

[2(1 x2) (x2 1)] dx = 11

(3 3x2) dx = (3x x3)11= 2 (2) = 4

3

3

31. A =

31(x x2) dx =

1

2x2 +

1

x

31

=29

6 3

2=

10

3


2 4

3

6

9

32. A =

91

y 1

y

dy =

91(y1/2 y1/2) dy =

2

3x3/2 2y1/2

91

= 1243

=

40

3

-3 3

-4

433. A =

13

[(x2 + 6) (x2 + 4x)] dx = 13

(6 4x 2x2) dx

=

6x 2x2 2

3x31

3=

10

3 (18) = 64

3

-1 1 2

-1

1

2

34. A =

3/20

[(x2 + 3x) x2] dx = 3/20

(3x 2x2) dx

=

3

2x2 2

3x33/2

0

=9

8

-8 8

435. A =

88

(4 x2/3) dx =4x 3

5x5/3

88

=64

564

5

=

128

5

-2 2

-2

2

36. A =

11

[(1 x2/3) (x2/3 1)] dx = 11

(2 2x2/3) dx

=

2x 6

5x5/3

11

=4

545

=

8

5

-2 2 4 6

5

10

15

20

37. A =

51

[(2x+ 2) (x2 2x 3)] dx+ 65[(x2 2x 3) (2x+ 2)] dx

=

51

(5 + 4x x2) dx+ 65(x2 4x 5) dx

=

5x+ 2x2 +

1

3x35

1+

1

3x3 2x2 5x

65

=100

383

+ (30)

100

3

=

118

3

-2 2 4

-2

2

4

38. A =

5/20

(x2 + 4x) 3

2x

dx =

5/20

5

2x x2

dx

=

5

4x2 1

3x35/2

0

=125

48

-4 -2 2 4

2

4

6

8

39. A =

04

x+ 6 +

1

2x

dx+

20(x+ 6 x3) dx

=

3

4x2 + 6x

04

+

1

2x2 + 6x 1

4x42

0

= (0 + 12) + (10 0) = 22


-1 1 2

-1

1

2

40. A =

10y2 dy =

1

3y310

=1

3

-2 -1 1 2

-2

-1

1

2

41. A =

21

[(2 y2) (y)] dy = 21

(2 + y y2) dy

=

2y +

1

2y2 1

3y32

1=

10

376

=

9

2

3 6

-3

3

42. A =

33

[(6 y2) y2] dy = 33

(6 2y2) dy =6y 2

3y33

3= 43 (43) = 83

2 4

-2

43. A =

02

[(y2 2y + 2) (y2 + 2y + 2)] dy = 02

(2y2 4y) dy

=

23y3 2y2

02

= 083

=

8

3

-8 -4

-4

444. A =

40[(y2 + 2y + 1) (y2 6y + 1)] dy =

40(8y 2y2) dy

=

4y2 2

3y34

0

=64

3

-2 2

2

445. A =

11

[(x+ 4) (x3 x)] dx = 11

(4 + 2x x3) dx

=

4x x2 1

4x41

1=

19

413

4

= 8

-2 2

-2

2

46. A =

01

(y3 y) dy + 10(y3 y) dy

=

1

4y4 1

2y20

1+

14y4 +

1

2y21

0

= 014

+

1

4 0 = 1

2

!

-2

2

47. A =

/40

(cosx sinx) dx+ /2/4

(sinx cosx) dx

= (sinx+ cosx)]/40 + ( cosx sinx)]/2/4=2 1 + (1) (2) = 22 2


!

-2

2

48. A =

/20

[2 sinx (x)] dx = /20

(2 sinx+ x) dx =

2 cosx+ 1

2x2/2

0

=2

8 (2) = 16 +

2

8

!

2

4

49. A =

5/6/6

(4 sinx 2) dx = (4 cosx 2x)]5/6/6

= 23 5

323

3

=

123 43

-! !

-2

2

50. A =

/2/2

[2 cosx ( cosx)] dx = /2/2

3 cosx dx

= 3 sinx]/2/2 = 3 (3) = 6

-2 2 4 6

-4

4

51. Region 1: y =x, y = x, x = 0, x = 4

-2 2 4 6

-4

4Region 2: y = x, y = x, x = 0, x = 4

-2 2 4 6

-4

4

52. Region 1: y =1

2x2, y = x 3, x = 1, x = 2

-2 2 4 6

-4

4Region 2: y = 3 x, y = 12x2, x = 1, x = 2

-2 2

2

4

6

53.

20

3x+ 1 4x dx = 1/2

0

3

x+ 1 4x

dx+

21/2

4x 3

x+ 1

dx

=3 ln |x+ 1| 2x21/2

0+2x2 3 ln |x+ 1|2

1/2

= 3 ln3

2 1

2+ 8 3 ln 3 1

2+ 3 ln

3

2

= 7 + 3 ln3

4 6.1370


-2 2

4

54.

11

ex 2ex dx = ln21

2ex ex dx+ 1

ln2

ex 2ex dx

=2ex exln21 + ex + 2ex1ln2

= 22 + 2e+ e1 + e+ 2e1 22= 3e+ 3e1 42

-3 3

4

55.

30

9 x2 dx = 1

4(3)2 =

9

4

-5 5

6

56.

55

25 x2 dx = 1

2(5)2 =

25

2

-2 2

2

4

57.

22

(1 +4 x2) dx =

22

1 dx+

22

4 x2 dx = 4 + 1

2(2)2 = 4 + 2

-2 2

4

58.

11

(2x+ 31 x2) dx =

11

(2x+ 3) dx 11

1 x2 dx

=x2 + 3x

11

1

2(1)2

= 1 + 3 1 3(1) 2= 6

2

59. The area of the ellipse is four times the area in the first quadrant portion ofthe ellipse. Thus,

A = 4

a0

b2 b2x2/a2 dx = 4b

a

a0

a2 x2 dx = 4b

a

1

4a2

= ab.

b

a


60. A =

21

(3x 2)

1

2x+

1

2

dx+

32

(2x+ 8)

1

2x+

1

2

dx

=

21

5

2x 5

2

dx+

32

52x+

15

2

dx

=

5

4x2 5

2x

21

+

54x2 +

15

2

32

= 054

+

45

4 10 = 5

22 4

2

4 (2, 4)

(1, 1)

(3, 2)

61. A =

26

(2x 2) dx+ 26

[x 2 (2)] dx

+

02

[2 (2)] dx+ 20[2 (2x 2)] dx

= 2

26

(2x 2) dx+ 02

4 dx+

20(4 2x) dx

= 2

2x+

2

3(x 2)3/2

26

+ 4x ]02 + (4x x2)20

= 2

4

20

3

+ 8 + 4 0 = 52

3

-6

-4

4

62. A =

22

1

2y + 1

(y2 2)

dy =

22

y2 +

1

2y + 3

dy =

1

3y3 +

1

4y2 + 3y

22

=29

323

3

=

52

3

63. The area with respect to x is Ax =

ln 3/20

(ex 1) dx+ ln 2ln 3/2

(2 ex) dx.

The area with respect to y is Ay =

21

ln y ln y + 1

2

dy.

If integration with respect to x is chosen, we get

Ax =

ln 3/20

(ex 1) dx+ ln 2ln 3/2

(2 ex) dx = (ex x)]ln 3/20 + (2x ex)]ln 2ln 3/2

=3

2 ln 3

2 1 + 2 ln 2 2 2 ln 3

2+

3

2= 3 ln 3 + 5 ln 2 0.1699.

If integration with respect to y is chosen, we get

Ay =

21

ln y ln y + 1

2

dy =

y ln y y (y + 1) ln y + 1

2+ (y + 1)

21

=

y ln y (y + 1) ln y + 1

2+ 1

21

= 2 ln 2 3 ln 32+ 1 ln 1 + 2 ln 1 1

= 3 ln 3 + 5 ln 2 0.1699


(see Problem 5.1.39 for the antiderivative of lnx)

64. Using Mathematica the numbers at which the curves intersect are approximately

0.4077767094044803 and 0.7148059123627778.The area is then

A =

0.71480591236277780.4077767094044803

(ex 4x2) dx 0.801284.

