SOLUCIONARIO

Chapter 6

Applications of the Integral 6.1

Rectilinear Motion Revisited

1. s(t) = 2. s(t) =

Z

6 dt = 6t + c;

5 = s(2) = 6(2) + c;

Z

(2t + 1) dt = t2 + t + c;

Z

(t2 − 4t) dt =

c = −7;

s(t) = 6t − 7

0 = s(1) = 12 + 1 + c = 2 + c;

c = −2;

s(t) = t2 + t − 2 3. s(t) =

1 3 t − 2t2 + c; 3

4. s(t) =

Z

√

7. v(t) =

Z

−5 dt = −5t + c;

6 = s(3) = −9 + c;

c = 15;

s(t) =

1 3 t − 2t2 + 15 3

1 9 5 (4t + 5)3/2 + c; 2 = s(1) = + c; c = − ; 6 2 2 1 5 s(t) = (4t + 5)2/3 − 6 2

Z

π 5 π 5 5 1 5 5. s(t) = −10 cos 4t + dt = sin 4t + + c; = s(0) = − + c = − + c; 6 2 6 4 2 2 4

5 π 5 5 c = ; s(t) = − sin 4t + + 2 2 6 2 Z 2 2 2 2 2 6. s(t) = 2 sin 3t dt = − cos 3t + c; 0 = s(π) = + c; c = − ; s(t) = − cos 3t − 3 3 3 3 3 Z

4t + 5 dt =

4 = v(1) = −5 + c;

5 (−5t + 9) dt = − t2 + 9t + c; 2 5 2 9 s(t) = − t + 9t − 2 2 s(t) =

c = 9;

2 = s(1) =

335

v(t) = −5t + 9;

13 + c; 2

9 c=− ; 2

336

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

8. v(t) = s(t) =

Z

6t dt = 3t2 + c;

0 = v(2) = 12 + c;

Z

(3t2 − 12) dt = t3 − 12t + c;

Z

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

c = −12;

v(t) = 3t2 − 12;

−5 = s(2) = −16 + c;

c = 11;

s(t) = t3 − 12t + 11 9. v(t) = s(t) = s(t) =

Z

(t3 − 2t2 + 5t − 3) dt =

1 4 2 3 5 2 t − t + t − 3t + 10 4 3 2

10. v(t) =

Z

11. v(t) =

Z

−3 = v(0) = c;

1 3 2 3 5 2 t − t + t − 3t + c; 4 3 2

v(t) = t3 − 2t2 + 5t − 3;

10 = s(0) = c;

1 1 (t − 1)2 dt = (t − 1)3 + c; 4 = v(1) = c; c = 4; v(t) = (t − 1)3 + 4; 3 3

Z 1 1 s(t) = (t − 3)3 + 4 dt = (t − 1)4 + 4t + c; 6 = s(1) = 4 + c; c = 2; 3 12 1 s(t) = (t − 1)4 + 4t + 2 12 (7t1/3 − 1) dt =

21 4/3 t − t + c; 4

50 = v(8) = 76 + c;

21 4/3 t − t + 26;

Z4 9 1 21 4/3 t − t − 26 dt = t7/3 − t2 − 26t + c; s(t) = 4 4 2 9 7/3 1 2 s(t) = t − t − 26t − 48 4 2

c = −26;

v(t) =

12. v(t) =

Z

100 cos 5t dt = 20 sin 5t + c;

0 = s(8) = 48 + c;

−20 = v(π/2) = 20 + c;

v(t) = 20 sin 5t − 40; Z s(t) = (20 sin 5t − 40) dt = −4 cos 5t − 40t + c;

c = −48;

c = −40;

15 = s(π/2) = −20π + c;

c = 15 + 20π;

s(t) = −4 cos 5t − 40t + 15 + 20π

13. v(t) = 2t − 2 = 2(t − 1) Z 5 Z dist. = |2(t − 1)| dt = 2 0

0

1

−(t − 1) dt + 2

Z

1

5

(t − 1) dt

1

5

1 2 1 2 1 15 1 = 2 − t +t +2 t −t =2 −0 +2 − − = 17 cm 2 2 2 2 2 0 1

337

6.1. RECTILINEAR MOTION REVISITED 14. v(t) = −2t + 4 = −2(t − 2) Z 6 Z dist. = | − 2(t − 2)| dt = 2 0

2

0

−(t − 2) dt + 2

Z

6

2

(t − 2) dt

2

6 1 2 1 2 = 2 − t + 2t +2 t − 2t = 2(2 − 0) + 2[6 − (−2)] = 20 cm 2 2 0 2

15. v(t) = 3t2 − 6t − 9 = 3(t + 1)(t − 3) Z 4 Z 3 Z 4 2 2 dist. = |13t − 6t − 9| dt = 3 −(t − 2t − 3) dt + 3 (t2 − 2t − 3) dt 0

1 = 3 − t3 + t2 + 3t 3

3

0

+3

0

1 3 t − t2 − 3t 3

4 3

3

20 = 3(9 − 0) + 3 − − (−9) = 34 cm 3

16. v(t) = 4t − 64t = 4t(t + 4)(t − 4) Z 5 Z 4 Z 5 dist. = |4t3 − 64t| dt = 4 −(t3 − 16t) dt + 4 (t3 − 16t) dt 3

1

1 = 4 − t4 + 8t2 4

4

1

+4

1

17. v(t) = 6π cos πt Z 3 Z dist. = |6π cos πt| dt = 6 1

1 4 t − 8t2 4

3/2

1

3/2

= 6(− sin πt)]1

5 4

4

31 175 = 4 64 − +4 − − (−64) = 306 cm 4 4

−π cos πt dt + 6 5/2

Z

5/2

π cos πt dt + 6

3/2 3

+ 6(sin πt)]3/2 + 6(− sin πt)]5/2

Z

3

5/2

−π cos πt dt

= 6[−(−1) − 0] + 6[1 − (−1)] + 6[0 − (−1)] = 24 cm 18. v(t) = 2(t − 3) Z 7 Z dist. = |2(t − 3)| dt = 2 2

2

3

−(t − 3) dt + 2

Z

3

7

(t − 3) dt

3 7

1 2 9 7 9 1 +2 =2 t − 3t −4 +2 − − = 17 cm = 2 − t2 + 3t 2 2 2 2 2 2 3

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

0

2

60 60 dt = t 3600 3600

2 0

=

60 1 mi = mi × 5280 ft/mi = 176 ft.

1800 30

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

338

CHAPTER 6. APPLICATIONS OF THE INTEGRAL Z

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

Z

a(t) = −32; v(0) = 0; s(0) = h; v(t) = −32 dt = −32t + c; 0 = v(0) = c; v(t) = −32t; Z 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 the positive root, we obtain √ √ −135 20, 385 −135 + 18, 225 + 2, 160 = ≈ 1.9440 s. tr = 4 4

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

Z 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; Z 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 + 24.5t

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

Z 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; Z 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 + 24.5t

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

Z 25. a(t) = −32; v(0) = 32; s(0) = 384; v(t) = −32 dt = −32t + c; 32 = v(0) = c; Z v(t) = −32t + 32; s(t) = (−32t + 32) dt = −16t2 + 32t + c; 384 = s(0) = c; s(t) = −16t2 + 32t + 384

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

339

6.1. RECTILINEAR MOTION REVISITED

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

Z 27. a(t) = −32; v(0) = −16; s(0) = 102; v(t) = −32 dt = −32t + c; −16 = v(0) = c; Z v(t) = −32t − 16; s(t) = (−32t − 16) dt = −16t2 − 16t + c; 102 = s(0) = c; s(t) = −16t2 − 16t + 102

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

Z 28. a(t) = −32; v(0) = 96; s(0) = 22; v(t) = −32 dt = −32t + c; 96 = v(0) = c; Z v(t) = −32t + 96; s(t) = (−32t + 96) dt = −16t2 + 96t + c; 22 = s(0) = c; s(t) = −16t2 + 96t + 22

Solving 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 impact velocity 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 we obtain v(t) = −1.8t + v0 and s(t) = −0.9t2 + v0 t. If the rock attains its maximum height at time t1 , then v(t1 ) = 0 = −1.8t + v0 and t1 = v0 /1.8. Solving 200, 000 = −0.9t21 + v0 t1 = −0.9 gives v0

p

3.6(200, 000) ≈ 848.5 m/s.

v 2 0

1.8

+ v0

v 0

1.8

= 0.9

v 2 0

1.8

30. (a) Taking a(t) = −32, v(0) = −2, and s(0) = 25, we have v(t) = −32t−2 25 and s(t) = −16t2 − 2t + 25. Using similar triangles, we obtain = s x 750 , 25(x − 30) = sx and x = . Then x − 30 25 − s

=

v02 3.6

s 30

x – 30 x

dx 750 ds 750 750 = = v(t) = (−32t − 2) dt (25 − s)2 dt (25 − s)2 (25 − s)2 1500(16t + 1) 1500(16t + 1) 375(16t + 1) =− =− =− 2 .

(25 − s)2 (16t2 + 2t)2 t (8t + 1)2 dx 375(8 + 1) (b) When t = 1/2, =− 1 = −540 ft/s.

2 dt 4 (4 + 1)

Z Z dv dv dv 31. From the hint, a = = v, and integrating with respect to s gives a ds = v ds. dt ds ds 1 Then as = v 2 + c, and solving for v we have v 2 = 2as − 2c. Since v = v0 when s = 0, 2 v02 = −2c and v 2 = 2as + v02 .

340

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

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

Z v(t) = a dt = at + c; v0 = v(0) = c; v(t) = at + v0 ; Z 1 1 s(t) = (at + v0 ) dt = at2 + v0 t + c; 0 = s(0) = c; s(t) = at2 + v0 t 2 2 2v0 1 . Then v(−2v0 /a) = −v0 , and Solving s(t) = at2 + v0 t = 0, we obtain t = 0 and t = − 2 a 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.

Z v(t) = a dt = at + c; v0 = v(0) = c; v(t) = at + v0 ; Z 1 1 s(t) = (at + v0 ) dt = at2 + v0 t + c 0 = s(0) = c; s(t) = at2 + v0 t 2 2 To find the maximum height reached on earth, we solve v(t) = at + v0 = 0. The maximum height is reached when t = −v0 /a and is s(−v0 /a) = v02 /2a − v02 /a = −v02 /2a. On the planet, the acceleration of gravity is a/2. Proceeding as on the earth, we obtain v(t) = 1 1 at + v0 , and s(t) = at2 + v0 t. To find the maximum height reached on the planet, we 2 4 1 solve v(t) = at + v0 = 0. The maximum height is reached when t = −2v0 /a and is 2 s(−2v0 /a) = v02 /a − 2v02 /a = −v02 /a. Thus, the maximum height reached on the planet is twice that reached on earth.

34. Let a be the acceleration due to gravity on earth. Then, with initial velocity 2v0 , we have 1 ve (t) = at + 2v0 and se (t) = at2 + 2v0 t. On the planet, with acceleration due to gravity 2 1 1 a/2 and initial velocity v0 , we have vp (t) = at + v0 and sp (t) = at2 + v0 t. To find the 2 4 2v0 . The maximum height reached on earth, we solve ve (t) = at + 2v0 = 0 and obtain t = − a 2 2 2 2v 4v 2v maximum height is se (−2v0 /a) = 0 − 0 = − 0 . To find the maximum height reached a a a 1 2v0 on the planet, we solve vp (t) = at + v0 = 0 and obtain t = − . The maximum height 2 a 2 2 2 v 2v v is sp (−2v0 /a) = 0 − 0 = − 0 . Thus, the maximum height reached on earth is twice a a a that reached on the planet. We want to find the initial velocity ϑ0 on the earth so that the v2 maximum height reached on earth is − 0 , the maximum height reached on the planet. With a 1 initial velocity ϑ0 , we have ve (t) = at + ϑ0 and se (t) = at2 + ϑ0 t. Solving ve (t) = at + ϑ0 = 0 2 ϑ0 ϑ2 ϑ2 ϑ2 v2 we obtain t = − . Then, we want s(−ϑ0 /a) = 0 − 0 = − 0 to be equal to − 0 . Solving a 2a a√ 2a a for ϑ0 we see that the initial velocity on earth must be 2v0 .

6.2

Area Revisited

341

6.2. AREA REVISITED

1

1 3 2 2 4 1. A = −(x − 1) dx = − x + x = − − = 3 3 3 3 −1 −1 Z

1

2

2

1 -2

-1

1

2

2

3

-1 -2

1 2

Z 2 1 3 1 + x −x −(x2 − 1) dx + (x2 − 1) dx = − x3 + x 3 3 0 1 0 1

2 2 2 = −0 + − − =2 3 3 3

2. A =

3. A =

Z

Z

1

1 −x3 dx = − x4 4 −3 0

0

−3

81 81 = =0− − 4 4

3 2 1 -1

1 -1

10 -3

-2

-1

1

2

-10 -20 -30

4. A =

Z

Z

1

2

(1 − x3 ) dx + −(1 − x3 ) dx = 0 1

3 3 7 −0 + 2− − = = 4 4 2

5. A =

Z

3

0

−(x2 − 3x) dx =

1 x − x4 4

2 1 + −x + x4 4 0 1

1

-1

1

2

-5

3

3 9 9 1 = −0= − x3 − x2 3 2 2 2 0

3

3

-3

6. A =

Z

0

(x + 1)2 dx =

−1

1 (x + 1)3 3

0

−1

=

1 1 −0= 3 3

-2

-1

1

-3

7. A =

Z

0

−1

(x − 6x) dx + 3

Z

0

1

6

−(x − 6x) dx 3

0

1 1 4 1 = x − 3x2 + − x4 − 3x2 4 4 −1 0

11 11 11 = 0− − + −0 = 4 4 2 Z 1 Z 2 8. A = (x3 − 3x2 + 2) dx + −(x3 − 3x2 + 2) dx