65. At P (x0, 1/x0) the slope of the line segment is 1/x20. The equation of the line through Qand R is then y = x/x20 + 2/x0. Setting y = 0 we see that the x-intercept is 2x0. The areais

A =

2x00

1x20x+

2

x0

dx =

12x20

x2 +2

x0x

2x00

= 2 + 4 = 2,

which does not depend on x0.

66. A =

ba(Ax+B) dx =

1

2Ax2 +Bx

ba

=1

2Ab2 +Bb

1

2Aa2 +Ba

=A

2(b2 a2) +B(b a) =

A

2(b+ a) +B

(b a)

=Aa+B +Ab+B

2(b a) = f(a) + f(b)

2(b a)

a b

f (a)f (b)

67. By symmetry with respect to the line y = x,

A = 2

a0(cosx x) dx = 2

sinx 1

2x2a

0

= 2 sin a a2

(Using Mathematica it is easily shown that a 0.739085.)68. The areas are the same. In Figure 6.2.16(b), the area of the straight swath of paint is k(ba).

Now, if y = f(x) describes the lower edge of the swath in Figure 6.2.16(a), then an equationfor the upper edge is y = f(x) + k. The area between the two graphs is then b

a{[f(x) + k] f(x)} dx =

bak dx = k(b a).

69. The areas are the same. Let w be the length of the line segments AB and CD, and without lossof generality, let AB reside on y = 0, with CD residing on y = h. Thus, in Figure 6.2.17(a),the area of the rectangle is wh. Since Figure 6.2.17(b) describes a parallelogram, the linedefined by AD can be written as x = f(y). Thus, the line defined by BC is x = f(y) + w.The area of the parallelogram is therefore h

0{[f(y) + w] f(y)} dy =

h0w dy = wh.

70. This project involves a research report, and thus a preset solution is not applicable. It is noted,however, that Cavalieris Principle relates directly to the situations presented in Problems 68and 69.

6.3. VOLUMES OF SOLIDS: SLICING METHOD 351

6.3 Volumes of Solids: Slicing Method

yx

(x, y)4

y

yy31. x2 + y2 = 16; y =

16 x2; A(x) = 3y2 = 3(16 x2)

V =

44

3(16 x2) dx = 3

16x 1

3x34

4

=3

128

3128

3

=

2563

3ft3

yx

(x, y)4

yy

2. x2 + y2 = 16; y =16 x2; A(x) = 1

2y2 =

8 1

2x2

V =

44

8 1

2x2dx =

8x 1

6x34

4

=

64

364

3

=

128

3ft3

4

-2

2 (x, y)y

3. x = y2; A(x) = 2y(8y) = 16y2 = 16x; V =

4016x dx = 8x2

40= 128

4. y = 4 x2; A(x) =3y2

4=

3

4(4 x2)2 = 3

4 2x2 + 1

4x4

V =

22

3

4 2x2 + 1

4x4dx =

3

4x 2

3x3 +

1

20x52

2

=3

64

156415

=

1283

15-2 2

4 (x, y)

yy/2

3y/2

5

-2

2

yx

5. y = 25x+ 2; A(x) =

1

2y2 =

2

25(x 5)2

V =

50

2

25(x 5)2 dx = 2

75(x 5)2

50

=10

3ft3

-3 3

-3

3

y1

3

1x

2

6. y =4 x2; A(x) = y2 (12) = (3 x2)

V =

33

(3 x2) dx = 3x 1

3x33

3= [2

3 (23)] = 43 ft3


3

3

x(x, y)

7. x = y + 3; A(y) = x2 = (y 3)2

V =

30(y 3)2 dy = 1

3(y 3)3

30

= 9

8. Let b denote the length of one side of the square base. Thus, B = b2.

Using similar triangles, we haveh

b=

h y2x

and x =b(h y)

2h. Thus,

A(y) = (2x)2 =b2(h y)2

h2, and

V =

h0

b2(h y)2h2

dy =

h0

b2 2b

2

hy +

b2

h2y2dy

=

b2y b

2

hy2 +

b2

3h2y3h

0

= b2h b2h+ b2h

3=

1

3hB.

b

hx

(x, y)y

h y

C

O A

B(1, 1)

x

9. x =y

V =

10[12 (y)2] dy =

10(1 y) dy =

y 1

2y21

0

=

2

C

O A

B(1, 1)

y

10. y = x2

V =

10(x2)2 dx =

10x4 dx =

5x510=

5

C

O A

B(1, 1)

y

11. y = x2

V =

10(12 x4) dx =

x 1

5x51

0

=4

5

C

O A

B(1, 1)

x

12. x =y

V =

10(y)2 dy =

10y dy =

2y210=

2


B(1, 1)

x

C

O A

13. x =y

V =

10(1y)2 dy =

10(1 2y + y) dy

=

y 4

3y3/2 +

1

2y21

0

=

6

C

O A

B(1, 1)

x

14. x =y

V =

10[12 (1y)2] dy =

10(2y y) dy

=

4

3y3/2 1

2y21

0

=5

6

-3 3

9

y

15. y = 9 x2

V =

33

(9 x2)2 dx = 2 30(81 18x2 + x4) dx

= 2

81x 6x3 + 1

5x53

0

=1296

5

2

5

x

16. x =y 1

V =

51(y 1)2 dy =

51(y 1) dy

=

1

2y2 y

51

=

15

212

= 8

2

1

2

6.3.17

x

17. x =1

y

V =

11/2

1

y

2 12

dy =

11/2

(y2 1) dy

=

1y y

11/2

=

2

52

=

2

1 2 3

1

2

3

y

18. y =1

x

V =

31/2

1

x

2dx =

1x

31/2

=

13 (2)

=

5

3


2 4

2

4

y

19. y = (x 2)2

V =

20(x 2)4 dx =

5(x 2)5

20=

32

5

-1

1

x

20. x =y 1

V =

10(y 1)2 dy =

10(y 2y + 1) dy

=

1

2y2 4

3y3/2 + y

10

=

6

-2 2

4

y1

y2

21. y1 = 4 x2; y2 = 1 14x2

V = 2

20

(4 x2)2

1 1

4x22

dx = 2

20

15 15

2x2 +

15

16x4dx

= 2

15x 5

2x3 +

3

16x52

0

= 32

2

2

x

22. x =1 y

V =

10(1 y)2 dy =

10(1 y) dy =

y 1

2y21

0

=

2

1 2

1

2

x1 x1 x2

23. x1 = y; x2 = y 1

V =

10y2 dy +

21[y2 (y 1)2] dy

=

1

3y31

0

+ (y2 y)21=

1

3+ 2

=

7

3

1 2

1

2

y1 y2

24. y1 = 1; y2 = 2 x

V =

1012 dx+

21(2 x)2 dx =

10dx+

21(4 4x+ x2) dx

= x]10 +

4x 2x2 + 1

3x32

1

= +

8

3 7

3

=

4

3


5

2

4

5 x

25. x = y2 + 1

V =

20[5 (y2 + 1)]2 dy =

20(4 y2)2 dy

=

20(16 8y2 + y4) dy =

16y 8

3y3 +

1

5y52

0

=256

15

1

-1

1

1 x

26. x = y2

V =

11

(1 y2)2 dy = 11

(1 2y2 + y4) dy

=

y 2

3y3 +

1

5y51

1=

8

15 815

=

16

15

1 2

1

2

1 2 y27. y = x1/3

V =

10[(2 x1/3)2 12] dx =

10(3 4x1/3 + x2/3) dx

=

3x 3x4/3 + 3

5x5/3

10

=3

5

1 2

1

2 2 x28. x = y2 + 2y

V =

20

22 2 (y2 + 2y)2 dy = 2

0(8y 8y2 + 4y3 y4) dy

=

4y2 8

3y3 + y4 1

5y52

0

=64

15

3

-3

3

x

29. x =y2 + 16

V =

33

52

y2 + 16

2dy =

33

(9 y2) dy

=

9y 1

3y33

3= [18 (18)] = 36

3 6 9

3

6

9

y1

y2

30. y1 = 9 12x2; y2 = x2 6x+ 9 = (x 3)2

V =

40

9 1

2x22 (x 3)4

dx

=

40

108x 63x2 + 12x3 3

4x4dx

=

54x2 21x3 + 3x4 3

20x54

0

=672

5


6

-5

5 x1x2

31. x1 = y + 6; x2 = y2

V =

32

[(y + 6)2 y4] dy = 32

(36 + 12y + y2 y4) dy

=

36y + 6y2 +

1

3y3 1

5y53

2=

612

5664

15

=

500

3

-4 -2 2 4

5

x32. x = (y 1)1/3

V =

91(y 1)2/3 dy = 3

5(y 1)5/3

91

=96

5

-1 1

1

y

33. y = x3 xV =

11

(x3 x)2 dx = 11

(x6 2x4 + x2) dx

=

1

7x7 2

5x5 +

1

3x31

1=

8

105 8105

=

16

105

-2 2

2

y

34. y = x3 + 1

V =

11

(x3 + 1)2 dx =

11

(x6 + 2x3 + 1) dx

=

1

7x7 +

1

2x4 + x

11

=

23

14 914

=

16

7

2

2

2 y1

35. y = ex

V =

10[(2 ex)2 12] dx =

10(3 4ex + e2x) dx

=

3x+ 4ex 1

2e2x

10

=

4e1 1

2e2 1

2

3

2

4

6

8

y

1

36. y = ex

V =

20[(ex)2 12] dx =

20(e2x 1) dx

=

1

2e2x x

20

=

1

2e4 5

2


! 2!

-1

1

y

37. y = | cosx|

V =

20

| cosx|2 dx = 20

1 + cos 2x

2dx

=

2

20

(1 + cos 2x) dx =

2

x+

1

2sin 2x

20

= 2

1

2

y

! /4 ! /4

38. y = secx

V =

/4/4

sec2 x dx = tanx]/4/4 = [1 (1)] = 2

1

y! /4

39. y = tanx

V =

/40

tan2 x dx =

/40

(sec2 x 1) dx = (tanx x)]/40

= 1

4 0

=

4 24

1

! /4

y1y2

40. y1 = cosx; y2 = sinx

V =

/40

(cos2 x sin2 x) dx = /40

cos 2x dx =

2sin 2x

/40

=

2

41. The volume of the right circular cylinder is r2h. Placing the center of the red circularcylinders base in Figure 6.3.19 on the origin, we see that A = r2 for every slice from y = 0to h. Thus, the volume V of the cylinder is

V =

h0

r2 dy = r2yh0= r2h.

yx

(x, y)a

y

42. Take the cross-sections to be rectangles perpendicular to the base ofthe cylinder and parallel to the diameter.

x2 + y2 = a2; y =a2 x2

z

2yz = xx

z(a) A(x) = 2yz = (2

a2 x2)x = 2xa2 x2

V =

a0

2xa2 x2 dx u = a2 x2, du = 2x dx

=

0a2u1/2 du = 2

3u3/2

0a2

= 23(0 a3) = 2

3a3


z = 3x

z

2yx

z(b) A(x) = 2yz = (2

a2 x2)3x = 23xa2 x2

V =

a0

23xa2 x2 dx u = a2 x2, du = 2x dx

=

0a23u1/2 du = 2

3

3u3/2

0a2

= 23

3(0 a3) = 2

3

3a3

43. (a) Using Mathematica, we obtain with the disk method

V =

11

[P (x)]2(1 x2) dx = 4315

(5a2 + 9b2 + 21c2 + 105d2 + 18ac+ 42bd).

(b) Setting a = 0.07, b = 0.02, c = 0.2, and d = 0.56 we obtain V 1.32 cubic units.(c)

-1 1

1

(d) Setting a = 0.06, b = 0.04, c = 0.1, and d = 0.54 we obtain V 1.26 cubic units.

x

r r

r

(0, h r)

h

44. (a) Using x =r2 y2 and the disk method, we obtain

V =

hrr

r2 y2

2dy

=

hrr

(r2 y2) dy = r2y 1

3y3hr

r

=

r2(h r) 1

3(h r)2

r3 + 1

3r3

= r2h 13h3.

(b) The weight of the ball is4

3r3ball and the weight of water displaced is

3(3rh2h3)water.

Using Archimedes principle andballwater

= 0.4 we have4

3(3)2(0.4) =

3(9h2 h3) or

h3 9h2 + 43.2 = 0. Solving for h we obtain h 2.5976 in.45. (a) Each eighth of the bicylinder can be sliced into squares whose sides follow the perimeter

of a quadrant of the cylinders base; that is, x2 + y2 = r2, one side of the square isy =

r2 x2, and its area is y2 = r2 x2. Using symmetry, the volume common to the

cylinders is thus

V = 8

r0(r2 x2) dx = 8

r2x x

3

3

r0

=16r3

3.

(b) This item involves a research report, and thus a preset solution is not applicable.

6.4. VOLUMES OF SOLIDS: SHELL METHOD 359

6.4 Volumes of Solids: Shell Method

C

O A

B(1, 1)

y

1. y =x

V = 2

10xx dx =

4

5x5/2

10

=4

5

C

O A

B(1, 1)

x

2. x = y2

V = 2

10y(1 y2) dy = 2

1

2y2 1

4y41

0

=

2

C

O A

B(1, 1)

x3. x = y2

V = 2

10(1 y)y2 dy = 2

1

3y3 1

4y41

0

=

6

C

O A

B(1, 1)

x4. x = y2

V = 2

10y y2 dy =

2y410=

2

C

O A

B(1, 1)

y

5. y =x

V = 2

10(1 x)x dx = 2

2

3x3/2 2

5x5/2

10

=8

15

C

O A

B(1, 1)