0

-2

-1

1

2

-3 -6

-1

1

1

1 2 1 4 1 x − x3 + 2x + − x4 + x3 − 2x 4 4 1

0 5 5 5 = −0 + 0− − = 4 4 2

=

3

-5

2

342

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

9. A =

Z

1

−(x3 − 6x2 + 11x − 6) dx +

0

+

Z

3

2

Z

2

1

(x3 − 6x2 + 11x − 6) dx

2

4

-2 -4

−(x3 − 6x2 + 11x − 6) dx

-6

1 2 1 11 11 1 4 − x4 + 2x3 − x2 + 6x x − 2x3 + x2 − 6x + 4 2 4 2 0 1

3 1 4 11 + − x + 2x3 − x2 + 6x 4 2 2

9 9 11 9 − 0 + (−2) − − + −2 = = 4 4 4 4 =

10. A =

Z

0

Z

(x − x) dx + 3

−1

1

1

−(x3 − x) dx

0

1 1 1 4 1 2 1 + − x4 + x2 = x − x 4 2 4 2 −1 0

1 1 1 + −0 = = 0− − 4 4 2

1 3

Z 1 Z 3 1 1 + x+ 11. A = −(1 − x−2 ) dx + (1 − x−2 ) dx = −x − x 1/2 x 1 1/2 1

5 11 10 = −2 − − −2 = + 2 3 6

12. A =

Z

2

1

0

(1 − x

−2

-1

1

-1

2

-2

2 1 1 5 ) dx = x + = −2= x 1 2 2

2

-2

13. A =

Z

1

−(x

0

=

14. A =

Z

=

− 1) dx +

2 − x3/2 + x 3 4

0

1/2

(2 − x

1/2

1

+

0

) dx +

2 2x − x3/2 3

4 0

Z

Z

+

1

4

4

9

2

(x1/2 − 1) dx

2 3/2 x −x 3

4

=

1

1 4 1 −0 + − − =2 3 3 3

2

4

-2

2

−(2 − x1/2 ) dx

2 −2x + x3/2 3

9 4

=

8 8 16 −0 + 0− − = 3 3 3

4

-2

8

343

6.2. AREA REVISITED 0 3 3 4/3 3 4/3 15. A = −x dx + x dx = − x + x 4 4 −2 0 −2 0

3 4/3 3 4/3 3 4/3 = 0 + (2 ) + (3 ) − 0 = (2 + 34/3 ) 4 4 4 Z

16. A =

Z

Z

0

1/3

8

−1

3

2

1/3

3 -2

8

3 27 11 2x − x4/3 = =4− − 4 4 4 −1

(2 − x1/3 ) dx =

3

4

17. A =

Z

0

− sin x dx +

−π

Z

π

0

π

0

sin x dx = cos x]−π − cos x]0

1

-π

= [1 − (−1)] − (−1 − 1) = 4 18. A =

Z

3π

0

8

π -1

3

3π

(1 + cos x) dx = (x + sin x)]0 = 3π − 0 = 3π

2 1 π

2π

3π

-1

19. A =

Z

π/2

−3π/2

−(−1 + sin x) dx = (x +

π/2 cos x)]−3π/2

π 3π = − − = 2π 2 2

2

-2

20. A =

Z

π/3

0

π/3

sec2 x dx = tan x]0

=

√

3−0=

√

3

4 2

-π

21. A =

Z

0

−2

−x dx +

Z

0

1

1 x2 dx = − x2 2

0

1 3 x 3

+

−2

1 0

= −(0 − 2) +

1 −0 3

=

π

7 3

2

-2

2

-2

22. A =

Z

−2

−3

−(x + 2) dx + −2

Z

0

−2

(x + 2) dx +

Z

0

0

√

2

(2 − x2 ) dx +

Z

2

√

√

2

2

−(2 − x2 ) dx -2

2

2

1 2 1 2 1 =− x + 2x + x + 2x + 2x − x3 2 2 3 −3 −2 0

2 1 3 − 2x − x √ 3 2

3 4√ 4 4√ 7 8√ = − −2 + + (0 + 2) + 2−0 − − 2 = + 2 2 3 3 3 6 3

-2

344

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

23. A =

Z

3

0

[x − (−2x)] dx =

Z

3

0

3 3x dx = x2 2

3

=

0

27 2

3

-4

4 -3 -6

24. A =

Z

2

0

(4x − x) dx =

Z

2

3 2 x 2

3x dx =

0

2

9

=6

6

0 3 -4

25. A =

Z

2

(4 − x ) dx = 2

−2

1 4x − x2 3

2

4

16 16 32 = − − = 3 3 3

−2

6

3 -4

26. A =

Z

1

0

(x − x2 ) dx =

1 2 1 3 x − x 2 3

1

=

0

4

1 6

2

1

-1

27. A =

Z

2

−1

(8 − x3 ) dx =

1 8x − x4 4

1

33 81 = 12 − − = 4 4

2

−1

5

-2

28. A =

Z

1

(x

0

1/3

− x ) dx = 3

3 4/3 1 4 x − x 4 4

1 0

=

2

1 2

1

-1

1 -1

29. A =

Z

1

−1

[4(1 − x2 ) − (1 − x2 )] dx =

= 2 − (−2) = 4

Z

1

−1

1 (3 − 3x2 ) dx = (3x − x3 ) −1

4

2

-2

30. A =

Z

1

−1

[2(1 − x2 ) − (x2 − 1)] dx =

= 2 − (−2) = 4 31. A =

Z

1

3

(x − x−2 ) dx =

Z

1

−1

2

1 (3 − 3x2 ) dx = (3x − x3 ) −1

3

-2

1 2 1 x + 2 x

3 1

=

29 3 10 − = 6 2 3

2

3

3

345

6.2. AREA REVISITED

√

9 Z 9 1 2 3/2 1/2 −1/2 1/2 32. A = y− √ dy = (y x − 2y −y ) dy = y 3 1 1 1

40 4 = = 12 − − 3 3 Z 1 Z 1 2 2 33. A = [(−x + 6) − (x + 4x)] dx = (6 − 4x − 2x2 ) dx Z

9

−3

9 6 3

2

4

−3

1 2 10 64 6x − 2x2 − x3 = − (−18) = 3 3 3 −3 Z 3/2 Z 3/2 34. A = [(−x2 + 3x) − x2 ] dx = (3x − 2x2 ) dx

-3

3

=

0

= 35. A =

Z

3 2 2 3 x − x 2 3 8

−8

3/2

-4

2 1

0

=

0

(4 − x2/3 ) dx =

9 8

4

-1

1

2

-1

8

64 64 3 128 = − − 4x − x5/3 = 5 5 5 5 −8

4

-8

36. A =

Z

1

−1

[(1 − x

2/3

) − (x

2/3

− 1)] dx =

Z

1

−1

=

−1 5

−1

(5 + 4x − x2 ) dx +

Z

5

2

(2 − 2x2/3 ) dx

1

4 6 4 8 = − − = 2x − x5/3 = 5 5 5 5 −1 Z 5 Z 6 37. A = [(2x + 2) − (x2 − 2x − 3)] dx + [(x2 − 2x − 3) − (2x + 2)] dx Z

-2

(x2 − 4x − 5) dx

5

6 1 1 3 5x + 2x2 + x3 + x − 2x2 − 5x 3 3 −1 5

8 100 118 100 − − + (−30) − − = = 3 3 3 3

Z 5/2 Z 5/2 3 5 38. A = (−x2 + 4x) − x dx = x − x2 dx 2 2 0 0

5/2 5 2 1 3 125 = x − x = 4 3 48 0

Z 0 Z 2 1 x + 6 + x dx + (x + 6 − x3 ) dx 39. A = 2 −4 0 0

2

3 2 1 2 1 4 = x + 6x + x + 6x − x = (0 + 12) + (10 − 0) = 22 4 2 4 −4 0

2

-2

20 15

5

6

8

10 5 -2

2

4

6

=

4 2

-2

2

4

-2

8 6 4 2 -4

-2

2

4

346

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

40. A =

Z

1 y dy = y 3 3

1

2

0

1

=

0

1 3

2 1

-1

1

2

-1

41. A =

Z

2

−1

=

[(2 − y 2 ) − (−y)] dy =

1 1 2y + y 2 − y 3 2 3

2

−1

Z

2

−1

2

(2 + y − y 2 ) dy

1

7 9 10 − − = = 3 6 2

-2

43. A =

Z

2

0

= 44. A =

Z

4

45. A =

Z

= 46. A =

=0− −

−2

1

=

8 3

4

0

−1

64 3

=

0

(y 3 − y) dy +

0

−2

1 4 1 2 y − y 4 2

0

-3

(−2y 2 − 4y) dy

Z

0

4

2

(8y − 2y 2 ) dy

Z

1

-8

-4

−1

4 2

19 13 = − − =8 4 4

-2

1

2

2

−(y 3 − y) dy

-2

2

-2

2

π/4

π/4 cos x)]0

4

4

(4 + 2x − x3 ) dx

1

1 4 1 2 1 1 1 + − y + y =0− − + −0= = 4 2 4 4 2 −1 0 Z π/4 Z π/2 47. A = (cos x − sin x) dx + (sin x − cos x) dx

0

Z

−1

0

6

-4

[(x + 4) − (x3 − x)] dx = 1

Z

3

-2

2

1 4x − x2 − x4 4

Z

8 3

[(−y + 2y + 1) − (y − 6y + 1)] dy =

2 4y 2 − y 3 3

3

2

0

2

-2

3

[(−y − 2y + 2) − (y + 2y + 2)] dy =

2

−1

√

2

2 − y 3 − 2y 2 3

0

=

2

2

−2

Z

3

1 -1

√3

2 3 42. A = √ [(6 − y ) − y ] dy = √ (6 − 2y ) dy = 6y − y √ 3 − 3 − 3 − 3 √ √ √ = 4 3 − (−4 3) = 8 3 √

Z

-1

!

π/2

= (sin x + + (− cos x − sin x)]π/4 √ √ √ = 2 − 1 + (−1) − (− 2) = 2 2 − 2

-2

347

6.2. AREA REVISITED

48. A =

Z

π/2

0 2

[2 sin x − (−x)] dx =

Z

π/2

(2 sin x + x) dx =

0

π 16 + π = − (−2) = 8 8 Z 5π/6 5π/6 49. A = (4 sin x − 2) dx = (−4 cos x − 2x)]π/6 2

π/2 1 −2 cos x + x2 2 0

2

π

-2

4

π/6

√ √ 5π √ π 12 3 − 4π = =2 3− − −2 3 − 3 3 3 Z π/2 Z π/2 3 cos x dx [2 cos x − (− cos x)] dx = 50. A = =

−π/2

π/2 3 sin x]−π/2

51. Region 1: y =

√

= 3 − (−3) = 6

−π/2

2

!

2

-π

π -2

4

x, y = −x, x = 0, x = 4 -2

2

4

6

2

4

6

2

4

6

2

4

6

-4

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

4

-2

-4

52. Region 1: y =

1 2 x , y = x − 3, x = −1, x = 2 2

4

-2

-4

1 Region 2: y = 3 − x, y = − x2 , x = −1, x = 2 2

4

-2

-4

53.

Z

0

2

Z

3

x + 1 − 4x dx =

Z 2 3 3 − 4x dx + 4x − dx x+1 x+1 0 1/2 1/2 2 = 3 ln |x + 1| − 2x2 0 + 2x2 − 3 ln |x + 1| 1/2 1/2

3 1 1 3 = 3 ln − + 8 − 3 ln 3 − + 3 ln 2 2 2 2 3 = 7 + 3 ln ≈ 6.1370 4

6 4 2 -2

2

348

54.

CHAPTER 6. APPLICATIONS OF THE INTEGRAL Z

1

−1

55.

Z

3

0

56.

Z

57.

Z

ln

√

2

2e

−x

−1

−e

x

ln √2

dx +

Z

1

ln

√

2

ex − 2e−x dx

4

1 = −2e − e −1 + e + 2e−x ln √2 √ √ √ √ = − 2 − 2 + 2e + e−1 + e + 2e−1 − 2 − 2 √ = 3e + 3e−1 − 4 2 −x

x

x

-2

2

4

p 1 9π 9 − x2 dx = π(3)2 = 4 4

-3

3

6

5

−5

Z

x

e − 2e−x dx =

2

−2

p

25 − x2 dx =

(1 +

25π 1 π(5)2 = 2 2 -5

p

Z

Z

5

4

p 1 4 − x2 ) dx = 1 dx + 4 − x2 dx = 4 + π(2)2 = 4 + 2π 2 −2 −2 2

2

2

-2

58.

Z

1

−1

(2x + 3 −

p

1 − x2 ) dx =

Z

1

−1

(2x + 3) dx −

Z

1

−1

p

4

1 − x2 dx

1 = x2 + 3x −1 − π(1)2 2 π π = 1 + 3 − 1 − 3(−1) − = 6 − 2 2 1

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

Z ap Z 4b a p 2 4b 1 2 b2 − b2 x2 /a2 dx = a − x2 dx = πa = πab.

A=4 a 0 a 4 0

2

-2

2

b

a

349

6.2. AREA REVISITED

Z 3 1 1 1 1 (3x − 2) − (−2x + 8) − x+ dx + x+ dx 2 2 2 2 1 2

Z 2 Z 3 5 5 5 15 = x− dx + − x+ dx 2 2 2 2 1 2

2 3 5 2 5 5 2 15 = x − x + − x + 4 2 4 2 1 2

45 5 5 + − 10 = =0− − 4 4 2 Z −2 Z −2 √ √ 61. A = (2 − −x − 2) dx + [− −x − 2 − (−2)] dx 60. A =

Z

2

−6

+ =2

Z

−2

−6

Z

0

−2

(2 −

−6

[2 − (−2)] dx +

√

−x − 2) dx +

Z

Z

2

0 0

4

(2, 4)

(3, 2)

2 (1, 1)

2

4

4

[2 − (2x − 2)] dx

4 dx +

−2

Z

0

2

-6

(4 − 2x) dx

−2

-4

2 2 0 = 2 2x + (−x − 2)3/2

+ 4x ]−2 + (4x − x2 ) 0 3 −6

20 52 = 2 −4 − − +8+4−0= 3 3

2 Z 2 Z 2 1 1 1 3 1 2 2 2 62. A = y + 1 − (−y − 2) dy = y + y + 3 dy = y + y + 3y 2 2 3 4 −2 −2 −2

29 23 52 = − − = 3 3 3 Z ln 3/2 Z ln 2 x 63. The area with respect to x is Ax = (e − 1) dx + (2 − ex ) dx.

0 2

ln 3/2

y+1 dy.

The area with respect to y is Ay = ln y − ln 2 1 If integration with respect to x is chosen, we get Z ln 3/2 Z ln 2 ln 3/2 ln 2 Ax = (ex − 1) dx + (2 − ex ) dx = (ex − x)]0 + (2x − ex )]ln 3/2 Z

0

ln 3/2

3 3 3 3 = − ln − 1 + 2 ln 2 − 2 − 2 ln + = −3 ln 3 + 5 ln 2 ≈ 0.1699.

2 2 2 2 If integration with respect to y is chosen, we get

2 Z 2

y+1 y+1 Ay = ln y − ln dy = y ln y − y − (y + 1) ln + (y + 1)

2 2 1 1

2

y+1 3 = y ln y − (y + 1) ln + 1

= 2 ln 2 − 3 ln + 1 − ln 1 + 2 ln 1 − 1 2 2 1 = −3 ln 3 + 5 ln 2 ≈ 0.1699

350

CHAPTER 6. APPLICATIONS OF THE INTEGRAL (see Problem 5.1.39 for the antiderivative of ln x)

64. Using Mathematica the numbers at which the curves intersect are approximately The area is then

−0.4077767094044803 A=

Z

and

0.7148059123627778

−0.4077767094044803

0.7148059123627778.

(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 Q and R is then y = −x/x20 + 2/x0 . Setting y = 0 we see that the x-intercept is 2x0 . The area is

2x0 Z 2x0 2 1 1 2 2 = −2 + 4 = 2, A= − 2x + dx = − 2 x + x x0 x0 2x0 x0 0 0

which does not depend on x0 .

b

Z b 1 2 1 1 2 66. A = (Ax + B) dx = Ax + Bx = Ab2 + Bb − Aa + Ba 2 2 2 a a

A 2 A = (b − a2 ) + B(b − a) = (b + a) + B (b − a) 2 2 Aa + B + Ab + B f (a) + f (b) = (b − a) = (b − a) 2 2

f (b)

f (a)

a

b

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

a Z a 1 A=2 (cos x − x) dx = 2 sin x − x2 = 2 sin a − a2 2 0 0 (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(b − a).

Now, if y = f (x) describes the lower edge of the swath in Figure 6.2.16(a), then an equation for the upper edge is y = f (x) + k. The area between the two graphs is then Z b Z b {[f (x) + k] − f (x)} dx = k dx = k(b − a). a

a

69. The areas are the same. Let w be the length of the line segments AB and CD, and without loss of 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 line defined by AD0 can be written as x = f (y). Thus, the line defined by BC 0 is x = f (y) + w.

The area of the parallelogram is therefore Z h Z h {[f (y) + w] − f (y)} dy = w dy = wh.

0

0

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

351

6.3. VOLUMES OF SOLIDS: SLICING METHOD

6.3

Volumes of Solids: Slicing Method

√ √ √ 1. x2 + y 2 = 16; y = 16 − x2 ; A(x) = 3y 2 = 3(16 − x2 ) 4

Z 4√ √ 1 3 2 V = 3(16 − x ) dx = 3 16x − x 3 −4 −4 √

√ 128 128 256 3 3 = 3 − − = ft 3 3 3

√ 1 1 2. x2 + y 2 = 16; y = 16 − x2 ; A(x) = πy 2 = π 8 − x2 2 2

4 Z 4 1 1 V = π 8 − x2 dx = π 8x − x3 2 6 −4 −4

64 128π 3 64 − − ft = =π 3 3 3 Z 4 4 3. x = y 2 ; A(x) = 2y(8y) = 16y 2 = 16x; V = 16x dx = 8x2 0 = 128

(x, y) y√3 y

y x y

4

(x, y) y x y

4

2

y

(x, y) y

0

4 -2

√

√ 1 4 3y 2 3 2 2 2 4. y = 4 − x ; A(x) = = (4 − x ) = 3 4 − 2x + x 4 4 4

2 Z 2√ √ 1 2 1 V = 3 4 − 2x2 + x4 dx = 3 4x − x3 + x5 4 3 20 −2 −2 √

√ 64 64 128 3 − − = = 3 15 15 15 2

√

-2

2

√3y/2

2

x

y 5

-2

√

4 − x2 ; A(x) = πy 2 − π(12 ) = π(3 − x2 )

√3 Z √3 1 3 2 V = √ π(3 − x ) dx = π 3x − x √ 3 − 3 − 3 √ √ √ = π[2 3 − (−2 3)] = 4π 3 ft3

y/2

y

1 2π 2 (x − 5)2 5. y = − x + 2; A(x) = πy 2 = 5 2 25 5 Z 5 2π 2π 10π 3 2 2 V = (x − 5) dx = (x − 5) = ft 25 75 3 0 0

6. y =

(x, y)

4

3

1 -3

-3

√3

x

y

2 3

1

352

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

7. x = −y + 3; A(y) = x2 = (y − 3)2 3 Z 3 1 2 3 =9 (y − 3) dy = (y − 3) V = 3 0 0

3

(x, y)

x

3

8. Let b denote the length of one side of the square base. Thus, B = b2 . h−y b(h − y) h = and x = . Thus, Using similar triangles, we have b 2x 2h 2 2 b (h − y) A(y) = (2x)2 = , and h2

Z h Z h 2 b (h − y)2 2b2 b2 2 2 dy = b y V = − y + dy h2 h h2 0 0

h b2 3 1 b2 2 b2 h 2 = hB. = b y − y + 2y = b2 h − b2 h + h 3h 3 3 0 9. x =

√

y 1

Z 1 Z 1 π 1 2 √ 2 2 = V =π [1 − ( y) ] dy = π (1 − y) dy = π y − y 2 2 0 0 0

h –y h

(x, y)

x

y b

B(1, 1)

C

x O

10. y = x2 V =π

Z

0

1

(x2 )2 dx = π

Z

1

x4 dx =

0

π π 5 i1 x = 5 5 0

A

B(1, 1)

C

y O

11. y = x2

1

1 4π 1 5 2 4 = V =π (1 − x ) dx = π x − x 5 5 0 0 Z

A

B(1, 1)

C

y O

12. x =

√

y Z 1 Z √ V =π ( y)2 dy = π 0

0

A

B(1, 1)

C

1

y dy =

π 2 i1 π y = 2 2 0

x O

A

353

6.3. VOLUMES OF SOLIDS: SLICING METHOD 13. x =

√

y Z 1 Z 1 √ 2 √ V =π (1 − y) dy = π (1 − 2 y + y) dy 0

4 1 = π y − y 3/2 + y 2 3 2

1 0

π = 6

O

√

y Z 1 Z 1 √ √ [12 − (1 − y)2 ] dy = π V =π (2 y − y) dy 0

=π

x

0

14. x =

B(1, 1)

C

4 3/2 1 2 y − y 3 2

1

A

B(1, 1)

C

x

0

=

0

5π 6

O

A

9

15. y = 9 − x Z 3 Z 3 V =π (9 − x2 )2 dx = 2π (81 − 18x2 + x4 ) dx 2

−3

1 = 2π 81x − 6x3 + x5 5

16. x =

3 0

1296π = 5

√

y−1 Z 5 Z 5 p (y − 1) dy V =π ( y − 1)2 dy = π 1

=π

17. x =

1 2 y −y 2

5 1

V =π

3

5

x

15 1 =π − − = 8π 2 2

2

2

" # Z 1 2 1 2 V =π − 1 dy = π (y −2 − 1) dy y 1/2 1/2

1

1 5 π = π − −y = π −2 − − = y 2 2 1/2 1 x

-3

1

1 y

Z

18. y =

y

0

1

1

x 2 6.3.17

3

2

3

1 1 1 5π dx = π − = π − − (−2) = x x 3 3 1/2 1/2

Z

3

2 1

y 1

2

3

354

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

19. y = (x − 2)2 Z 2 V =π (x − 2)4 dx

4

=

0

π (x − 2)5 5

i2 0

=

32π 5

2

y 2

20. x =

√

y−1 Z 1 Z 1 √ √ V =π ( y − 1)2 dy = π (y − 2 y + 1) dy 0

=π

1

0

1 2 4 3/2 y − y +y 2 3

1 0

x

π = 6

-1

1 21. y1 = 4 − x2 ; y2 = 1 − x2 4

2 #

Z 2" Z 2 1 2 15 15 2 2 V = 2π (4 − x ) − 1 − x dx = 2π 15 − x2 + x4 dx 4 2 16 0 0 2

3 5 = 32π = 2π 15x − x3 + x5 2 16 0 22. x =

4

√

1−y

1 Z 1 Z 1 p 1 π (1 − y) dy = π y − y 2 V =π ( 1 − y)2 dy = π = 2 2 0 0 0

4

y1 y2 -2

2

2

x 2

23. x1 = y; x2 = y − 1 Z 1 Z 2 V =π y 2 dy + π [y 2 − (y − 1)2 ] dy 0

=π 24. y1 = 1; V =π

Z

0

1 3 y 3

1 0

1

+ π(y − y)

y2 = 2 − x

1

12 dx + π

2

Z

1

2

2 1

=π

1 1 = πx]0 + π 4x − 2x2 + x3 3

1

1

1 +2 3

(2 − x)2 dx = π 2

2

Z

1

7π = 3

dx + π

0

=π+π

x1 – x2

x1 1

Z

2

2

2

1

8 7 − 3 3

(4 − 4x + x2 ) dx

=

4π 3

1

y1

y2 1

2

355

6.3. VOLUMES OF SOLIDS: SLICING METHOD 25. x = y 2 + 1 Z 2 Z 2 2 2 (4 − y 2 )2 dy [5 − (y + 1)] dy = π V =π 0

0

=π

Z

2

0

26. x = y 2 V =π

Z

8 1 (16 − 8y 2 + y 4 ) dy = π 16y − y 3 + y 5 3 5

1

−1

(1 − y ) dy = π 2 2

1 2 = π y − y3 + y5 3 5

Z

1

2 0

2

256π = 15

0

3 = π 3x − 3x4/3 + x5/3 5

4

8 16π 8 − − = =π 15 15 15

1

0

=

0

0

8 1 = π 4y 2 − y 3 + y 4 − y 5 3 5

2 0

1 = π 9y − y 3 3

1 30. y1 = 9 − x2 ; 2 Z "

3

−3

2

2–y

1 1

0

2

2–x

2

1

64π = 15

p y 2 + 16 Z 3 Z 2 p V =π y 2 + 16 52 − dy = π

-1

1

1

29. x =

−3

1

1–x

3π 5

28. x = −y 2 + 2y Z 2n Z 2

2 o 2 2 V =π 2 − 2 − (−y + 2y) dy = π (8y − 8y 2 + 4y 3 − y 4 ) dy

5

(1 − 2y + y ) dy 2

27. y = x1/3 Z 1 Z 1 V =π [(2 − x1/3 )2 − 12 ] dx = π (3 − 4x1/3 + x2/3 ) dx

5–x

1

1

−1

−1

4

2

3

3

−3

x

(9 − y 2 ) dy

= π[18 − (−18)] = 36π

y2 = x2 − 6x + 9 = (x − 3)2 # 2 4 1 2 4 V =π 9− x − (x − 3) dx 2 0

Z 4 3 4 2 3 =π 108x − 63x + 12x − x dx 4 0

4 3 672π = π 54x2 − 21x3 + 3x4 − x5 = 20 5 0

3

-3

9

y1

6 3

y2 3

6

9

356

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

31. x1 = y + 6; x2 = y 2 Z 3 Z V =π [(y + 6)2 − y 4 ] dy = π −2

1 1 = π 36y + 6y 2 + y 3 − y 5 3 5

5

3

−2

3

(36 + 12y + y 2 − y 4 ) dy

−2

664 500π 612 − − = =π 5 15 3

x1

x2

6

-5

32. x = (y − 1)1/3 9 Z 9 3π 96π 2/3 5/3 (y − 1) dy = V =π (y − 1) = 5 5 1 1

5

-4

33. y = x3 − x Z 1 Z 3 2 V =π (x − x) dx = π −1

=π

1 7 2 5 1 3 x − x + x 7 5 3

−1

=π

1 7 1 4 x + x +x 7 2

1

(x6 − 2x4 + x2 ) dx

−1

1

−1

8 8 16π =π − − = 105 105 105

= π 3x + 4e

36. y = ex V =π

Z

0

=π

2

−x

1 − e−2x 2

23 9 16π =π − − = 14 14 7

1

1 2x e −x 2

2

2 0

=π

0

= π 4e

0

[(e ) − 1 ] dx = π x 2

4

Z

y -1

1

2

35. y = e−x Z 1 Z 1 V =π [(2 − e−x )2 − 12 ] dx = π (3 − 4e−x + e−2x ) dx 0

2

(x6 + 2x3 + 1) dx

−1

1

-2 1

−1 1

34. y = x3 + 1 Z 1 Z V =π (x3 + 1)2 dx = π

x

−1

1 1 − e−2 − 2 2

y -2

2

2

1

2–y

2

8

2

(e

2x

0

1 4 5 e − 2 2

− 1) dx

6

4

y 2

1 3

357

6.3. VOLUMES OF SOLIDS: SLICING METHOD 37. y = | cos x| Z Z 2π 2 | cos x| dx = π V =π

1 + cos 2x dx 2 0 0

2π Z π π 2π 1 (1 + cos 2x) dx = = x + sin 2x = π2 2 0 2 2 0 2π

1

y !

2!

-1

38. y = sec x Z π/4 π/4 V =π sec2 x dx = π tan x]−π/4 = π[1 − (−1)] = 2π

2

−π/4

y

1

–π/4

39. y = tan x Z π/4 Z 2 V =π tan x dx = π 0

0

π/4

4π − π 2

π =π 1− −0 = 4 4

π/4

1

(sec x − 1) dx = π(tan x − 2

π/4 x)]0

y π/4

40. y1 = cos x; y2 = sin x Z π/4 Z 2 2 V =π (cos x − sin x) dx = π 0

π/4

0

cos 2x dx =

iπ/4 π π sin 2x = 2 2 0

1

y1 y2 π/4

41. The volume of the right circular cylinder is πr2 h. Placing the center of the red circular cylinder’s base in Figure 6.3.19 on the origin, we see that A = πr2 for every slice from y = 0 to h. Thus, the volume V of the cylinder is Z h ih V = πr2 dy = πr2 y = πr2 h.

0

0

42. Take the cross-sections to be rectangles perpendicular to the base of the cylinder and parallel to the diameter. p x2 + y 2 = a2 ; y = a2 − x2 √ √ (a) A(x) = 2yz = (2 a2 − x2 )x = 2x a2 − x2 Z a p V = 2x a2 − x2 dx u = a2 − x2 , du = −2x dx 0

=

Z

0

a2

2 −u1/2 du = − u3/2 3

0

a2

2 2 = − (0 − a3 ) = a3 3 3

(x, y) a

z=x

x

y x y

2y

z z

358

CHAPTER 6. APPLICATIONS OF THE INTEGRAL √ √ √ √ (b) A(x) = 2yz = (2 a2 − x2 ) 3x = 2 3x a2 − x2 Z a √ p V = 2 3x a2 − x2 dx u = a2 − x2 , du = −2x dx 0

=

Z

0

a2

√

− 3u

1/2

z = √3x

x

z

2y z

#0 √ √ √ 2 3 3/2 2 3 2 3 3 du = − u =− (0 − a3 ) = a 3 3 3 2 a

43. (a) Using Mathematica, we obtain with the disk method Z 1 4π [P (x)]2 (1 − x2 ) dx = V =π (5a2 + 9b2 + 21c2 + 105d2 + 18ac + 42bd).

315 −1 (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.

p 44. (a) Using x = r2 − y 2 and the disk method, we obtain Z h−r p 2 r2 − y 2 dy V =π −r

Z

(0, h – r) –r

r

h−r

1 x h (r2 − y 2 ) dy = π r2 y − y 3 3 –r −r −r

1 3 1 2 3 2 = π r (h − r) − (h − r) − −r + r 3 3 1 = πr2 h − πh3 .

3 4 π (b) The weight of the ball is πr3 ρball and the weight of water displaced is (3rh2 −h3 )ρwater .

3 3 ρball π 4 2 Using Archimedes’ principle and = 0.4 we have π(3) (0.4) = (9h2 − h3 ) or ρwater 3 3 h3 − 9h2 + 43.2 = 0. Solving for h we obtain h ≈ 2.5976 in. =π

h−r

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 + y 2 = r2 , one side of the square is y = r2 − x2 , and its area is y 2 = r2 − x2 . Using symmetry, the volume common to the cylinders is thus

r Z r x3 16r3 V =8 (r2 − x2 ) dx = 8 r2 x − = .

3 3 0 0 (b) This item involves a research report, and thus a preset solution is not applicable.