y

6. y =x

V = 2

10(1 x)(1x) dx = 2

10(1x x+ x3/2) dx

= 2

x 2

3x3/2 1

2x2 +

2

5x5/2

10

=7

15


5

5

x

7. x = y

V = 2

50y y dy = 2

3y350

=250

3

1

-2

-1

1

x

8. x = 1 y

V = 2

10(y + 2)(1 y) dy = 2

10(2 y y2) dy

= 2

2y 1

2y2 1

3y31

0

=7

3

-1 1 2

1

2

3

x

9. x =y

V = 2

30yy dy =

4

5y5/2

30

=363

5

2 4

4

y

10. y = x2

V = 2

20x x2 dx =

2x420= 8

1 2 3

1

y

11. y = x2

V = 2

10(3 x)x2 dx = 2

x3 1

4x41

0

=3

2


-3 3

5

x1 x212. x1 =

y; x2 = y

V = 2

90y[y (y)] dy = 2

902y3/2 dy =

8

5y5/2

90

=1944

5

3

5

y

13. y = x2 + 4

V = 2

20x(x2 + 4 2) dx = 2

20(x3 + 2x) dx

= 2

1

4x4 + x2

20

= 16

2 4

-2

2

y

14. y = x2 5x+ 4

V = 2

41x(x2 + 5x 4) dx = 2

41(x3 + 5x2 4x) dx

= 2

14x4 +

5

3x3 2x2

41

= 2

32

3+

7

12

=

135

6

1 2

2

x1

x2

15. x1 = 1 +y; x2 = 1y

V = 2

10y[1 +

y (1y)] dy = 2

102y3/2 dy

=8

5y5/2

10

=8

5

2 4

2

4

y

16. y = (x 2)2

V = 2

40(4 x)[4 (x 2)2] dx = 2

40(x3 8x2 + 16x) dx

= 2

1

4x4 8

3x3 + 8x2

40

=128

3


2

-1

1x17. x = y3

V = 2

10(y + 1)(1 y3) dy = 2

10(1 + y y3 y4) dy

= 2

y +

1

2y2 1

4y4 1

5y51

0

=21

10

2

1

2

y1y2

18. y1 = x1/3 + 1; y2 = x+ 1

V = 2

10(1 x)[x1/3 + 1 (x+ 1)] dx

= 2

10(x1/3 + x x4/3 x2) dx

= 2

3

4x4/3 +

1

2x2 3

7x7/3 1

3x31

0

=41

42

1

1

y1y2

19. y1 = x; y2 = x2

V = 2

10x(x x2) dx = 2

1

3x3 1

4x41

0

=

6

1 2 3

1

y1y2

20. y1 = x; y2 = x2

V = 2

10(2 x)(x x2) dx

= 2

10(2x 3x2 + x3) dx

= 2

x2 x3 + 1

4x41

0

=

2

2 4

2

4

y

21. y = x3 + 3x2

V = 2

30x(x3 + 3x2) dx = 2

30(x4 + 3x3) dx

= 2

15x5 +

3

4x43

0

=243

10


-1

1

y

22. y = x3 x

V = 2

01x(x3 x) dx = 2

15x5 +

1

3x30

1=

4

15

-2 2

-2

2

y1y2

23. y1 = 2 x2; y2 = x2 2

V = 2

02

x[2 x2 (x2 2)] dx = 2 02

(2x3 4x) dx

= 2

1

2x4 2x2

02

= 4

3 6

-3

3

y1y2

24. y1 = 4x x2; y2 = x2 4x

V = 2

40(x+ 1)[4x x2 (x2 4x)] dx = 2

40(8x+ 6x2 2x3) dx

= 2

4x2 + 2x3 1

2x44

0

= 128

-5

5

x

25. x = y2 5y

V = 2

50y(y2 + 5y) dy = 2

14y4 +

5

3y35

0

=625

6

2 4 6

2

x1x2

26. x1 = y + 4; x2 = y2 + 2

V = 2

21y[y + 4 (y2 + 2)] dy = 2

21(2y + y2 y3) dy

= 2

y2 +

1

3y3 1

4y42

1

= 2

8

3 13

12

=

19

6


2

6

y1

y2

27. y1 = x+ 6; y2 = x3

V = 2

20x(x+ 6 x3) dx = 2

1

3x3 + 3x2 1

5x52

0

=248

15

1

1x1

x2 2/2

28. x1 = 1 y2; x2 = y2

V = 2

2/20

y(1 y2 y2) dy = 2 2/20

(y 2y3) dy

= 2

1

2y2 1

2y42/2

0

=

4

1

y!" /2

29. y = sinx2

V = 2

/20

x(1 sinx2) dx = 2 /20

(x x sinx2) dx

= 2

1

2x2 +

1

2cosx2

/20

= 2

4 1

2

=

2 22

2

2

y

30. y = ex2

V = 2

10x(ex

2

) dx = 2

1

2ex

2

10

= e

31. We use the shell method.

V = 2

r0(r x)

h

rx

dx = 2

r0

hx h

rx2dx = 2

1

2hx2 h

3rx3r

0

=1

3r2h

32. The equation of the line through (r1, h) and (r2, 0) is x =1

h(r1 r2)y + r2. We use the disk


method.

V =

h0

1

h(r1 r2)y + r2

2dy =

h0

1

h2(r1 r2)2y2 + 2

hr2(r1 r2)y + r22

dy

=

1

3h2(r1 r2)2y3 + 1

hr2(r1 r2)y2 + r22y

h0

= h

1

3(r21 2r1r2 + r22) + r2(r1 r2) + r22

=

h

3(r21 + r1r2 + r

22)

33. We use the disk method.

V =

rr(r2 y2)2 dy =

rr(r2 y2) dy

=

r2y 1

3y3r

r=

2

3r3

23r3

=4

3r3

34. The equation of the line is y =1

a

r2 a2x and the equation of the circle is y = r2 x2.

We use the disk method.

V =

ba

r2 a2a

x

2dx+

ba

r2 x2

2dx

= r2 a2a2

a0x2 dx+

ba(r2 x2) dx =

r2 a2a2

1

3x3a0

+

r2x 1

3x3b

a

=

3(r2 a2)a+

br2 1

3b3ar2 1

3a3

=

3(3br2 2ar2 b3)

35. The equation of the ellipse is y = b

1 x

2

a2. We use the disk method.

V =

aa

b

1 x

2

a2

2dx = b2

aa

1 x

2

a2

dx = b2

x 1

3a2x3a

a

= b22a

32a

3

=

4ab2

3

36. The equation of the ellipse is y = b

1 x

2

a2. Since the solid is symmetric with respect to

the x-axis, we will find the volume of the upper hemispheroid and multiply by 2. We use theshell method.

V = 2

2

a0x

b2 b

2

a2x2 dx

= 4

a

2

3b2

b2 b

2

a2x23/2a

0

=a2b

3


y1

y2

h

r

6.5. LENGTH OF A GRAPH 365

37. y1 =2x2

2g. The depth of the liquid below the x-axis is y2 = h

2r2

2g. So the volume is

V = 2

r0

x

2x2

2g+ h

2r2

2g

dx = 2

r0

2

2gx3 +

2hg 2r22g

x

dx

= 2

2

8gx4 +

2hg 2r24g

x2r

0

=2r4

4g+

4hgr2 22r44g

= r2h 2r4

4g

38. The liquid will touch the bottom of the bucket when y2 = h 2r2

2g= 0, or =

2hg

r. The

volume of the liquid is then

V = r2h 2r4

4g= r2h (2hg/r

2)r4

4g= r2h 1

2r2h =

1

2r2h

6.5 Length of a Graph

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

37. y1 =2x2

2g. The depth of the liquid below the x-axis is y2 = h

2r2

2g.

So the volume is

V = 2

r0x

2x2

2g+ h

2r2

2g

dx

= 2

r0

2

2gx3 +

2hg 2r22g

x

dx

= 2

2

8gx4 +

2hg 2r24g

x2r

0

=2r4

4g+

4hgr2 22r44g

= r2h 2r4

4g.