359

6.4. VOLUMES OF SOLIDS: SHELL METHOD

6.4

Volumes of Solids: Shell Method

1. y =

√

x

V = 2π

Z

1

0

√ 4π 5/2 x x dx = x 5

B(1, 1)

C

1

=

0

4π 5

y O

2. x = y 2 V = 2π

Z

1

0

y(1 − y 2 ) dy = 2π

1 2 1 4 y − y 2 4

1

A

B(1, 1)

C

=

0

π 2

x O

3. x = y 2 V = 2π

Z

1

0

(1 − y)y 2 dy = 2π

1 3 1 4 y − y 3 4

1

C

=

0

π 6

A

B(1, 1)

x

O

4. x = y 2 V = 2π

Z

1

0

y · y 2 dy =

π 4 y 2

i1 0

C

=

π 2

A

B(1, 1)

x

O

5. y =

√

x

V = 2π

Z

1

0

√ (1 − x) x dx = 2π

2 3/2 2 5/2 x − x 3 5

1 0

A

B(1, 1)

C

8π = 15

y O

6. y =

√

x

V = 2π

Z

0

1

(1 − x)(1 −

√

x) dx = 2π

2 1 2 = 2π x − x3/2 − x2 + x5/2 3 2 5

Z

1

0

1 0

(1 −

√

7π = 15

A

B(1, 1)

C

x−x+x

3/2

) dx

y O

A

360

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

7. x = y

5

V = 2π

Z

5

0

y · y dy =

2π 3 y 3

5

=

0

250π 3

x

5

8. x = 1 − y Z 1 Z 1 V = 2π (y + 2)(1 − y) dy = 2π (2 − y − y 2 ) dy 0

1

x

0

1 1 1 7π = 2π 2y − y 2 − y 3 = 2 3 3 0

1

-1

-2

9. x =

√

y

V = 2π

3

Z

3

0

4π 5/2 √ y y y dy = 5

3 0

=

√

36π 3 5

2 1

x

-1

10. y = x2 V = 2π

Z

2

0

1

2

4

π i2 x · x2 dx = x4 = 8π 2 0

y 2

4

11. y = x2 V = 2π

Z

0

1

1 1 4 3π 3 (3 − x)x dx = 2π x − x = 4 2 0

1

2

y 1

2

3

361

6.4. VOLUMES OF SOLIDS: SHELL METHOD 12. x1 =

√

y; Z V = 2π

0

9

√ x2 = − y √

√ y[ y − (− y)] dy = 2π

Z

8π 5/2 2y 3/2 dy = y 5

9

0

9 0

x1

1944π = 5

x2 5

-3

3

13. y = x2 + 4 Z 2 Z 2 (x3 + 2x) dx x(x2 + 4 − 2) dx = 2π V = 2π 0

= 2π

1 4 x + x2 4

2

0

= 16π

5

0

y

3

14. y = x2 − 5x + 4 Z 4 Z 4 V = 2π x(−x2 + 5x − 4) dx = 2π (−x3 + 5x2 − 4x) dx 1

5 1 = 2π − x4 + x3 − 2x2 4 3

4 1

1

= 2π

7 32 + 3 12

√ √ 15. x1 = 1 + y; x2 = 1 − y Z Z 1 √ √ V = 2π y[1 + y − (1 − y)] dy = 2π 0

8π 5/2 y = 5

1 0

135π = 6

1

1 4 8 3 x − x + 8x2 4 3

0

0

=

128π 3

4

x1

2y 3/2 dy

0

4

y

-2

x2

8π = 5

0

= 2π

2

2

16. y = (x − 2)2 Z 4 Z 4 V = 2π (4 − x)[4 − (x − 2)2 ] dx = 2π (x3 − 8x2 + 16x) dx

2

1

2

2

4

4

2

y

362

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

17. x = y 3 V = 2π

Z

1

0

(y + 1)(1 − y 3 ) dy = 2π

Z

0

(1 + y − y 3 − y 4 ) dy

2

1 1 1 1 21π = 2π y + y 2 − y 4 − y 5 = 2 4 5 10 0

-1

18. y1 = x1/3 + 1; y2 = −x + 1 Z 1 V = 2π (1 − x)[x1/3 + 1 − (−x + 1)] dx

2

= 2π

1

0

= 2π

19. y1 = x; V = 2π

(x1/3 + x − x4/3 − x2 ) dx

3 4/3 1 2 3 7/3 1 3 x + x − x − x 4 2 7 3

y2 = x2 Z

1

x(x − x ) dx = 2π 2

0

y1

1

0

Z

x

1

1

y2 2

1

=

0

1 3 1 4 x − x 3 4

41π 42

1 0

1

π = 6

y2

y1

1

20. y1 = x; V = 2π

Z

y2 = x2 1

0

= 2π

Z

0

1

(2 − x)(x − x ) dx 2

1

y1

y2

(2x − 3x2 + x3 ) dx

1

2

3

1

π 1 4 2 3 = = 2π x − x + x 4 2 0 21. y = −x3 + 3x2 Z 3 Z 3 V = 2π x(−x3 + 3x2 ) dx = 2π (−x4 + 3x3 ) dx 0

3

243π 1 5 3 4 = = 2π − x + x 5 4 10 0

0

4

2

y 2

4

363

6.4. VOLUMES OF SOLIDS: SHELL METHOD 22. y = x3 − x

0 Z 0 1 5 1 3 4π 3 V = 2π −x(x − x) dx = 2π − x + x = 5 3 15 −1 −1

1

y -1

23. y1 = 2 − x2 ; y2 = x2 − 2 Z 0 Z 2 2 V = 2π √ −x[2 − x − (x − 2)] dx = 2π − 2

= 2π

1 4 x − 2x2 2

2

0 √

− 2

0

√ − 2

y1

(2x3 − 4x) dx

= 4π

0

1 = 2π 4x2 + 2x3 − x4 2

4

2

-2

24. y1 = 4x − x2 ; y2 = x2 − 4x Z 4 Z 4 V = 2π (x + 1)[4x − x2 − (x2 − 4x)] dx = 2π (8x + 6x2 − 2x3 ) dx

y2

-2

3

y1 y2

0

= 128π

3

6

-3

0

25. x = y 2 − 5y

5 Z 5 1 4 5 3 625π 2 V = 2π y(−y + 5y) dy = 2π − y + y = 4 3 6 0 0

5

x -5

26. x1 = y + 4; x2 = y 2 + 2 Z 2 Z 2 V = 2π y[y + 4 − (y 2 + 2)] dy = 2π (2y + y 2 − y 3 ) dy 1

1 1 = 2π y + y 3 − y 4 3 4 2

2 1

= 2π

1

8 13 − 3 12

19π = 6

2

x1 x2 2

4

6

364

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

27. y1 = x + 6; y2 = x3

2 Z 2 1 3 1 248π V = 2π x(x + 6 − x3 ) dx = 2π x + 3x2 − x5 = 3 5 15 0 0

6

y1

y2 2

28. x1 = 1 − y 2 ; x2 = y 2 Z √2/2 Z V = 2π y(1 − y 2 − y 2 ) dy = 2π 0

= 2π

1 2 1 4 y − y 2 2

√2/2 0

0

√

1

2/2

(y − 2y 3 ) dy

x1 √2/2

π = 4

1

29. y = sin x2 Z √π/2 Z √π/2 2 V = 2π x(1 − sin x ) dx = 2π (x − x sin x2 ) dx 0 0

√π/2

1 2 1 π 1 π 2 − 2π = 2π = 2π x + cos x2 − = 2 2 4 2 2 0 30. y = ex

x2

1

y √π /2

2

V = 2π

Z

0

1

2

x(ex ) dx = 2π

1 x2 e 2

1 0

= πe − π

2

y 2

31. We use the shell method.

r Z r Z r h h 1 2 h 1 V = 2π (r − x) x dx = 2π hx − x2 dx = 2π hx − x3 = πr2 h r r 2 3r 3 0 0 0 32. The equation of the line through (r1 , h) and (r2 , 0) is x =

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

365

6.4. VOLUMES OF SOLIDS: SHELL METHOD method. h

1 (r1 − r2 )y + r2 h

2

1 2 2 2 2 dy = π (r1 − r2 ) y + r2 (r1 − r2 )y + r2 dy V =π h2 h 0 0

h

1 1 (r1 − r2 )2 y 3 + r2 (r1 − r2 )y 2 + r22 y

=π 2 3h h

0 1 2 πh 2 2 2 (r + r1 r2 + r22 ) = πh (r1 − 2r1 r2 + r2 ) + r2 (r1 − r2 ) + r2 = 3 3 1 Z

Z

h

33. We use the disk method.

Z r p (r2 − y 2 ) dy ( r2 − y 2 )2 dy = π −r −r

r

1 2 2 4 = π r2 y − y3 = π r3 − − r3 = πr3 3 3 3 3 −r

V =π

Z

r

√ 1√ 2 34. The equation of the line is y = r − a2 x and the equation of the circle is y = r2 − x2 . a We use the disk method. !2 Z b √ 2 Z b p 2 r − a2 V =π x dx + π r2 − x2 dx a a a

a

b 2 Z Z b 1 3 r2 − a2 a 2 r − a2 1 3 2 2 2 x dx + π x + π r x − x =π (r − x ) dx = π a2 a2 3 3 0 a 0 a

π 2 1 1 π = (r − a2 )a + π br2 − b3 − ar2 − a3 = (3br2 − 2ar2 − b3 ) 3 3 3 3 r

35. The equation of the ellipse is y = b 1 −

x2 . We use the disk method. a2

!2 a

Z a x2 x2 1 V =π b 1− 2 dx = πb2 1 − 2 dx = πb2 x − 2 x3 a a 3a −a −a −a

2 2a 2a 4πab = πb2 − − = 3 3 3 Z

a

r

r

x2 . Since the solid is symmetric with respect to a2 the x-axis, we will find the volume of the upper hemispheroid and multiply by 2. We use the shell method. !

3/2 #a Z a r b2 2 a2 b2 2 πa2 b 2 2 V = 2 2π x b − 2 x dx = 4π − 2 b − 2x = a 3b a 3 0

36. The equation of the ellipse is y = b

1−

0

366

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

36

6.5. LENGTH OF A GRAPH

ω 2 x2 ω2 r2 . The depth of the liquid below the x-axis is y2 = h − .

2g 2g ω 2 x2 ω 2 r2 37. y1 = . The depth of the liquid below the x-axis is y2 y =1 h − . So the volume is So the volume is 2g 2g h

Z r # # ! r " 22 22 ! r" 2 ω 2 x2 ωω rx ω 3 y 2hg − ω 2 r2 ω 2 r2 x dx = 2π +dx h− x +2 x dx V = 2π x V = 2π +h− 2g 2g 2g 2g 0 0 2g 2g 0 " 2 #$r

Z r 2 2 2 2 4 2r 2 4 ω 2hg − ω r πω r 4πhgr − 2πω r πω 2 r4 2 2 2 ω 3= 2π2hg −x4ω+ r = πr2 h − = 2π x + x 4gdx x 0 = 4g + 8g 4g 4g 2g 2g 0 2 r √ ω 4 2hg − ω 2 r2 2 ω2 r2 2hg liquid will touch xthe bottom of the bucket when y2 = h − = 0, or ω = . Th = 2π 38. The x + 2g r 8g 4g 0

37. y1 =

volume of the liquid is then

=

πω 2 r4 4πhgr2 − 2πω 2 r4 πω 2 r4 2 2πr 4 h− 2 4 + = . πω r π(2hg/r )r 1 1 4g V4g= πr2 h − = πr2 h 4g − = πr2 h − πr2 h = πr2 h 4g

6.5

Length of a Graph ω2 r2

38. The liquid will touch the bottom of the bucket when y2 = h − 1.

volume of the liquid is then

4g

2g

= 0, or ω =

2

√

2hg . The r

2.

V = πr2 h −

πω r 3. π(2hg/r2 )r4 1 1 = πr2 h − = πr2 h − πr2 h = πr2 h 4g 4g 2 2 2 4

4.

5.

6.5

Length of a Graph 6.

1. y = 1; 0

s=

Z

1

−1

2. y = 2; 0

s=

Z

0

3

7. p √ 1 + 12 dx = 2 2 8.

p

9.

1+

22

√ dx 10.= 3 5 11.

3 12.

3. y 0 = x1/2 2 " # 13.

3/2 #1 Z 1r 3/2 8 9 8 13 133/2 − 8 9 114. + x = −1 = ≈ 1.4397 s= 1 + x dx = 4 27 4 27 4 27 0 15.

0

16.

4. y 0 = 2x−1/3 s Z 8p Z 17.

Z 8 8 p x2/3 + 4 −1/3 −2/3 s= 1 + 4x dx = 18. dx = x x2/3 + 4 dx x2/3 1 1 1

2 −1/3 x dx 3

Z 8 i8 3 1/2 = u du = u3/2 = 83/2 − 53/2 ≈ 11.4471 2 5 5 u = x2/3 + 4, du =

2

367

6.5. LENGTH OF A GRAPH 5. y 0 = 2x(x2 + 1)1/2 Z 4p Z 2 2 s= 1 + 4x (x + 1) dx = 1

=

1

4

4 2 3 140 5 x +x = − = 45 3 3 3 1

p

(2x2

+

1)2

dx =

Z

4

(2x2 + 1) dx

1

6. y = 2(x + 1)3/2 − 1; y 0 = 3(x + 1)1/2 0 Z 0 Z 0p √ 2 3/2 1 + 9(x + 1) dx = 9x + 10 dx = (9x + 10) s= 27 −1 −1 −1 2 = (103/2 − 1) ≈ 2.2684 27

x−1 1 1/2 1 −1/2 x − x = 1/2 2 r 2 2x Z 4r Z 4r Z 4 (x − 1)2 4x + x2 − 2x + 1 (x + 1)2 dx = dx = dx s= 1+ 4x 4x 4x 1 1 1

4

Z Z 1 28 8 1 4 x+1 1 4 1/2 1 2 3/2 10 −1/2 1/2 = = dx = (x + x ) dx = x + 2x − = 2 1 x1/2 2 1 2 3 2 3 3 3 1

7. y 0 =

1 2 1 x4 − 1 x − 2 = 2 2x 2x2 Z 4r Z 4r 4 (x4 − 1)2 4x + x8 − 2x4 + 1 s= 1+ dx = dx 4 4x 4x4 2 2

4

Z Z 4r 4 (x + 1)2 1 4 2 1 253 13 1 1 3 1 227 −2 dx = x − = − = (x + x ) dx = = 4 4x 2 2 2 3 x 2 2 12 6 24 2

8. y 0 =

1 4x6 − 1 9. y 0 = x3 − 3 = 4x 4x3 Z 3r Z 3r Z 3r (4x6 − 1)2 16x6 + 16x12 − 8x6 + 1 (4x6 + 1)2 1+ s= dx = dx = dx 6 6 16x 16x 16x6 2 2 2

3 Z 3 6 Z 4x + 1 1 3 1 1 3 −3 4 dx = (4x + x ) dx = x − 2 = 4x3 4 2 4 2x 2 2

1 1457 127 4685 = − = ≈ 16.2674 4 18 8 288

368

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

4x8 − 1 1 10. y 0 = x4 − 4 = 4x 4x4 Z 2r Z 2r Z 2r 8 (4x − 1)2 16x8 + 16x16 − 8x8 + 1 (4x8 + 1)2 s= 1+ dx = dx = dx 16x8 16x8 16x8 1 1 1 2

Z Z 1 2 4x8 + 1 1 2 1 4 5 1 4 −4 = dx = (4x + x ) dx = x + 4 1 x4 4 1 4 5 3x3 1

1 3067 56 3011 = − = ≈ 6.2729 4 120 120 480 (4 − x2/3 )1/2 ; 11. y 0 = − x1/3      2

Z

s=

8

1

1,

s

4 − x2/3 1+ dx = x2/3

Z

8

1

2 x1/3

dx = 3x2/3

i8 1

2<x<3

(x − 2)−1/3 , 3 < x < 10 3      3 (x − 6)1/2 , 10 < x < 15 4 Z 3 Z 15 r Z 10 r √ 4 9 s= 1 + 1 dx + 1 + (x − 2)−2/3 dx + 1 + (x − 6) dx 9 16 2 10 3 Z 15 Z 10 r √ √ 1 4 = 2+ (x − 2)2/3 + (x − 2)−1/3 dx + 9x − 38 dx 9 4 10 3

3/2 10

15 √ 1 1 2 4

2/3 3/2 = 2 + (x − 2) + (9x − 38)

+

9 4 9 3 10 3 3/2 3/2 √ 40 13 1 = 2+ − + (973/2 − 523/2 ) ≈ 19.7954 9 9 54

12. y 0 =

13. y 0 = 2x;

s=

Z

3

−1

14. y = (x + 1) 0

−1/2

p 1 + 4x2 dx

;

s=

Z

3

−1

15. y = cos x; 0

s=

Z

π

0

16. y 0 = sec2 x;

s=

Z

dx 2 = − y −1/3 dy 3

1 + (x +

p 1 + cos2 x dx

π/4

−π/4

17.

p

p

1 + sec4 x dx

1)−1

dx =

Z

3

−1

r

x+2 dx x+1

=9

369

6.5. LENGTH OF A GRAPH

s=

Z

0

8

r

4 1 + y −2/3 dy = 9

Z

8

0

s

Z

9y 2/3 + 4 1 dy = 2/3 3 9y

8

y −1/3

0

u = 9y 2/3 + 4, du = 6y −1/3 dy 40 Z 40 1 1 3/2 1 = u1/2 du = u (403/2 − 8) ≈ 9.0734 = 18 4 27 27 4 18.

p

9y 2/3 + 4 dy

dx 1 1 y3 − 1 = y 3/2 − y −3/2 = dy 2 2 2y 3/2 s Z 9 Z 9s 3 Z 9s 3 (y 3 − 1)2 4y + y 6 − 2y 3 + 1 (y + 1)2 s= 1+ dy = dy = dy 4y 3 4y 3 4y 3 4 4 4

9 Z Z 1 9 y3 + 1 1 2 5/2 1 9 3/2 −3/2 −1/2 = (y +y ) dy = dy = y − 2y 2 4 y 3/2 2 4 2 5 4

1271 1 1448 59 − = ≈ 42.3667 = 2 15 5 30

3 2 −1/3 (1 − x2/3 )1/2 2/3 1/2 19. (a) y = (1 − x ) ; y = (1 − x ) − x =− 2 3 x1/3 s r Z 1 Z 1 Z 1 1 1 − x2/3 1 s= 1+ dx = dx = dx 2/3 2/3 1/3 x x x 0 0 0 1 At x = 0, 1/3 is discontinuous. x (b) The graph is symmetric with respect to both coordinate axes, so 2/3 3/2

0

s=4

Z

0

1

1

x1/3

dx = 4

3 2/3 x 2

1

= 6.

0

1/2

−1/2 x2 bx x2 0 20. y = b 1 − 2 ; y =− 2 1− 2 a a a s

Z a Z as −1 b2 x 2 x2 b2 x2 a2 s=4 1+ 4 1− 2 dx = 4 1+ 4 dx a a a a2 − x2 0 0 s Z as Z b2 x 2 4 a a4 − a2 x2 + b2 x2 =4 1+ 2 2 dx = dx a (a − x2 ) a 0 a2 − x2 0 21. y =

√

r 2 − x2 ; Z 2πr = s = 4

0

r

p

y 0 = −x(r2 − x2 )−1/2 1 + x2 (r2 − x2 )−1 dr = 4r

Letting r = 1 we have 2π = 4

Z

0

1

Z

r

0

1 √ dx or 1 − x2

√

Z

0

1 dx dr r 2 − x2

1

√

1 π dx = .

2 2 1−x

370

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

22. y = x . Let s(x) = 0

3

differentials, we have Z

2.1

2

6.6

p

Z

2

x

p

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

1 + x6 dx = s(2 + 0.1) ≈ s(2) + s0 (2) dx = 0 +

p √ 1 + 26 (0.1) = 0.1 65 ≈ 0.8062.

Area of a Surface of Revolution

1. y 0 = x−1/2 Z 8 Z √ p −1 2 x 1 + x dx = 4π S = 2π 0

0

8

√

8π x + 1 dx = (x + 1)3/2 3

8 0

=

208π 8π (27 − 1) = 3 3

1 (x + 1)−1/2 2 r

3/2 #5 Z 5r Z 5 √ 4π 1 1 5 S = 2π x + 1 1 + (x + 1)−1 dx = 2π x + 1 + dx = x+ 4 4 3 4 1 1 1 " # 3/2 3/2 25 9 π 4π − = (253/2 − 93/2 ) ≈ 51.3127 = 3 4 4 6

2. y 0 =

3. y 0 = 3x2 Z S = 2π

p 1 + 9x4 dx u = 1 + 9x4 , du = 36x3 dx 0

Z 10 √ π 3/2 i10 π 1 du = u = (103/2 − 1) ≈ 3.5631 = 2π u 36 27 27 1 1 1

x3

1 −2/3 x 3 s Z 8 1/3 p Z 8 x x−4/3 dx = 2π S = 2π x 1+ 9x4/3 + 1 dx u = 9x4/3 + 1, du = 12x1/3 dx 9 3 1 1

Z π 3/2 i145 2π 145 √ 1 π du = u (1453/2 − 103/2 ) ≈ 199.4805 = u = 3 10 12 27 27 10 Z 3 p i3 π π 0 5. y = 2x; S = 2π x 1 + 4x2 dx = (1 + 4x2 )3/2 = (373/2 − 1) ≈ 117.3187 6 6 0 0 Z 2 p i2 π π 6. y 0 = −2x; S = 2π x 1 + 4x2 dx = (1 + 4x2 )3/2 = (173/2 − 1) ≈ 36.1769 6 6 0 0 4. y 0 =

7. y 0 = 2

Z

√ Z 7 √ i 7 S = 2π (2x + 1) 1 + 4 dx = 2 5π (2x + 1) dx = 2 5π x2 + x 2 2 √ 2 √ = 2 5π(56 − 6) = 100 5π 7

√

371

6.6. AREA OF A SURFACE OF REVOLUTION 8. y 0 = −x(16 − x2 )−1/2 Z √7 p Z x 1 + x2 (16 − x2 )−1 dx = 2π S = 2π 0

√

7

0

x

r

16 dx = 8π 16 − x2

Z

√

0

7

x(16 − x2 )−1/2 dx

u = 16 − x , du = −2x dx

Z 9 Z 16 √ 16 1 −1/2 = 8π u u−1/2 du = 4π(2 u) 9 = 4π(8 − 6) = 8π − du = 4π 2 16 9 2

1 4x6 − 1 9. y 0 = x3 − x−3 = 4 4x3 Z 2 r Z 2 r 16x12 − 8x6 + 1 16x6 + 16x12 − 8x6 + 1 dx = 2π x dx S = 2π x 1+ 16x6 16x6 1 1

2 Z p Z π 2 (4x6 + 1)2 π 2 π 4 5 1 4 −2 = dx = (4x + x ) dx = x − 2 1 x2 2 1 2 5 x 1

π 251 1 253π − − = = 2 10 5 20

1 4x4 − 1 10. y 0 = x2 − x−2 = 4 4x2 r

r Z 2 Z 1 3 1 16x8 − 8x4 + 1 π 2 4x4 + 3 (4x4 + 1)2 S = 2π x + 1+ dx = dx 3 4x 16x4 6 1 x 16x4 1

Z Z π 2 (4x4 + 3)(4x4 + 1) π 2 3 −3 5 = dx = x dx 4x + 4x + 6 1 4x3 6 1 4

2 3 4635π π 4855 55 π 2 6 2 x + 2x − 2 − = = = 6 3 8x 6 96 24 576 1 r 4h2 − 4hx + r2 x −1/2 ; 1 + [f 0 (x)]2 = 1− 2h h 4h2 (1 − x/h) √ 2 p p 4h − 2hx + r2 r √ 2 √ 4h − 4hx + r2 f (x) 1 + [f 0 (x)]2 = r 1 − x/h = 2 2h 2h 1 − x

h Z h r p 2 πr 2 1 S = 2π r + 4h2 − 4hx dx = − (r2 + 4h2 − 4hx)3/2 h 3 4h 0 2h 0 πr 2 2 3/2 3 = 2 [(r + 4h ) − r ] 6h (b) With h = 0.1r we have

11. (a) f 0 (x) = −

πr S= [r3 (1.04)3/2 − r3 ] ≈ πr2 6(0.01)r2 The approximate percentage error is or approximately 1%.

1.043/2 − 1 0.06

≈ πr2

0.060596 ≈ πr2 .

0.06

(0.060596/0.06)πr2 − πr2 0.060596 = −1 ≈ 0.0099, 2 πr 0.06

372

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

12. y =

√

r 2 − x2 ; y 0 = −x(r2 − x2 )−1/2 Z bp Z p 2 2 2 2 2 −1 S = 2π r − x 1 + x (r − x ) dx = 2π a

= 2πr

Z

a

b

b

p

r 2 − x2

s

r2 dx r 2 − x2

dx = 2πr(b − a)

a

13. For x < −2, y = −x − 2 and y 0 = −1. For x > 2, y = x + 2 and y 0 = 1.

Z −2 Z 2 √ √ (−x − 2) 1 + 1 dx + 2π S = 2π (x + 2) 1 + 1 dx −4

√ Z = 2 2π

−2

−4

√

(−x − 2) dx + 2 2π

Z

−2 2

(x + 2) dx

−2

−2

2 √ √ 1 1 2 = 2 2π − x2 − 2x + 2 2π x + 2x 2 2 −4 −2 √ √ √ = 2 2π[(−2 + 4) − (−8 + 8)] + 2 2π[(2 + 4) − (2 − 4)] = 20 2π 14. Since the graph is symmetric with respect to the y-axis, we will find the area on [0, a] and multiply by 2. y = (a2/3 − x2/3 )3/2 ; S = 4π

Z

0

a

y0 = − s

(a2/3 − x2/3 )3/2 Z

(a2/3 − x2/3 )1/2 x1/3

1+

(a

2/3

−x x2/3

2/3

)

dx = 4π

Z

0

a

(a2/3 − x2/3 )3/2

s

a2/3 dx x2/3

2 u = a2/3 − x2/3 , du = − x−1/3 dx 3 0

0

Z 0 1/3 3 12πa 12πa2 3 =− (0 − a5/3 ) = = 4πa1/3 u3/2 − du = 4πa1/3 − u5/2 2 5 5 5 a2/3 a2/3 = 4πa1/3

a

x−1/3 (a2/3 − x2/3 )3/2 dx

15. Let θ be the angle formed when the cone is cut and flattened out. The length of the arc of the sector is 2πr, the circumference of the base of the cone. Then θ/2πr = 2π/2πL (the angle subtended by the sector is to the length of the sector as 2π radians is to the circumference of the circle of radius L), and θ = 2πr/L. 1 2 2πr Using the hint in the text, the lateral surface area is L = πrL. 2 L 16. If L is the slant height, then L2 = r2 + h2 and the surface area is πrL = √ r πr r2 + h2 . The surface can also be obtained by revolving the line y = x h about the x-axis: r r Z h Z r 2 r 2πr h r2 + h2 S = 2π x 1+ dx = x dx h h 0 h2 0 h √

# h p 2πr r2 + h2 1 2 x = πr r2 + h2 = h2 2 0

θ L

2πr

L h

r

373

6.6. AREA OF A SURFACE OF REVOLUTION r1 r2 = , r1 L2 = r2 L1 , and r1 L2 − r2 L1 = 0.

L1 L2 From Problem 15, the lateral surface area of the frustum is

17. By similar triangles,

L2

0

L

L1

}| { z S = πr2 L2 − πr1 L1 = π(r2 L2 − r1 L1 ) = π[r2 L2 + (r1 L2 − r2 L1 ) − r1 L1 ]

r1

r2

= π[(r2 + r1 )L2 − (r2 + r1 )L1 ] = π(r1 + r2 )(L2 − L1 ) = π(r1 + r2 )L.

18. By the Pythagorean Theorem, L2 = h2 + (r2 − r1 )2 or L = Section 6.6, p S = π(r1 + r2 )L = π(r1 + r2 ) h2 + (r2 − r1 )2 .

p

h2 + (r2 − r1 )2 . From (1) in L r1

r2 – r1 h

19. We need to extend (3) in Section 6.6 to include functions which are not necessarily nonZ b p negative. In this case we have S = 2π |f (x)| 1 + [f 0 (x)]2 dx. Next, we require the fact a

that the surface area obtained by revolving f around y = L is the same as that obtained by revolving F (x) = f (x) − L around the x-axis. Then F 0 (x) = f 0 (x) and S = 2π

Z

a

20. y 0 =

2 −1/3 x ; 3

b

Z p |F (x)| 1 + [F 0 (x)]2 dx = 2π

S = 2π

a

Z

1

8

b

p |f (x) − L| 1 + [f 0 (x)]2 dx.

r Z 4 2π 8 4 − x2/3 p 2/3 |x2/3 − 4| 1 + x−2/3 dx = 9x + 4 dx 9 3 1 x1/3

yB R = , which 21. (a) Since 4BCT is similar to 4T CS we have CB/T C = CT /CS or R R+h 2 p R gives yB = . Now, revolving x = R2 − y 2 for yB ≤ y ≤ R around the y-axis, we R+h obtain the surface area v !2 u Z Rp Z R u −y t 2 2 AS = 2π R −y 1+ p dy = 2π R dy R2 R2 − y 2 yB R+h

R2 2πR2 h = 2πR R − = .

R+h R+h AS h = .

AE 2(R + h) 2000 AS = ≈ 0.119332 ≈ 11.9%. (b) With h = 2000 and R = 6380, AE 2(6380 + 2000) h 1 (c) Setting = we obtain 2h = R + h or h = R = 6380 km.

2(R + h) 4 Since AE = 4πR2 , we have

374

CHAPTER 6. APPLICATIONS OF THE INTEGRAL AS h 1 = lim = h→∞ 2R + 2h AE 2 From a large distance we would expect to see half of the earth’s surface.

(d) lim

h→∞

(e)

6.7

AS 3.76 × 105 = ≈ 0.4917 = 49.17% AE 2(6380 + 3.76 × 105 )

Average Value of a Function

1. fave =

2. fave =

3. fave =

4. fave

5. fave

6. fave

1 5 − (−2)

Z

1 2−0

Z

2

5

1 (2x2 ) 4

0

Z

2

1 2 − (−2) Z

1 9−0

Z

1 = 3−0

Z

1 = 3−0

Z

1

0

−3

9

3

0 3

1 (2 − 18) = −4 4

1 3 x + 10x 3

2

3

3

2

1 1 (x + 1) dx = · (x + 1)3 2 3 2

x3 dx =

−2

1 4

x(3x − 1)2 dx = x1/2 dx =

(5x + 1)

1/2

1 9

68 −0 3

=

34 3

1 4 x − x3 + 2x2 − x 2

1 1 9 = − = −2 2 2 2

2

1 (3x − 4x) dx = (x3 − 2x2 ) 4 −1

Z

1 2

=

0

1 (2x − 3x + 4x − 1) dx = 2 −1 1

0

0

=

2

0

1 1−0

1 2

1

5 1 2 42 1 (x + 3x) = [40 − (−2)] = 7 7 7 −2

(2x + 3) dx =

(x2 + 10) dx =

Z

9. fave =

4x dx =

−2

1 = 3 − (−1) 1 = 2−0

1

−3

1 = 1 − (−1)

8. fave =

11. fave

Z

Z

7. fave =

10. fave

1 1 − (−3)

1 4 x 4 Z

0

1

2

2

=

−2

0

3

=

−1

1 = 2

9

−1

1 [9 − (−3)] = 3 4

1 9− 3

=

9 4 1 x − 2x3 + x2 4 2

13 3

1 (4 − 4) = 0 4

(9x3 − 6x2 + x) dx =

2 3/2 x 3

1

=2

0

1 2 dx = · (5x + 1)3/2 3 15

3 0

=

2 14 (64 − 1) = 45 5

3 p 1 1 2 1 61 3/2 2 x x + 16 dx = · (x + 16) = (125 − 64) = 3 3 9 9 0

1 0

=

3 4

6.7. AVERAGE VALUE OF A FUNCTION 12. fave

13. fave 14. fave

15. fave 16. fave

17. fave 18. fave 19. fave

375

1/3 Z 1 1 1 1 1 1 = 1+ dx u = 1 + , du = − 2 dx 1 − 1/2 1/2 x x2 x x 2 Z 2 3 3 =2 −u1/3 du = 2 · (−u4/3 ) = − (24/3 − 34/3 ) ≈ 2.7104 4 2 3 3

1/2 1/2 Z 1/2 1 1 1 = x−3 dx = 4 − x−2 = −2(4 − 16) = 24 = −2 1/2 − 1/4 1/4 2 x2 1/4 1/4

4

Z 4 1 1 3 5/3 1 3 5/3 12 = (x2/3 − x−2/3 ) dx = x − 3x1/3 4 − 3 · 41/3 − − = 4−1 1 3 5 3 5 5 1 1 5/3 4 = 4 − 41/3 + ≈ 1.2285 5 5 5

5 Z 5 1 1 1 1 1 1 −2 −1 2(x + 1) dx = [−2(x + 1) ] = − = =− − = 5−3 3 2 x + 1 6 4 12 3 3 Z 9 √ 3 √ 1 ( x − 1) 1 √ = dx u = x − 1, du = √ dx 9−4 4 x 2 x 2 Z 2 1 1 1 1 3 = 2u3 du = · u4 = (16 − 1) = 5 1 5 2 10 2 1 π Z π 1 1 1 sin x dx = (− cos x) = − [−1 − (−1)] = 0 = π − (−π) −π 2π 2π −π π/4 Z π/4 1 2 2 4 1 = = (1 − 0) = cos 2x dx = · sin 2x π/4 − 0 0 π 2 π π 0 √ π/2 Z π/2 √ 3 3 1 3 3 2 = = − (0 − 3) = csc x dx = (− cot x) π/2 − π/6 π/6 π π π π/6

Z 1/3 sin πx 1 dx u = cos πx, du = −π sin πx dx 1/3 − (−1/3) −1/3 cos2 πx Z 3 1/2 1 = − 2 du = 0 2 1/2 πu

1

Z 1 1 1 1 3 1 4 2 1 2 2 21. fave = (x + 2x) dx = x +x = − = 1 − (−1) −1 2 3 2 3 3 3 −1 √ 1 −6 ± 36 + 12 2√ Setting f (c) = c2 + c = , we obtain 3c2 + 6c − 1 = 0. Then c = = −1 ± 3.