38. The liquid will touch the bottom of the bucket when y2 = h 2r2

2g= 0, or =

2hg

r. The

volume of the liquid is then

V = r2h 2r4

4g= r2h (2hg/r

2)r4

4g= r2h 1

2r2h =

1

2r2h

6.5 Length of a Graph

1. y = 1; s = 11

1 + 12 dx = 2

2

2. y = 2; s = 30

1 + 22 dx = 3

5

3. y =3

2x1/2

s =

10

1 +

9

4x dx =

8

27

1 +

9

4x

3/210

=8

27

13

4

3/2 1

=

133/2 827

1.4397

4. y = 2x1/3

s =

81

1 + 4x2/3 dx =

81

x2/3 + 4

x2/3dx =

81x1/3

x2/3 + 4 dx

u = x2/3 + 4, du =2

3x1/3 dx

=

85u1/2

3

2du

= u3/2

85= 83/2 53/2 11.4471


5. y = 2x(x2 + 1)1/2

s =

41

1 + 4x2(x2 + 1) dx =

41

(2x2 + 1)2 dx =

41(2x2 + 1) dx

=

2

3x3 + x

41

=140

3 5

3= 45

6. y = 2(x+ 1)3/2 1; y = 3(x+ 1)1/2

s =

01

1 + 9(x+ 1) dx =

01

9x+ 10 dx =

2

27(9x+ 10)3/2

01

=2

27(103/2 1) 2.2684

7. y =1

2x1/2 1

2x1/2 =

x 12x1/2

s =

41

1 +

(x 1)24x

dx =

41

4x+ x2 2x+ 1

4xdx =

41

(x+ 1)2

4xdx

=1

2

41

x+ 1

x1/2dx =

1

2

41(x1/2 + x1/2) dx =

1

2

2

3x3/2 + 2x1/2

41

=1

2

28

3 8

3

=

10

3

8. y =1

2x2 1

2x2=x4 12x2

s =

42

1 +

(x4 1)24x4

dx =

42

4x4 + x8 2x4 + 1

4x4dx

=

42

(x4 + 1)2

4x4dx =

1

2

42(x2 + x2) dx =

1

2

1

3x3 1

x

42

=1

2

253

12 13

6

=

227

24

9. y = x3 14x3

=4x6 14x3

s =

32

1 +

(4x6 1)216x6

dx =

32

16x6 + 16x12 8x6 + 1

16x6dx =

32

(4x6 + 1)2

16x6dx

=

32

4x6 + 1

4x3dx =

1

4

32(4x3 + x3) dx =

1

4

x4 1

2x2

32

=1

4

1457

18 127

8

=

4685

288 16.2674


10. y = x4 14x4

=4x8 14x4

s =

21

1 +

(4x8 1)216x8

dx =

21

16x8 + 16x16 8x8 + 1

16x8dx =

21

(4x8 + 1)2

16x8dx

=1

4

21

4x8 + 1

x4dx =

1

4

21(4x4 + x4) dx =

1

4

4

5x5 +

1

3x3

21

=1

4

3067

120 56

120

=

3011

480 6.2729

11. y = (4 x2/3)1/2

x1/3; s =

81

1 +

4 x2/3x2/3

dx =

81

2

x1/3dx = 3x2/3

81= 9

12. y =

1, 2 < x < 3

2

3(x 2)1/3, 3 < x < 103

4(x 6)1/2, 10 < x < 15

s =

32

1 + 1 dx+

103

1 +

4

9(x 2)2/3 dx+

1510

1 +

9

16(x 6) dx

=2 +

103

(x 2)2/3 + 4

9(x 2)1/3 dx+ 1

4

1510

9x 38 dx

=2 +

(x 2)2/3 + 4

9

3/210

3

+1

4

1

9

2

3

(9x 38)3/2

1510

=2 +

40

9

3/213

9

3/2+

1

54(973/2 523/2) 19.7954

13. y = 2x; s = 31

1 + 4x2 dx

14. y = (x+ 1)1/2; s = 31

1 + (x+ 1)1 dx =

31

x+ 2

x+ 1dx

15. y = cosx; s = 0

1 + cos2 x dx

16. y = sec2 x; s = /4/4

1 + sec4 x dx

17.dx

dy= 2

3y1/3


s =

80

1 +

4

9y2/3 dy =

80

9y2/3 + 4

9y2/3dy =

1

3

80y1/3

9y2/3 + 4 dy

u = 9y2/3 + 4, du = 6y1/3 dy

=1

18

404

u1/2 du =1

27u3/2

404

=1

27(403/2 8) 9.0734

18.dx

dy=

1

2y3/2 1

2y3/2 =

y3 12y3/2

s =

94

1 +

(y3 1)24y3

dy =

94

4y3 + y6 2y3 + 1

4y3dy =

94

(y3 + 1)2

4y3dy

=1

2

94

y3 + 1

y3/2dy =

1

2

94(y3/2 + y3/2) dy =

1

2

2

5y5/2 2y1/2

94

=1

2

1448

15 59

5

=

1271

30 42.3667

19. (a) y = (1 x2/3)3/2; y = 32(1 x2/3)1/2

23x1/3

= (1 x

2/3)1/2

x1/3

s =

10

1 +

1 x2/3x2/3

dx =

10

1

x2/3dx =

10

1

x1/3dx

At x = 0,1

x1/3is discontinuous.

(b) The graph is symmetric with respect to both coordinate axes, so

s = 4

10

1

x1/3dx = 4

3

2x2/3

10

= 6.

20. y = b

1 x

2

a2

1/2; y = bx

a2

1 x

2

a2

1/2

s = 4

a0

1 +

b2x2

a4

1 x

2

a2

1dx = 4

a0

1 +

b2x2

a4

a2

a2 x2dx

= 4

a0

1 +

b2x2

a2(a2 x2) dx =4

a

a0

a4 a2x2 + b2x2

a2 x2 dx

21. y =r2 x2; y = x(r2 x2)1/2

2r = s = 4

r0

1 + x2(r2 x2)1 dr = 4r

r0

1

r2 x2 dxdr

Letting r = 1 we have 2 = 4

10

11 x2 dx or

10

11 x2 dx =

2.


22. y = x3. Let s(x) = x2

1 + t6 dt. We want s(2.1) or s(2 + 0.1). Using approximation by

dierentials, we have 2.12

1 + x6 dx = s(2 + 0.1) s(2) + s(2) dx = 0 +

1 + 26(0.1) = 0.1

65 0.8062.

6.6 Area of a Surface of Revolution

1. y = x1/2

S = 2

802x1 + x1 dx = 4

80

x+ 1 dx =

8

3(x+ 1)3/2

80

=8

3(27 1) = 208

3

2. y =1

2(x+ 1)1/2

S = 2

51

x+ 1

1 +

1

4(x+ 1)1 dx = 2

51

x+ 1 +

1

4dx =

4

3

x+

5

4

3/251

=4

3

25

4

3/29

4

3/2=

6(253/2 93/2) 51.3127

3. y = 3x2

S = 2

10x31 + 9x4 dx u = 1 + 9x4, du = 36x3 dx

= 2

101

u

1

36du

=

27u3/2

101

=

27(103/2 1) 3.5631

4. y =1

3x2/3

S = 2

81x

1 +

x4/3

9dx = 2

81

x1/3

3

9x4/3 + 1 dx u = 9x4/3 + 1, du = 12x1/3 dx

=2

3

14510

u

1

12du

=

27u3/2

14510

=

27(1453/2 103/2) 199.4805

5. y = 2x; S = 2 30x1 + 4x2 dx =

6(1 + 4x2)3/2

30=

6(373/2 1) 117.3187

6. y = 2x; S = 2 20x1 + 4x2 dx =

6(1 + 4x2)3/2

20=

6(173/2 1) 36.1769

7. y = 2

S = 2

72(2x+ 1)

1 + 4 dx = 2

5

72(2x+ 1) dx = 2

5x2 + x

72

= 25(56 6) = 1005

6.6. AREA OF A SURFACE OF REVOLUTION 371

8. y = x(16 x2)1/2

S = 2

70

x1 + x2(16 x2)1 dx = 2

70

x

16

16 x2 dx = 8 70

x(16 x2)1/2 dx

u = 16 x2, du = 2x dx

= 8

916u1/2

12du

= 4

169

u1/2 du = 4(2u)169

= 4(8 6) = 8

9. y = x3 14x3 =

4x6 14x3

S = 2

21x

1 +

16x12 8x6 + 116x6

dx = 2

21x

16x6 + 16x12 8x6 + 1

16x6dx

=

2

21

(4x6 + 1)2

x2dx =

2

21(4x4 + x2) dx =

2

4

5x5 1

x

21

=

2

251

1015

=

253

20

10. y = x2 14x2 =

4x4 14x2

S = 2

21

1

3x3 +

1

4x

1 +

16x8 8x4 + 116x4

dx =

6

21

4x4 + 3

x

(4x4 + 1)2

16x4dx

=

6

21

(4x4 + 3)(4x4 + 1)

4x3dx =

6

21

4x5 + 4x+

3

4x3

dx

=

6

2

3x6 + 2x2 3

8x2

21

=

6

4855

96 55

24

=

4635

576

11. (a) f (x) = r2h

1 x

h

1/2; 1 + [f (x)]2 =

4h2 4hx+ r24h2(1 x/h)

f(x)1 + [f (x)]2 = r

1 x/h

4h2 2hx+ r22h1 x2 =

r

2h

4h2 4hx+ r2

S = 2

h0

r

2h

r2 + 4h2 4hx dx = r

h

2

3

14h

(r2 + 4h2 4hx)3/2

h0

=r

6h2[(r2 + 4h2)3/2 r3]

(b) With h = 0.1r we have

S =r

6(0.01)r2[r3(1.04)3/2 r3] r2

1.043/2 1

0.06

r2 0.060596

0.06 r2.