3 6 3 √ 2 The only solution on [−1, 1] is −1 + 3.

3 6 Z 6 1 1 2 2 38 22. fave = (x + 3)1/2 dx = · (x + 3)3/2 = (27 − 8) = 6−1 1 5 3 15 15 1 √ Setting f (c) = c + 3 = 38/15, we obtain c + 3 = 1444/225. Thus, c = 769/225.

20. fave =

376

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

23. We are given

1 5−1

Z

5

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

Z

5

f (x) dx = 12.

1

1

b Z √ 1 b 1 2 3/2 2√ 24. Solving (1 − x) dx = b = 0 gives b = 9/4. At b = 9/4 the x− x =1− b 0 b 3 3 0 area bounded by the graph above the x-axis is the same as the area below the x-axis.

6 Z 6 1 1 3 1 1 100 + 3t − t2 dt = 25. Tave = 100t + t2 − t3 = 103◦ 6−0 0 2 6 2 6 0 5 Z 5 1 1 1 (50 + 4x + 3x2 ) dx = (50x + 2x2 + x3 ) = (425 − 53) = 93 26. Rave = 5−1 1 4 4 1 5

1 1X R(k) = (57 + 70 + 89 + 114 + 145) = 95 5 5 k=1

27. Using s0 (t) = v(t) we have vave

1 = t2 − t1

Z

t2

t1

t2 s(t2 ) − s(t1 ) 1 = v(t) dt = s(t) = v. t2 − t1 t2 − t1 t1

Z 2π/ω Z 1 2 ωk 2π/ω 2 1 kx dt = A cos2 (ωt + φ) dt 2π/ω − 0 0 2 4π 0

2π/ω

Z

ωkA2 2π/ω 1 ωkA2 1 = [1 + cos 2(ωt + φ)] dt = sin(2ωt + 2π)

t+ 4π 2 8π 2ω 0 0

ωkA2 2π 1 1 kA2 = + sin(4π + 2φ) − sin 2φ = 8π ω 2ω 2ω 4 Z 2π/ω Z 2π/ω Z 1 1 ωm ωm 2π/ω 2 2 2 2 0 2 Kave = mv dt = [x (t)] dt = ω A sin (ωt + φ) dt 2π/ω − 0 0 2 4π 0 4π 0

2π/ω

Z

A2 ωk A2 ω(mω 2 ) 2π/ω 1 1 [1 − cos 2(ωt + φ)] dt = sin(2ωt + 2φ)

= t− 4π 2 8π 2ω 0 0

2 2 ωkA 2π 1 1 kA = − sin(4π + 2φ) − − sin 2φ = 8π ω 2ω 2ω 4 " #

2

Z t1 Z t1 t1 2t 4 4 29. mv1 − mv0 = (t1 − 0)F = −1 1 − 2 t2 + t − 1 dt k 1− dt = k t1 − 0 0 t1 t1 t1 0

t1

4 2 4 2kt1 = k − 2 t3 + t2 = k − t1 + 2t1 = 3t1 t1 3 3 0

R

Z R 1 P P 1 P 2 3 P R2 30. vave = (R2 − r2 ) dr = R2 r − r 3 = R = R − 0 0 4vl 4vlR 3 4vlR 3 6vl 0 Z a 31. 0, since f (x) dx = 0.

28. Uave =

−a

377

6.7. AVERAGE VALUE OF A FUNCTION

32. Intuitively, the average value of the linear function f (x) = ax + b on [x1 , x2 ] should be the x1 + x2 value of f at the midpoint of that interval, or X = . This can be proven as follows: 2 Z x2 a x2 1 1 (ax + b) dx = fave = x2 + bx x2 − x1 x1 x2 − x1 2 x 2

1 2

2 1 ax2 ax1 1 a(x2 − x21 ) = + bx2 − − bx1 = + b(x2 − x1 ) x2 − x1 2 2 x2 − x1 2

x2 + x1 =a + b = aX + b 2 0 33. fave =

34. fave

Z

x+h

f 0 (x) dx =

x

1 = a−1

Z

a

x+h 1 f (x + h) − f (x) f (x) = h h x

1 (n + 1)x dx = xn+1 a−1 n

1

a 1

=

an+1 − 1 = an + an−1 + · · · + a + 1 a−1

Z b Z b Z b [f (x) − fave ] dx = 0, then f (x) dx = fave dx. Thus, f (x) dx = fave (b − a) a a a a Z b 1 and therefore fave = f (x) dx. b−a a

35. If

Z

1 h

b

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.

1 The average value of f on the interval [0, n] appears to be (n − 1), which can be proven as 2 follows: Z 1

Z n Z 2 Z n 1 1 fave = f (x) dx = 0 dx + 1 dx + · · · + (n − 1) dx n−0 0 n 0 1 n−1

n−1 1X 1 (n − 1)n 1 1 k= = (n − 1) = [0 + 1 + · · · + (n − 1)] = n n n 2 2 k=1

37. There is no unique answer to this question; among several possible approaches, here is probably the simplest. Suppose the circle is centered at the origin and that one of the points on the circle is (−1, 0). If (x, y) is any other point on the circle, then the length of the chords between (−1, 0) and (x, y) is the distance between the points: p p √ (x + 1)2 + y 2 = x2 + y 2 + 2x + 1 = 2x + 2.

The average chord length Lave is then Lave =

1 1 − (−1)

Z

1

−1

√

2x + 2 dx =

1 1 (2x + 2)3/2 · · 2 2 3/2

1

−1

=

1 3/2 8 4 ·4 = = .

6 6 3

378

CHAPTER 6. APPLICATIONS OF THE INTEGRAL 6.8. WORK

38. (a) The surface area is S = 2Ď&#x20AC;

Z

a

b

Z b ! b p " 0 (x)]2 dx. If f 0 (x) = 0, then S = 2Ď&#x20AC; ! (x)]2 dx. If f ! (x) = f (x) 38.

1 + [f (x) (a) The surface area is S = 2Ď&#x20AC; f (x) 1f+ [fdx. a

If 0 â&#x2030;¤ |f 0 (x)| < for a â&#x2030;¤ x â&#x2030;¤ b, then

a

If 0 â&#x2030;¤ |f ! (x)| < " for a â&#x2030;¤ x â&#x2030;¤ b, then Z b Z b Z b !p ! b p b " " 2 2 f (x) 1 + dx =2Ď&#x20AC;2Ď&#x20AC; f1(x) f (x) dx â&#x2030;¤ S â&#x2030;¤ 2Ď&#x20AC; 2Ď&#x20AC; dx. f (x) 1 + "2 dx = 2Ď&#x20AC; 1 + dx â&#x2030;¤ fS(x) â&#x2030;¤ 2Ď&#x20AC; a

a

a

a

a

(b) At any x in [a, b] the circumference of a (b) circular cross-section is 2Ď&#x20AC;f (x). The average At any x in [a, b] the circumference of a circular cross-sectio Z b ! b 1 1 circumference is then C = 2Ď&#x20AC;f (x) dx.

Letting L = b â&#x2C6;&#x2019; a Cbe=the length2Ď&#x20AC;f of the circumference is then (x) dx. Letting L = bâ&#x2C6;&#x2019;a a bâ&#x2C6;&#x2019;a a ! b Z b â&#x2C6;&#x161; limb, we (a), haveCL CLâ&#x2030;¤=S 2Ď&#x20AC; limb, we have CL = 2Ď&#x20AC; f (x) dx. Thus, from part â&#x2030;¤ 1 f+(x) 2 dx.

CL.Thus, from part (a), CL a

a

6.8

6.8

Work

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

Work

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

â&#x2C6;&#x161; â&#x2C6;&#x161; 2. The horizontal component of force is 50 32.N.The horizontal component of force is 50 3 N. W = 50 3 ¡ 8 = 400 â&#x2C6;&#x161; â&#x2C6;&#x161; # $ 30° W = 50 3 ¡ 8 = 400 3 joules.

1 3. Since 10 = k , k = 20 and F 50 = 20x. Solving 8 = 20x we obtai 100 2

â&#x2C6;&#x161;

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

1 2 , k = 20 and F = 20x. Solving = 20xxwe obtain = N.ft. (b) 8When = 0.5, F =x250 2 5 (c) Solving 200 = 500x, we obtain x = 0.4 m. The length of the s 4. (a) Since 50 = k(0.1), k = 500 and F = 500x.

5. (b) When x = 0.5, F = 250 N. (c) Solving 200 = 500x, we obtain x = 0.46.m. The length of the spring is 0.5 + 0.4 = 0.9 m.

Z 0.2 7.

0.2 5. (a) W = 500x dx = 250x2 0 = 10 joules 0 8.

Z 0.6

0.6 (b) W = 500x dx = 250x2 0.5 = 27.59.joules

3. Since 10 = k

0.5

6. (a) W =

Z

0

6

3 3 x dx = x2 2 4

6 0

10.

9 27 = 27 in-lb = 11.12 ft-lb = 4 ft-lb

16 12.

3 2 3 (b) W = x dx = x = 192 in-lb = 16 ft-lb 2 4 0 0 13.

2 7. Since 10 = k , k = 15 and F = 15x. 14.

3 15.

1 Z 1 15 2 15 (a) W = 15x dx = x = ft-lb 16.

2 2 0 0 3 Z 3 17.

15 2 75 (b) W = 15x dx = x = ft-lb 2 2 2 2 Z

16

379

6.8. WORK 50 50 and F = x. W = 3 3

8. Since 50 = k·3, k =

9. We use 500 km = 0.5 × 106 m.

W = (6.67 × 10

−11

)(6.0 × 10 )(10 ) 24

4

= 453.1 × 108 joules

Z

0

10

50 25 2 x dx = x 3 3

1 1 − 6.4 × 106 6.9 × 106

10. We use 200 km = 0.2 × 106 m.

W = (6.67 × 10−11 )(7.3 × 1022 )(5 × 104 )

10 0

=

2500 625 in-lb = ft-lb 3 9

≈ 4.531 × 1010

1 1 − 6 1.7 × 10 1.9 × 106

≈ 1.507 × 1010

= 150.7 × 108 joules 12 Z 12 1 11. W = 62.4π(3)2 x dx = 9(62.4π) x2 = 9(62.4π)(72) ≈ 127, 030.9 ft-lb 2 0 0

y

12

3 x

1 12. y = x 5 Z W =

x 4

Z

1 2 x (20 − x) dx 25 0 0

10 Z 10 1 4 3 1 4 4 2 x − x3 dx = 62.4π x − x = 62.4π 5 25 15 100 0 0

500 = 62.4π ≈ 32, 672.6 ft-lb 3

Z 10 Z 10 1 2 1 13. W = 62.4π x (25 − x) dx = 62.4π x2 − x3 dx 25 25 0 0

10

700 1 3 1 4 = 62.4π = 62.4π x − x ≈ 45, 741.6 ft-lb 3 100 3 0 √ 14. x2 + y 2 = 9; y = 9 − x2 Z 3 Z 3p W = 9 − x2 (5 − x) dx 62.4(2y · 12)(5 − x) dx = 1497.6 10

−3

62.4πy (20 − x) dx = 62.4π 2

Z = 1497.6 5

3

−3

p

9 − x2 dx −

Z

3

−3

x

10

p

−3

9 − x2 dx

20

y

10 y

x 2

3

y

y

1 The first integral represents the area of the semicircle and is thus π(3)2 . The second integral 2

1 2 has an odd integrand and is thus 0. Therefore W = 1497.6 5 · π(3) = 33, 696π ft-lb.

2

380

CHAPTER 6. APPLICATIONS OF THE INTEGRAL 3 x 4 Z W =

x

15. y =

Z

3 x(x + 5) dx 0 0 4

4

Z 4 1 3 5 2 184 2 = 936 (x + 5x) dx = 936 x + x = 57, 408 ft-lb = 936 3 2 3 0 0 √ 16. x2 + y 2 = 25; y = 25 − x2 Z 5 Z 5 p W = 80(2y · 25)x dx = 4000 x 25 − x2 dx 4

62.4(2y · 10)(x + 5) dx = 62.4(20)

2

2

4

5 5

1 4000 4000 = 4000 − (25 − x2 )3/2

= − (25 − x2 )3/2 = − (0 − 213/2 ) 3 3 3 2 2

4

3 y

y

5 5 3

y

y

x

≈ 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.

40

Z 40 1 2 = 64, 000 ft-lb W = 20(100 − x) dx = 20 100x − x 2 0 0 18. The weight of the system is F (x) = 3000 + 40(200 − x) lb when x feet of chain have been pulled up.

100 Z 100

1 = 900, 000 ft-lb W = [3000 + 40(200 − x)] dx = 3000x + 40 200x − x2

2 0 0 19. (a) W = 80 · 65 = 5200 ft-lb

1 (b) The weight of the system is F (x) = 80 + (65 − x) when the bucket has been lifted x 2 feet.

65 Z 65

1 1 1 W = 80 + (65 − x) dx = 80x + 65x − x2

= 6256.25 ft-lb 2 2 2 0 0

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

40

Z 40

x 1 ≈ 24, 960 ft-lb W = 62.4 20 − dx = 62.4 20x − x2 2 4 0 0 21. If x is the distance separating the electron and the nucleus, then the force is F (x) = k/x2 .

4

Z 4 k k 1 3k W = dx = − = −k − 1 = 2 x 1 4 4 1 x 22. (a) If x is the distance above the earth, then the weight of the system is F (x) = 2, 700, 000 − 100x.

381

6.8. WORK (b) W =

Z

1000

0

(2, 700, 000 − 100x) dx = 2, 700, 000x − 50x2

23. Since p = kv −γ ,

1000 0

= 2, 650, 000, 000 ft-lb

v2 k k 1−γ W = p dv = kv dv = = v (v 1−γ − v11−γ ) 1−γ 1−γ 2 v1 v1 v1 1 1 = (kv2−γ v2 − kv1−γ v1 ) = (p2 v2 − p1 v1 ), 1−γ 1−γ Z

Z

v2

v2

−γ

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

24. Using Newton’s second law F = ma = mg, we have Z y2 Z y2 y W = F (y) dy = mg dy = mgy]y21 = mgy2 − mgy1 . y1

y1

25. Since the distance moved is 0, no work is done.  2x, 0 ≤ x ≤ 1      2, 1 ≤ x ≤ 2   26. The force is F (x) = −x + 4, 2 ≤ x ≤ 4 .    0, 4 ≤ x ≤ 5     x − 5, 5 ≤ x ≤ 6 Z 1 Z 2 Z 4 Z 6 The work is W = 2x dx + 2 dx + (−x + 4) dx + 0 + (x − 5) dx 0

=

1 x2 0

1

+

2 2x]1

+

2

1 − x2 + 4x 2

4 2

+

5

1 2 x − 5x 2

6 5

= (1 − 0) + (4 − 2) + (8 − 6) + (−12 + 12.5) = 5.5 N-m.

27. W = 165 · 1350 = 222, 750 ft-lb

28. Since the water leaks out of the bucket at a constant rate and the weight x 180 20 of the rope is negligible, it is reasonable to approximate that the overall lifted weight is the midpoint or average of the bucket’s starting and ending 10 weights of 200 and 180, respectively, or (200 + 180)/2 = 190 lb. Moving y this 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 of the 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 0 and dx = x0 (t) dt = v dt, we have Z x2 Z t2 W = F (x) dx = mv 0 v dt u = v, du = v 0 dt =

x1 v2

Z

v1

t1

1 mu du = mu2 2

v2 v1

=

1 1 mv22 − mv12 .

2 2

382

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

30. From the figure in the text, we see that sin θ = x/30 so that x = 30 sin θ and dx = 30 cos θ dθ.

Also, when 6.9. x = 3, sin θ PRESSURE = 0.1 and θ AND ≈ 0.1.FORCE Then FLUID 381 Z 3 Z 0.1 Z 0.1 sin θ W = 30.mg tanthe θ dx = 550(9.8) (30 cos = 16500(9.8) θ dθdx = 30 cos θ dθ.

From figure in the text, wecos seeθthat sin θθ)=dθx/30 so that x = 30 sinsin θ and 0 0 0 Also, when x0.1 = 3, sin θ = 0.1 and θ ≈ 0.1. Then ! 3 ! 0.1 ! 0.1 sin θ 6.9. FLUID PRESSURE AND FORCE W = mg tan θ dx = 550(9.8) (30 cos θ) dθ = 16500(9.8) sin θ dθ 6.9. FLUID PRESSURE AND cos θ FORCE 0 0 0

= 161700(− cos θ)]0 = 161700(1 − cos 0.1) ≈ 807.8 N-m.

6.9

FLUID PRESSURE AND FORCE Fluid Pressure= and Force0.1 6.9.161700(1 From the in the see that sin θ = x/30 so that x = 30 sin θ and d 161700(− cos θ)] =30. −figure cos 0.1) ≈ text, 807.8weN-m.

0 the figure in the text, we see that sin θ = x/30 so that x = 30 sin θ and dx = 30 cos θ 30. From Also, when x = 3, sin θ = 0.1 and θ ≈ 0.1. Then 2 30.

From inand the θtext, we Then see that sin θ = x/30 so that x = 30 sin θ and d Also, when x = 3,the sin figure θ ==0.1 ≈ 0.1.

F = (1248)(25π) ! 3 31200π lb ! 0.1 ! 0.

Also, when 6.9. x= 3, sin θ PRESSURE = 0.1 and θ AND ≈ 0.1.

Then sin θ ! ! 0.1 ! 0.1 2 FLUID FORCE 3 W = mg tan θ dx = 550(9.8) (30 cos θ) dθ = 16500(9.8) sin θ ! 0.1 = 62.4(20) = 1248 lb/ft ; F = (1248)(4π) =! 4992π lb !

3 θθ = 16500(9.8) W = mg tan θ dx (300cos cos θ) sin θ dθ 0 0 0 = 550(9.8) sindθ 2 cos θ W = 30.mg tan θ dx = 550(9.8) (30 cos θ) dθ = 16500(9.8) 02 0 0 0.1 the θ)] figure in 161700(1 the text, see θ = x/30 so that x = 30 = = 1248 lb/ft ; = 1248 F = lb/ft (1248)(100π) = From 124800π = 161700(− cos = −cos cosθthat 0.1) sin ≈ 807.8 N-m.

0 0.1 0 1. 62.4(20) (a) pressure = 62.4(20) ; F = (1248)(25π) =0lb 31200π lb0 ≈we = 161700(− cos θ)]0Also, = 161700(1 807.8 N-m. when x0.1 =− 3, cos sin θ0.1) = 0.1 and θ ≈ 0.1. Then = 161700(− cos θ)]0 = 161700(1 − cos 0.1) ≈ 807.8 N-m.

2 3 2

1. (a) pressure = 62.4(20) = 1248 lb/ft ;

6.9 (b) pressure (c) pressure

Fluid Pressure and Force

= 62.4(20) = 1248 2. (a) pressureoil =(b) 55 pressure lb/ft · 96 ft = 5280 lb/ft lb/ft ;

F = (1248)(4π) =!4992π lb 3

!

sin θ 381 Fluid Pressure Force 6.9. FLUID PRESSURE AND 6.9 FORCE 6.9. FLUID PRESSURE AND FORCE W = and mg tan θ dx = 550(9.8) (30 cos θ) dθ = 165 3 2 6.9. FLUID PRESSURE AND cos θ FORCE Fluid Pressure and Force pressure 1248 lb/ft ; F = (1248)(100π) = 0 124800π lb 0 (b) pressurewater (c) = 62.4 lb/ft= 62.4(20) · 85 6.9 ft ==5304 lb/ft 6.9. 6.9 FLUID PRESSURE AND FORCE and Force Fluid Pressure 2

0.1

1. (a) pressure = 62.4(20) = 1248 lb/ftcos; θ)]F0.1==(1248)(25π) = 0.1) 31200π lb weN-m.

30.

From the in the see t 161700(− 161700(1 −figure cos ≈ text, 807.8

0 the figure in the text, we see that sin θ = x 3(a) 30.sin From (c) forceoil =2.5280 · 125 · 350 231, 000, 000 lbwe From the== figure in see sin2 θ== x/30 so2that 30 θ and dx 30lbcos θ dθ.

1.the = that 62.4(20) 1248 lb/ft ; F x==(1248)(25π) = 31200π Also,=when x = 3, sin θ = 0.1 and θ 2 (a)30. pressure 55 lb/ft · text, 96pressure ft = 5280 lb/ft oil Also, when x= 0.1 and dx θ ≈= 0.1.

1. (a) pressure = text, 62.4(20) = 1248sin = (1248)(25π) =4992π 31200π lb (b)figure pressure lb/ft ;x/30 F so = (1248)(4π) == lb 30.

From the in the we see that θ = that x 3, = sin 30θsin and 30Then cos ! θ 2 Also, when x= 3, sin θ =000 0.1 lb and=θ62.4(20) ≈ 0.1. Then 3 3 2 (b) pressure = 1248 lb/ft ; F = (1248)(4π) = 4992π lb (d) forcewater = 5304 · 125 · 350 = 232, 050, ! 3 ! 0.1 2 x = 3, sin θ = 0.1 and θ ≈ 0.1.

Then 6.9 Fluid Pressure and Force W = mg tan θ dx = 550(9. (b) pressure = 62.4 lb/ft ·Also, 85 ftwhen = 5304 lb/ft (b) pressure = 62.4(20) = 1248 lb/ft ; F = (1248)(4π) = 4992π lb sin θ

water

!

(c) pressure = 62.4(20) = 1248 lb/ft ; F =W(1248)(100π) = 124800π lb 2 mg tan θ dx ! !==0.1 0 = 550(9.8) 2 F! = (1248)(100π) !

0124800π lb 0

0.1 = 1248 lb/ft ; !

2 (c) pressure = 62.4(20)

0.1 sinFLUID θ 3 0.1 0.1 2 (c) pressure 62.4(20) =3θ) 1248 lb/ft ;θ=FORCE F =2lb/ft (1248)(100π) =0.1 124800π =dθ 161700(− cos θ)] = 16 (c) forceoil · 125 ==231, 000, 000 lb sin 6.9.= PRESSURE AND 1. = (a) pressure = 1248 ; sin F θ= (1248)(25π) =0lb 31200 3. (a) pressure = (62.4)(8) == 499.2 ;350θFdx = 224, 640 W5280 = lb/ft mg ·tan =2.(499.2)(30)(15) 550(9.8) (30 cos dθ62.4(20) = 16500(9.8) (a)=pressure 55 · 96 ftlb = 5280 lb/ft = dθ 161700(− cos θ)]0 = 161700(1 − co oil W mg tan =lb/ft 550(9.8) (30 cos θ) = 16500(9.8) sin θ dθ 3 θ dx 2 cos θ 2 3 2 0 0 0 θ= 1248 pressure 55 lb/ftlb·=96 ftpressure = 5280 lb/ft Z 8 water = 5304 · 2.

8pressure 0 000 0 = cos 0 3 62.4(20) 2; oil = (b) = lb/ft F = (1248)(4π) = 4992π 2.232, (a) 55 lb/ft · 96 ft 5280 lb/ft (d) force 125(a) · 350 =0.1 050, (b) pressureoil = 62.4 lb/ft · 85 ft = 5304 6.9 lb/ft Fluid Pressure and F water 0.1 3θ)] 2 0.1) 30.

From theft in N-m. the text, we ≈ see807.8 that sin θ = x/30 so that x = 30 2 161700(1 = 161700(− cospressure θ)]0 = − cos 0.1) ≈figure 807.8 2 N-m. = 161700(− cos = 161700(1 −6.9 cos = 62.4 lb/ft · 62.4 85 = 5304 lb/ft 0 Fluid and Force (c) pressure =3 62.4(20) 1248 lb/ftPressure ; F = (1248)(100π) = 1248 = 59, 904 (b) sidewall force = 62.4x(30) dx = (b) 62.4(15)x (b) =lb · 231, 85 ==0.1 5304 lb/ft water water (c)2 pressure force 5280 · 125 ·lb/ft 350 000 lb Also, when x ==3, sinft θ000, = and θ ≈(a) 0.1.

Then oil = 1. pressure = 62.4(20) = 1248 lb/ 3. (a) pressure = (62.4)(8) =(c)499.2 lb/ft ; F = (499.2)(30)(15) = 224, 640 lb 0 0force 3(a) 2 = 1248 lb/ft2 ; F = ( force = 5280 · 125 · 350 = 231, 000, 000 lb (c) = 5280 · 125 · 350 = 231, 000, 000 lb oil 1. pressure = 62.4(20) oil ! pressure (a) pressure 55232, lb/ft050, · 96000 ft = lb/ft = 62.4(20) = 1248 lb/ (d) Z 8 · 125 !· oil 350 lb5280 3 == ! 8 8 forcewater2.= 5304 (b) 0.1 sin θ 2 6.9 Fluid Pressure "8 and 15 6.9. lbFLUID PRESSURE AND FORCE 3 2 cos (b) pressure = 62.4(20) =(30 1248 lb/ft ; =F 165 =( force · 232, 125 ·050, 350 = 232, 050, 000 (d) force 5304 · 125=· 350 = 000 lbθ dx W = Force mg = 550(9.8) θ) dθ water water 6.962.4x(15) Fluid and 2(d)=Force 25304 (b) pressure =tan 62.4 85 ft(c) = 5304 6.9. 2 lb/ft FLUID· PRESSURE AND pressure 62.4(20) = 1248 lb/ water coslb/ft θ=FORCE x = 29, 952 lb end force = (b) dx Pressure = 62.462.4x(30) sidewall force dx = 62.4(15)x = 59, 904 lb 0 0 2640 lb 3. (a) pressure = (62.4)(8) = 499.2 lb/ft 2; (c)Fpressure = (499.2)(30)(15) = 224, 0 = 62.4(20) = 1248 lb/ft ; F =( 2 2 2 0.1 1. (a) pressure = 62.4(20) = 1248 lb/ft ; F = (1248)(25π) = 31200π lb 3 0 (c) force = 5280 · 125 · 350 = 231, 000, 000 lb 0 3. (a) pressure (62.4)(8) = 161700(− lb/ft ; 0 F ==30. = 640 oil !lb/ft 0 (62.4)(8) From the inlb the text, see 3. (a) pressure = ;499.2 F = (499.2)(30)(15) 224, 640 cos θ)] 161700(1 −figure cos 0.1) ≈224, 807.8 N-m.

2.(499.2)(30)(15) (a) pressure = 55 lb/ft ·we 96 ftlb=t 8= oil #2 ; $F ==499.2 %(1248)(25π) ! 8= 62.4(20) 3 we see that sin θ =2x "=in 1. (a) pressure = 1248 lb/ft lb pressure 8 30. From the figure thelb/ft text, √(b) 2= 31200π (a) 55 ·3, 96sin ft θ==5280 lb/ft 2oil"8= ! 8 62.4x(30) Also, when x904 = 0.1 and θ 8= !62.4(20) 3 force 125 · 350 = 232, 050, 000 lb 8 force 15 (b) pressure = ==1248 lb/ft =; 5304 Fdx =· 2.

(1248)(4π) = 4992π lb= sidewall = 62.4(15)x 59, lb water (b) pressure = 62.4 lb/ft · 85 " 2 (d) Also, when x = 3, sin θ 0.1 and θ ≈ 0.1.

Then water 0 8 4. The equation of the line through (1, 0) and (5/2, 3/2) is 3 2 2 dx1248 = (b) 62.4 xforce = 29, 952 lb end(b) force = ! lb/ft = 62.4 · 85 ft = 5304 lb 0dx62.4x(30) 3 y 904 sidewall = dx 2=lb0(b) 62.4(15)x = 59, 904 lb (b) sidewall force =;2 F62.4x(30) = 62.4(15)x = pressure 59, pressure =62.4x(15) 62.4(20) = lb/ft = (1248)(4π) = 4992π water 2(1, 0) (c) force · 125 · 350 = 231 2lb124800π ! ! # $ % √ √ oil = 5280 ! (c) pressure = 62.4(20) = 1248 lb/ft ; F = (1248)(100π) = lb 3 0.1 8 8 pressure Pressure and Force 0 == ·(499.2)(30)(15) mg=tan θ dx = 000 550(9.

3. (a) = (62.4)(8) =(c) 499.2 lb/ft ;0 W·F125 =lb2θ 0 0 Fluid sin 06.9 force 350 231, 000, oil = 5280 2 $= W tan θ952 dx =lb 550(9.8) #dx %8 # 15 ! 8 62.4x(15) 0 = 3 3 !

5304 · 125 · 350 = 2 !62.4 8mg 8; force = x2 "(d) =force 29,water end (c) pressure = 62.4(20) = 1248 lb/ft F == (1248)(100π) =$ 124800π lb θ √ 3 28 √ "8 =cos 02 = 5304 · 125 · 350 0 0.1 cos 15 (d)15 forcewater 232, 050, 000 y= x− . Using symmetry, 2 2 2.end (a)force pressure = 55 lb/ft · 96 ft = 5280 lb/ft 2 2 = 161700(− θ)] = 16 oil 62.4x(15) 0dx62.4x(15) (a) pressure 62.4(20) = 1248 ; = F29, = (1248)(25π) 0 31200 √ (b) 0= xdx lb==161700(1 end force1.= dx sidewall force 62.4x(30) 62.4(15)x 59, = 3/2 62.4 = = 952 lb 3=x62.4 329,lb/ft 3. (a) pressure = 0.1 (62.4)(8) ==904 499.2 lb = 161700(− cos952 θ)] −lb co 3 34. The equation of the line through3 (1, 0) = 0 0 2 2 2 2 3 2 0 and is· 85 y= x− .pressure Using symmetry, √ F= 0 0 (5/2, 2= ! !lb/ft 0 (62.4)(8) 3. 0 8 = ;4992π pressure = 62.43/2) lb/ft ft==62.4(20) lb/ft 2. (a) pressure lb/ft (b) · 96√ ft = 5280 $ = 499.2 %√ water lb/ft Z 5/2 Z 5/2 (b) pressure =(a) lb/ft ; # F = 83 oil = 55 √ √ √(b) 8 3!5304 312486.9 3 ! 8(1248)(4π) √ 15 sidewall force =x −62.4x(30) d Fluid Pressure and F 2 3 (c) 3 oilequation 4. 3force The equation of the line through (1, 0) and (5/2, 3/2) is = .lbUs 4.