The approximate percentage error is(0.060596/0.06)r2 r2

r2=

0.060596

0.061 0.0099,

or approximately 1%.


12. y =r2 x2; y = x(r2 x2)1/2

S = 2

ba

r2 x2

1 + x2(r2 x2)1 dx = 2

ba

r2 x2

r2

r2 x2 dx

= 2r

badx = 2r(b a)

13. For x < 2, y = x 2 and y = 1. For x > 2, y = x+ 2 and y = 1.

S = 2

24

(x 2)1 + 1 dx+ 2 22

(x+ 2)1 + 1 dx

= 22

24

(x 2) dx+ 22 22

(x+ 2) dx

= 22

12x2 2x

24

+ 22

1

2x2 + 2x

22

= 22[(2 + 4) (8 + 8)] + 22[(2 + 4) (2 4)] = 202

14. Since the graph is symmetric with respect to the y-axis, we will find the area on [0, a] andmultiply by 2.

y = (a2/3 x2/3)3/2; y = (a2/3 x2/3)1/2

x1/3

S = 4

a0(a2/3 x2/3)3/2

1 +

(a2/3 x2/3)x2/3

dx = 4

a0(a2/3 x2/3)3/2

a2/3

x2/3dx

= 4a1/3 a0x1/3(a2/3 x2/3)3/2 dx u = a2/3 x2/3, du = 2

3x1/3 dx

= 4a1/3 0a2/3

u3/232du

= 4a1/3

35u5/2

0a2/3

= 12a1/3

5(0 a5/3) = 12a

2

5

15. Let be the angle formed when the cone is cut and flattened out. The lengthof the arc of the sector is 2r, the circumference of the base of the cone. Then/2r = 2/2L (the angle subtended by the sector is to the length of the sectoras 2 radians is to the circumference of the circle of radius L), and = 2r/L.

Using the hint in the text, the lateral surface area is1

2L22r

L

= rL.

L

2r

16. If L is the slant height, then L2 = r2 + h2 and the surface area is rL =

rr2 + h2. The surface can also be obtained by revolving the line y =

r

hx

about the x-axis:

S = 2

h0

r

hx

1 +

rh

2dx =

2r

h

h0x

r2 + h2

h2dx

=2rr2 + h2

h2

1

2x2h

0

= rr2 + h2

L r

h

6.6. AREA OF A SURFACE OF REVOLUTION 373

17. By similar triangles,r1L1

=r2L2

, r1L2 = r2L1, and r1L2 r2L1 = 0.From Problem 15, the lateral surface area of the frustum is

S = r2L2 r1L1 = (r2L2 r1L1) = [r2L2 + (0

r1L2 r2L1) r1L1]= [(r2 + r1)L2 (r2 + r1)L1] = (r1 + r2)(L2 L1) = (r1 + r2)L.

r1r2

LL1

L2

18. By the Pythagorean Theorem, L2 = h2 + (r2 r1)2 or L =h2 + (r2 r1)2. From (1) in

Section 6.6,

r1

r2 r1LhS = (r1 + r2)L = (r1 + r2)

h2 + (r2 r1)2.

19. We need to extend (3) in Section 6.6 to include functions which are not necessarily non-

negative. In this case we have S = 2

ba

|f(x)|1 + [f (x)]2 dx. Next, we require the fact

that the surface area obtained by revolving f around y = L is the same as that obtained byrevolving F (x) = f(x) L around the x-axis. Then F (x) = f (x) and

S = 2

ba

|F (x)|1 + [F (x)]2 dx = 2

ba

|f(x) L|1 + [f (x)]2 dx.

20. y =2

3x1/3; S = 2

81

|x2/3 4|1 +

4

9x2/3 dx =

2

3

81

4 x2/3x1/3

9x2/3 + 4 dx

21. (a) Since BCT is similar to TCS we have CB/TC = CT/CS or yBR

=R

R+ h, which

gives yB =R2

R+ h. Now, revolving x =

R2 y2 for yB y R around the y-axis, we

obtain the surface area

AS = 2

RyB

R2 y2

1 + yR2 y2

2dy = 2

RR2R+h

Rdy

= 2R

R R

2

R+ h

=

2R2h

R+ h.

Since AE = 4R2, we haveASAE

=h

2(R+ h).

(b) With h = 2000 and R = 6380,ASAE

=2000

2(6380 + 2000) 0.119332 11.9%.

(c) Settingh

2(R+ h)=

1

4we obtain 2h = R+ h or h = R = 6380 km.


(d) limh

ASAE

= limh

h

2R+ 2h=

1

2From a large distance we would expect to see half of the earths surface.

(e)ASAE

=3.76 105

2(6380 + 3.76 105) 0.4917 = 49.17%

6.7 Average Value of a Function

1. fave =1

1 (3) 13

4x dx =1

4(2x2)

13

=1

4(2 18) = 4

2. fave =1

5 (2) 52

(2x+ 3) dx =1

7(x2 + 3x)

52

=1

7[40 (2)] = 42

7

3. fave =1

2 0 20(x2 + 10) dx =

1

2

1

3x3 + 10x

20

=1

2

68

3 0

=

34

3

4. fave =1

1 (1) 11

(2x3 3x2 + 4x 1) dx = 12

1

2x4 x3 + 2x2 x

11

=1

2

1

2 9

2

= 2

5. fave =1

3 (1) 31

(3x2 4x) dx = 14(x3 2x2)

31

=1

4[9 (3)] = 3

6. fave =1

2 0 20(x+ 1)2 dx =

1

2 13(x+ 1)3

20

=1

2

9 1

3

=

13

3

7. fave =1

2 (2) 22

x3 dx =1

4

1

4x42

2=

1

4(4 4) = 0

8. fave =1

1 0 10x(3x 1)2 dx =

10(9x3 6x2 + x) dx =

9

4x4 2x3 + 1

2x21

0

=3

4

9. fave =1

9 0 90x1/2 dx =

1

9

2

3x3/2

90

= 2

10. fave =1

3 0 30(5x+ 1)1/2 dx =

1

3 215

(5x+ 1)3/230

=2

45(64 1) = 14

5

11. fave =1

3 0 30xx2 + 16 dx =

1

3 13(x2 + 16)3/2

30

=1

9(125 64) = 61

9

6.7. AVERAGE VALUE OF A FUNCTION 375

12. fave =1

1 1/2 11/2

1 +

1

x

1/3 1x2

dx u = 1 +1

x, du = 1

x2dx

= 2

23u1/3 du = 2 3

4(u4/3)

23

= 32(24/3 34/3) 2.7104

13. fave =1

1/2 1/4 1/21/4

x3 dx = 412x2

1/21/4

= 2

1

x2

1/21/4

= 2(4 16) = 24

14. fave =1

4 1 41(x2/3 x2/3) dx = 1

3

3

5x5/3 3x1/3

41

=1

3

3

545/3 3 41/3

12

5

=

1

545/3 41/3 + 4

5 1.2285

15. fave =1

5 3 532(x+ 1)2 dx =

1

2[2(x+ 1)1]

53

= 1x+ 1

53

= 1

6 1

4

=

1

12

16. fave =1

9 4 94

(x 1)3x

dx u =x 1, du = 1

2xdx

=1

5

212u3 du =

1

5 12u421

=1

10(16 1) = 3

2

17. fave =1

()

sinx dx =1

2( cosx)

= 12

[1 (1)] = 0

18. fave =1

/4 0 /40

cos 2x dx =4

12sin 2x

/40

=2

(1 0) = 2

19. fave =1

/2 /6 /2/6

csc2 x dx =3

( cotx)

/2/6

= 3(03) = 3

3

20. fave =1

1/3 (1/3) 1/31/3

sinx

cos2 xdx u = cosx, du = sinx dx

=3

2

1/21/2

1u2

du = 0

21. fave =1

1 (1) 11

(x2 + 2x) dx =1

2

1

3x3 + x2

11

=1

2

4

3 2

3

=

1

3

Setting f(c) = c2+c =1

3, we obtain 3c2+6c1 = 0. Then c = 6

36 + 12

6= 1 2

3

3.

The only solution on [1, 1] is 1 + 23

3.

22. fave =1

6 1 61(x+ 3)1/2 dx =

1

5 23(x+ 3)3/2

61

=2

15(27 8) = 38

15

Setting f(c) =c+ 3 = 38/15, we obtain c+ 3 = 1444/225. Thus, c = 769/225.


23. We are given1

5 1 51f(x) dx = 3. The area under the graph is

51f(x) dx = 12.

24. Solving1

b

b0(1x) dx = 1

b

x 2

3x3/2

b0

= 1 23

b = 0 gives b = 9/4. At b = 9/4 the

area bounded by the graph above the x-axis is the same as the area below the x-axis.

25. Tave =1

6 0 60

100 + 3t 1

2t2dt =

1

6

100t+

3

2t2 1

6t36

0

= 103

26. Rave =1

5 1 51(50 + 4x+ 3x2) dx =

1

4(50x+ 2x2 + x3)

51

=1

4(425 53) = 93

1

5

5k=1

R(k) =1

5(57 + 70 + 89 + 114 + 145) = 95

27. Using s(t) = v(t) we have

vave =1

t2 t1 t2t1

v(t) dt =1

t2 t1 s(t)t2t1

=s(t2) s(t1)

t2 t1 = v.

28. Uave =1

2/ 0 2/0

1

2kx2 dt =

k

4

2/0

A2 cos2(t+ ) dt

=kA2

4

2/0

1

2[1 + cos 2(t+ )] dt =

kA2

8

t+

1

2sin(2t+ 2)

2/0

=kA2

8

2

+

1

2sin(4 + 2) 1

2sin 2

=kA2

4

Kave =1

2/ 0 2/0

1

2mv2 dt =

m

4

2/0

[x(t)]2 dt =m

4

2/0

2A2 sin2(t+ ) dt

=A2(m2)

4

2/0

1

2[1 cos 2(t+ )] dt = A

2k

8

t 1

2sin(2t+ 2)

2/0

=kA2

8

2

1

2sin(4 + 2)

12

sin 2

=kA2

4

29. mv1 mv0 = (t1 0)F = t1t1 0

t10

k

1

2t

t1 1

2dt = k

t10

1 4

t21t2 +

4

t1t 1

dt

= k

43t21

t3 +2

t1t2t1

0

= k

43t1 + 2t1

=

2kt13

30. vave =1

R 0 R0

P

4vl(R2 r2) dr = P

4vlR

R2r 1

3r3R

0

=P

4vlR

2

3R3=PR2

6vl

31. 0, since

aa

f(x) dx = 0.

6.7. AVERAGE VALUE OF A FUNCTION 377

32. Intuitively, the average value of the linear function f(x) = ax + b on [x1, x2] should be the

value of f at the midpoint of that interval, or X =x1 + x2

2. This can be proven as follows:

fave =1

x2 x1 x2x1

(ax+ b) dx =1

x2 x1a2x2 + bx

x2x1

=1

x2 x1ax222

+ bx2 ax21

2 bx1

=

1

x2 x1a(x22 x21)

2+ b(x2 x1)

= a

x2 + x1

2

+ b = aX + b

33. f ave =1

h

x+hx

f (x) dx =1

hf(x)

x+hx

=f(x+ h) f(x)

h

34. fave =1

a 1 a1(n+ 1)xn dx =

1

a 1xn+1

a1

=an+1 1a 1 = a

n + an1 + + a+ 1

35. If

ba[f(x) fave] dx = 0, then

baf(x) dx =

bafave dx. Thus,

baf(x) dx = fave(b a)

and therefore fave =1

b a baf(x) dx.

36. The average of f on [0, 1] is 0. On [0, 2], it is 1/2. On [0, 3], it is 1, and on [0, 4], it is 3/2.

The average value of f on the interval [0, n] appears to be1

2(n 1), which can be proven as

follows:

fave =1

n 0 n0f(x) dx =

1

n

100 dx+

211 dx+ +

nn1

(n 1) dx

=1

n[0 + 1 + + (n 1)] = 1

n

n1k=1

k =1

n

(n 1)n

2

=

1

2(n 1)

37. There is no unique answer to this question; among several possible approaches, here is prob-ably the simplest. Suppose the circle is centered at the origin and that one of the points onthe circle is (1, 0). If (x, y) is any other point on the circle, then the length of the chordsbetween (1, 0) and (x, y) is the distance between the points:

(x+ 1)2 + y2 =x2 + y2 + 2x+ 1 =

2x+ 2.

The average chord length Lave is then

Lave =1

1 (1) 11

2x+ 2 dx =

1

2 12 (2x+ 2)

3/2

3/2

11

=1

6 43/2 = 8

6=

4

3.


38. (a) The surface area is S = 2

baf(x)

1 + [f (x)]2 dx. If f (x) = 0, then S = 2

baf(x) dx.

If 0 |f (x)| < for a x b, then

2

baf(x) dx S 2

baf(x)

1 + 2 dx = 2

1 + 2

baf(x) dx.

(b) At any x in [a, b] the circumference of a circular cross-section is 2f(x). The average

circumference is then C =1

b a ba2f(x) dx. Letting L = b a be the length of the

limb, we have CL = 2

baf(x) dx. Thus, from part (a), CL S 1 + 2 CL.

6.8 Work

1. W = 55 20 = 1100 yd-lb = 3300 ft-lb

3050 100

6.8. WORK 377

38. (a) The surface area is S = 2

baf(x)

1 + [f (x)]2 dx. If f (x) = 0, then S = 2

baf(x) dx.

If 0 |f (x)| < for a x b, then

2

baf(x) dx S 2

baf(x)

1 + 2 dx = 2

1 + 2

baf(x) dx.

(b) At any x in [a, b] the circumference of a circular cross-section is 2f(x). The average

circumference is then C =1

b a ba

2f(x) dx. Letting L = b a be the length of the

limb, we have CL = 2

baf(x) dx. Thus, from part (a), CL S 1 + 2 CL.

6.8 Work

1. W = 55 20 = 1100 yd-lb = 3300 ft-lb2. The horizontal component of force is 50

3 N. W = 50

3 8 = 4003 joules.

3. Since 10 = k

1

2

, k = 20 and F = 20x. Solving 8 = 20x we obtain x =

2

5ft.

4. (a) Since 50 = k(0.1), k = 500 and F = 500x.

(b) When x = 0.5, F = 250 N.

(c) Solving 200 = 500x, we obtain x = 0.4 m. The length of the spring is 0.5+ 0.4 = 0.9 m.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

2. The horizontal component of force is 503 N.

W = 503 8 = 4003 joules.

3. Since 10 = k

1

2

, k = 20 and F = 20x. Solving 8 = 20x we obtain x =

2

5ft.

4. (a) Since 50 = k(0.1), k = 500 and F = 500x.

(b) When x = 0.5, F = 250 N.

(c) Solving 200 = 500x, we obtain x = 0.4 m. The length of the spring is 0.5+ 0.4 = 0.9 m.

5. (a) W =

0.20

500x dx = 250x20.20

= 10 joules

(b) W =

0.60.5

500x dx = 250x20.60.5

= 27.5 joules

6. (a) W =

60

3

2x dx =

3

4x260

= 27 in-lb =27

12ft-lb =

9

4ft-lb

(b) W =

160

3

2x dx =

3

4x2160

= 192 in-lb = 16 ft-lb

7. Since 10 = k

2

3

, k = 15 and F = 15x.

(a) W =

1015x dx =

15

2x210

=15

2ft-lb

(b) W =

3215x dx =

15

2x232

=75

2ft-lb

6.8. WORK 379

8. Since 50 = k3, k = 503

and F =50

3x. W =

100

50

3x dx =

25

3x2100

=2500

3in-lb =

625

9ft-lb

9. We use 500 km = 0.5 106 m.W = (6.67 1011)(6.0 1024)(104)

1

6.4 106 1

6.9 106 4.531 1010

= 453.1 108 joules10. We use 200 km = 0.2 106 m.

W = (6.67 1011)(7.3 1022)(5 104)

1

1.7 106 1

1.9 106 1.507 1010

= 150.7 108 joules

x

y12 3

11. W =

120

62.4(3)2x dx = 9(62.4)1

2x2120

= 9(62.4)(72) 127, 030.9 ft-lb

x

y

20 10

4

y

12. y =1

5x

W =

100

62.4y2(20 x) dx = 62.4 100

1

25x2(20 x) dx

= 62.4

100

4

5x2 1

25x3dx = 62.4

4

15x3 1

100x410

0

= 62.4

500

3

32, 672.6 ft-lb

13. W =

100

62.4

1

25x2(25 x) dx = 62.4

100

x2 1

25x3dx

= 62.4

1

3x3 1

100x410

0

= 62.4

700

3

45, 741.6 ft-lb

x

y

3

y

214. x2 + y2 = 9; y =

9 x2

W =

33

62.4(2y 12)(5 x) dx = 1497.6 33

9 x2(5 x) dx

= 1497.6

5

33

9 x2 dx

33

x9 x2 dx

The first integral represents the area of the semicircle and is thus

1

2(3)2. The second integral

has an odd integrand and is thus 0. Therefore W = 1497.6

5 1

2(3)2

= 33, 696 ft-lb.


x

y

3

4y

5

15. y =3

4x

W =

4062.4(2y 10)(x+ 5) dx = 62.4(20)

40

3

4x(x+ 5) dx

= 936

40(x2 + 5x) dx = 936

1

3x3 +

5

2x24

0

= 936

184

3

= 57, 408 ft-lb

x

y5

3 y16. x2 + y2 = 25; y =

25 x2

W =

5280(2y 25)x dx = 4000

52x25 x2 dx

= 4000

13(25 x2)3/2

52

= 40003

(25 x2)3/252

= 40003

(0 213/2) 128, 312.1 ft-lb

17. The weight of the chain is F (x) = 20(100 x) lb when x feet of chain have been pulled up.

W =

400

20(100 x) dx = 20100x 1

2x240

0

= 64, 000 ft-lb

18. The weight of the system is F (x) = 3000 + 40(200 x) lb when x feet of chain have beenpulled up.

W =

1000

[3000 + 40(200 x)] dx =3000x+ 40

200x 1

2x2100

0

= 900, 000 ft-lb

19. (a) W = 80 65 = 5200 ft-lb(b) The weight of the system is F (x) = 80 +

1

2(65 x) when the bucket has been lifted x

feet.

W =

650

80 +

1

2(65 x)

dx =

80x+

1

2

65x 1

2x265

0

= 6256.25 ft-lb

20. The weight of the bucket after it has been lifted x feet is 62.4(20 x/2) lb. The bucket willbecome empty when it has been lifted 40 feet.

W =

400

62.420 x

2

dx = 62.4

20x 1

4x240

0

24, 960 ft-lb

21. If x is the distance separating the electron and the nucleus, then the force is F (x) = k/x2.

W =

41

k

x2dx = k

x

41

= k1

4 1

=

3k

4

22. (a) If x is the distance above the earth, then the weight of the system is F (x) = 2, 700, 000100x.

6.8. WORK 381

(b) W =

10000

(2, 700, 000 100x) dx = 2, 700, 000x 50x210000

= 2, 650, 000, 000 ft-lb

23. Since p = kv ,

W =

v2v1

p dv =

v2v1

kv dv =k

1 v1

v2v1

=k

1 (v12 v11 )

=1

1 (kv2 v2 kv1 v1) =

1

1 (p2v2 p1v1),

where p1 and p2 are the pressures corresponding to volumes v1 and v2, respectively.

24. Using Newtons second law F = ma = mg, we have

W =

y2y1

F (y) dy =

y2y1

mg dy = mgy]y2y1 = mgy2 mgy1.

25. Since the distance moved is 0, no work is done.

26. The force is F (x) =

2x, 0 x 12, 1 x 2

x+ 4, 2 x 40, 4 x 5

x 5, 5 x 6

.

The work is W =

102x dx+

212 dx+

42(x+ 4) dx+ 0 +

65(x 5) dx

= x210+ 2x]21 +

12x2 + 4x

42

+

1

2x2 5x

65

= (1 0) + (4 2) + (8 6) + (12 + 12.5) = 5.5 N-m.27. W = 165 1350 = 222, 750 ft-lb

x

y10

180 2028. Since the water leaks out of the bucket at a constant rate and the weight

of the rope is negligible, it is reasonable to approximate that the overalllifted weight is the midpoint or average of the buckets starting and endingweights of 200 and 180, respectively, or (200 + 180)/2 = 190 lb. Movingthis average weight by 10 ft yields 1900 ft-lb of work done.

Without integration, it can be seen from the figure that the overall work done is the sum ofthe areas of a 180 10 rectangle and a right triangle of width 20 and height 10, or (180 10) + (20 10)/2 = 1800 + 100 = 1900 ft-lb.

29. Using F = ma = mv and dx = x(t) dt = v dt, we have

W =

x2x1

F (x) dx =

t2t1

mvv dt u = v, du = v dt

=

v2v1

mudu =1

2mu2

v2v1

=1

2mv22

1

2mv21 .


30. From the figure in the text, we see that sin = x/30 so that x = 30 sin and dx = 30 cos d.Also, when x = 3, sin = 0.1 and 0.1. Then

W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.

6.9 Fluid Pressure and Force

1. (a) pressure = 62.4(20) = 1248 lb/ft2; F = (1248)(25) = 31200 lb

(b) pressure = 62.4(20) = 1248 lb/ft2; F = (1248)(4) = 4992 lb

(c) pressure = 62.4(20) = 1248 lb/ft2; F = (1248)(100) = 124800 lb

2. (a) pressureoil = 55 lb/ft3 96 ft = 5280 lb/ft2

(b) pressurewater = 62.4 lb/ft3 85 ft = 5304 lb/ft2

(c) forceoil = 5280 125 350 = 231, 000, 000 lb(d) forcewater = 5304 125 350 = 232, 050, 000 lb

3. (a) pressure = (62.4)(8) = 499.2 lb/ft2; F = (499.2)(30)(15) = 224, 640 lb

(b) sidewall force =

8062.4x(30) dx = 62.4(15)x2

80

= 59, 904 lb

end force =

8062.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb

(1, 0)

x

y

6.9. FLUID PRESSURE AND FORCE 381


W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb

4. The equation of the line through (1, 0) and (5/2,3/2) is y =

3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.

3/2

4. The equation of the line through (1, 0) and (5/2,3/2) is

y =

3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

124.8[1.20 (0.10)] = 162.12 lb.

(5/2, 0)

x

y



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.

3/2



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280

= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb

(1, 0)

x

y



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

3

3x

3

3

dx

= 124.8

5/21

3

3x2

3

3x

dx = 124.8

3

9x3

3

6x25/2

1

= 124.8[1.20 (0.10)] = 162.12 lb.

5.

6.

7.

8.

9.

10.



W =

30mg tan dx = 550(9.8)

0.10

sin

cos (30 cos ) d = 16500(9.8)

0.10

sin d

= 161700( cos )]0.10 = 161700(1 cos 0.1) 807.8 N-m.










80

62.4x(30) dx = 62.4(15)x280= 59, 904 lb

end force =

80

62.4x(15) dx = 62.4

15

2

x280

= 29, 952 lb


3

3x

3

3. Using symmetry,

F =

5/21

62.4x(2y) dx = 124.8

5/21

x

Documents

CSM Chapters6