The (5/2, 3/2) is y√ 2 2end 62.4x(15) dx = 62.4 x = 29, 952 force = √ = 5280 · 125 · 350 = 231, 000, 000 lb 0 3 3 (b) sidewall force = 62.4x(30) dx = 6.9 Fluid Pressure and Force & ' (c)√pressure √ = 62.4(20) = 1248 lb/ft ; F 2= (1248)(100π) =62.4(1 1248 3 3 (1, 0 (b)124.8 pressure = 62.4 lb/ft · 85 ft = 5304 lb/ft √ √ F = x − dx 62.4x(2y) dx = x ! ! water !

8 0 line5/2 through (1, 0) and (5/2, 3/2) is y1.= (a) pressure x − 0 = .62.4(20) 0Using symmet 5/2 4. The equation of the = 1248 lb/ # $ 3 = 5304 3 · 125 3 (d) force 3 water &3√!end8 force ' 62.4x(15) 3 symmetry, · 350 = 232,3050, 000 lb 2 = 1 23 √ !

15 1. 3(a) = 62.4(20) =√ 1248 lb/ft ; Fdx == ( y= x − 000 .lbUsing (c) force · 125 · 350 231, 000, 2.x (a)5/2 pressure lb/ft · 96!pressure ft5/2 = 5280 lb/ft F =oil1= 5280 62.4x(2y) dx = = 124.8 x − oil = 55dx 62.4 end force 3 3 62.4x(15) 3 0dx = 3/2 (b)= =√62.4(20) = 1248 lb/ &√ 'pressure 3! 5/2 F34.= √ 2y = 2 x !

2the ! !# 3 3 The equation of line through (1, 0) and (5/2, 3/2) is x 0 62.4x(2y) dx = 124.8 x x − dx 5/2 % & 3 2 (b) pressure = 62.4(20) = 1248 lb/ft ; F √ √ √ √ 5/2 1 1 √ 3 · 85 √ 3. (a) pressure =050, (62.4)(8) = pressure 499.2 lb/ft !; =5/2 F = (499.2)(30)(15) = 224, 640 lb = 1248 3=lb/( Z 5/2 !

3 (b) 62.4 lb/ft ft = 5304 lb/ft 5/2 3 3 (d) force = 5304 · 125 · 350 = 232, 000 lb (c) pressure = 62.4(20) water 2 3= 62.4(20) equation the line through 62.4x(2y) dx =&124.8 x (c)3'( x1 −4. The dx =of1248 3 2 3 water !

3FF == 3 2! 8 1 dx & '62.4x(2y) pressure ; √ F = & √lb/ft '(1( √ √ 5/2 =oil 124.8 x x− dx 3"' 3(a) 8 ! 5/2 !&lb &5280 4.√ equation of the through (1,√ 0)3 and (5/2 force = · 125 · The 350 = 000 2 1 5/2 x − x dx = 5/2 124.8√3(b) sidewall x3√− x= (c)62.4x(30) = 124.8 √ √ 5 ! 5/2 2. pressure = 55 lb/ft · 96'( ft = force dx = 1162.4(15)x =231, 59,3000, 904 lbline 2 3 3 3 oil 3 3 3 1 3 2 !

0 2640 = (62.4)(8) = 499.2 = (499.2)(30)(15) 224, lb232, 33 − 3x 3 lb/ft 3 3. (a) pressure 3 = 124.8 2. (a) pressure = 55 √ lb/ft 3x ·'( 96 5/2 ft =x5280 F = =x' dx =oil124.8 − d &(d) &= 3 2=·62.4x(2y) 0 1 √ forcewater √ √ x92 − lb/ftx ;6!dxF = 124.8 5304 125 · 350 (b)050, pressure 62.4 lb/ft2 ·!85 ! 000 lb 3=5/2

3

dx√= 124.8 x − x 3 dx5/ = 124.8 5/2 3 62.4x(2y) %x8 % =&' 5/2 1x −$ √93 3&# √13Fwater 8 % √13 !! 5/2 3water 3 3(b) 92528031 ·5/2 85 = ft6= 5304 lb 3 = 62.4 15=6 124.8 1− F(c) 62.4x(2y) 124.8 2 &pressure 32x"228dx 362.4 3 =xlb/ft x(62.4)(8) dx x2= − F'oil= =force 124.8 force · dx 125 & · 350 = 231 1 3952 3√ 2 62.4x(15) = x = 29, lb end = √ √ !

3. (a) pressure = = 499.2 lb/ft ; (499.2)(30)(15) =1lb12 &000 1 = dx 124.8 dx2lb= 5/2 124.8 − 62.4x(30) = 62.4(15)x 3 x 0−= 59, 3 x904 9 =x5280 6 x· 350 = 231, ! 000, (c) force · 125 oil 10 5/2 3 3 3√' 0 5304 · 125 √ · 350 2 (d) force6 3 2 3= 1 3 3 (−0.10)] 9 ! water =& = 124.8[1.20 − = 162.12 lb.

2 1 8 = 124.8[1.20 −0 (−0.10)] = 162.12 lb. √ x − = 5304 x ·!dx = 124.8 x000 − = 124.8 = 124.8 " √ 5/2 1 (d) force 125 · 350 = 232, 050, 8 √ √ water # − $(b) %sidewall 3 dx =362.4(15)x2 3= 259, 93 3lbx !

8 = 124.8[1.20 (−0.10)] = force 162.12 = lb.1 62.4x(30) √ x 1−= 904 x lb d = pressure 124.8 3. (a) = (62.4)(8) 499.2 3 3 The equation of the line through (5/2, 0)8 and (1, 3/2) is 0 15 ≈ 124.8[1.20 − (−0.10)] = 0) 162.12 lb.

2 symm 03. (a)3/2) 1 = 499.2 4. The of the line through and (5/2, is yy = =(62.4)(8) x− . 3Using !lb/ft pressure ;3 (−0.1 F = 8 62.4x(15) dx = 62.4 x2 = (1, 29, = 952 end force = =3%124.8[1.20 − 5. equation # √ √ ! 8lb 124.8[1.20 − (−0.10)] 162.12 lb.

8 3 15!$8 force √(b)= sidewall 2 the line = 62.4x(30) = 124.8[1.20 = 162.12d 2− (−0.10)] 0 through 3 55. 3 5. The equation of (5/2, 0) and (1, 3/2) is 5.0 & ' 62.4x(15) dx = 62.4 x = 29, 952 lb end force = 0dx = 62.4(1 √ 5. force √ =2 62.4x(30) ! 5/2 (b)√sidewall (1, 0 √ ! 5/2√ !

6. y=− x+ . Using symmetry, 0 8 3 3 0 3/2 0 √ 5.

5. 3 √ # $ 3 5 !

3 F = 62.4x(2y) dx = 124.8 x x − dx 8 3 3 6.6 62.4x(15) dx end force = √ 6. line through 15 = − 0) x + (5/2,. Using !

3 .6.force 4. The equation of the (1, 3/2)symmetry, is y = 1 x − end Using √ 0dx = 3/2 √ 7. y =√ 62.4x(15) 62.4 =3 symmetry, 3 Z 5/2 Z 5/2 3 and1 4.

6.6The equation of the line 2 6.

3 3 through (1, 0) and (5/2, 3/2) is y = x 0 & √ '(5/2 √ 0) 3 5 !3 ! 5/2' % (5/2, √ 7.& √√ ! 5/2 & √ 3 7.

3 4. The 5 3equation 33 & √3 2 of line through (1 F = 62.4x(2y) dx = 124.8 x7. − x8.+F = 5/2dx 3 '7.

3 the √ 2 ! 5/2 !

7. & ' √ (5/2 − x8.+ of the dxThe = 124.8 x line − x √(1, 0) and 124.8 5/2 dx = 124.8 !dx through 3 6 =62.4x(2y) 5/2 3x − !31 5/2 3x x4.

1 1 3equation 96 6 ! 3 3 F = 8.

62.4x(2y) dx =1 124.8 8. 1 x 3 x 8.dx x 8.

F− = 62.4x(2y) dx = 124.8 x 5/21 x − d 9.

9. ! !# 3 5/2√= 1162.12 3& 3 62.4x(2y) 3 !dx5/ % √− (−0.10)] % &' 5/21 F =√ 1√ √ 5/2 √ √ ! 1 √ = !124.8[1.20 Z 5/2 lb.

5/2 9.

1 9. = 3 ' 5 3 dx2 =&124.8 & √ 10.

3 3 9.'

5 −3 23 x2 + 5& √3 x dx √ 1x3 62.4x(2y) & √ = 124.8 '(F− ! 5/2 5/2 9. − 3 x2 + 5 3 x dx = 124.8 =√124.8 + x ! 5/2√&3√ 1 ! 5/210. = 124.8 −3 2 x√ x3 33 − 6 124.8 12 3+ 3 310. 3 23 x2 − 93 x !dx5/2= &124.8 2 1 √ √ x' = = 5.

10.

124.8 3 6 10. 12 x9 − x dx = 124.8 x −1 x3 = 124.8 3 2 1 193 3 x 1 3 x − x d = 124.8 1 10.

3 3 9 6 1 ≈ 124.8(1.50 − 0.53) = 121.59 lb.

3 3 1 1 6. = 124.8[1.20 − (−0.1 = 124.8[1.20 − (−0.10)] = 162.12 lb. ≈ 124.8(1.50 − 0.53) = 121.59 lb.

1

5.

cos θ

≈ 124.8[1.20 − (−0.10)] 162.12force lb. = (b) =sidewall

!

8

3

= 124.8[1.20 − (−0.10)] = 162.12 lb.

7.

5.

6.

7.

8.

9.

10.

8.

9.

10.

5.

5.

6.

6.

7.

7.

8.

8.

9.

10.

9.

10.

5.

6.

7.

8.

9.

10.

= 124.8[1.20 − (−0.10)] = 162.12

30. From the figure in the text, we see that Also, when x = 3, sin θ = 0.1 and θ ≈ 0 ! 3 !

6.9. FLU W = mg tan θ dx = 550(9.8) 6.9. FLUID PRESSU 6.9. FLUID PRESSURE AND FORCE 0 0.1

Fro = 161700(− cos θ)]0 =30.

16170 30. From the figure Als 30. From the figure in the text, we see that Also, when x = Also, when x = 3, sin θ = 0.1 and θ ≈ 0 ! 3 6.9 383 Fluid !Pressure and For W = !

3 02 6.9. FLU W = =mg tan θ dxFLUID = 550(9.8) 1. (a) pressure 62.4(20) 1248 lb/ft ; 6.9. = PRESSU

6.9. FLUID PRESSURE AND FORCE

= 161 0 3 1 2 0.1 x − 6 and y2 = − x + 2. (b) pressure = 62.4(20) 30.

Fro; = 161700(− cos = θ)]1248 =lb/ft 16170 0 the 2 2 30.

From figure 6.9 382 CHAPTER 6.

2F Als 6.9=Also, Fluid (c) pressure = 62.4(20) 1248 lb/ft when xPr =;

Z 4 Z 4 1. ! (a) 1 3 3(a) 3 1. pressure = 6.9 Fluid Pressure and For 2. (a) pressureoil = 55 3 lb/ft · 96 ft = 528 F = 62.4x − x + 2 − x − 6 dx = 62.4 x(−2x + 8) dx W = 6. The equations of the lines are y1 = x −(b) 6 and y2 (b) = 2 2 3 pressure 02= 2 2 2 = pressurewater = 62.4 lb/ft 85 ft(c) = 1. (b) (a) pressure = 62.4(20) 1248· lb/ft ;

4

Z 4 = 161 (c)$% pressure = "# forceoil = $ # ! 4 2 (c) 5280 · 125 · 350 = 231, 00 64 32 2 2. (a) pressure = 62.4(20) 1 3 = 1248 lb/ft ; = 62.4 − =F665.60 lb (b) = 62.4 (−2x2 + 8x) dx = 62.4 − x3 + 4x2 pressure 6.9 = 62.4x + 2 =− x2. −(a) dx =(b) 6 oil (d) − forcexwater 53046.9 ·6Fluid 350 = 232, 2F 3 3 3 (c) pressure = 62.4(20) = 1248 lb/ftPr ; 2 2· 125 2 2 2

6. The equations of the lines are y1 =

7. F =

Z

4

√ 50x(2 x) dx = 100

0

Z

4

x3/2 dx = 40x5/2

0

i4 0

(b) pressurewa (c)2 !3.4 (a) pressure = (62.4)(8)# 1. (a) 499.2 lb/ft 3(a) 2pressure 1. = (c) force 5 oil == (a) pressure lb/ft · 96 2 (d) ! 8 − (b) = 62.4 2. (−2x + 8x)oil dx==5562.4 xft3 = + 528 4x (d)33force (b) pressure water== y pressure 2 (b) lb/ft · 85 dx ft(c) = sidewall force==62.4 62.4x(30) =

= 1, 280 lb

water

7. F =

!

0

4

50x(2 x) dx = 100

0

8.

3. (a) (c) pressure =

(a) pressure = (c) forceoil = = 231, 00 ! 8· 1253. · 350 2. & (b) (a) ! 5280 4 4 √ (d) force 2. (a) pressure 62.4x(15) dx = 62 end force = oil 3/2 5/2 = 5304 · 125 · 350 = 232, (b) sidewall(b) fo water 0

x 0 dx = 40x

=

(b) pressure0wa (c)2

lb/ft 4 3. (a) pressure = (62.4)(8) =(c)499.2 force end force oil = 5 (d)= ! 8through x The equation of the line 4.

(1, 0) (d) forcewater = (b) sidewall force = 62.4x(30)4. = The 3.dx (a) ! 5/2 04. o (a) equation pressure = ! 8 3. The 2 F = 62.4x(2y) dx =(b) 1 Solving x = y and y = −x + 2 simultaneously, we find that the graphs8. Solving x = y2 andy yend 62.4x(15) dx = we 62 force+=2 simultaneously, = −x (b) sidewall fo 1 intersect at (4, −2) and (1, 1). To compute the force we divide the plate intersect at (4, −2) and (1, 1). To0 compute F = the forc ! 5/2 & √ 3 force with1)the line x = 1.

For the upper end 2 pa= into two parts with the line x = 1. For the upper part we use symmetry. into two parts (1, x 0) − 124.8 4. The equation of=the line through (1, 1 ! ! 3

F =

Z

1

Z

√

62.4x(2 x) dx +

0

= 124.8

Z

4

1

1

x3/2 dx + 62.4

0

√ 62.4x[(−x + 2) − (− x)] dx

Z

1

4

(4, –2)

(x3/2 − x2 + 2x) dx

= 124.8

Z

0

= 124.8x2

2

2x dx + 124.8

Z

2

19.

4

(−x2 + 4x) dx

4 1 3 2 + 124.8 − x 2x = 499.2 + 665.6 = 1, 164.8 lb 0 3 2

2

0

8.

5.

9.

1

! 45. & √ 6.(xF3/= x3/2 dx!5. +5/2 62.4 1

3 1

10.

6.

6.

10.

7.

7.

8. 5) (0, 9. y

8.

10.

(4, 3) x

y (2, 2) x

1 dx = 1= 62.4x(2y)

F =

!

2 # = $ 124.8%6. #x − 1 3 7.2 2 5/2 7.1 = 124.8 x + 62.4 8. x5= 5= 124.8[1.20 5 − (−0.10)] 0 # 8. $ = 112 16 9.

5. 49 = = 49.92 + 62.4 9. − 155. 15 10.

7.

(4, 0)

=

4 4. The √ 4. + The o =!

124.8[1.20 − equation (−0.10)] 5/2dx 62.4x(2 x) 62.4x[(

= 124.8

6.

16.

10. The equation of the line through (4, 0) and (2, 2) is y = −x + 4. Using symmetry, 17.

Z 2 Z 4 18.

F = 62.4x(4) dx + 62.4x[2(−x + 4)] dx 2

0

5.

1

4 2 2 5/2 1 3 5/2 2 = 124.8 x + 62.4 x − x +x 5 5 3 1 0

112 16 = 49.92 + 62.4 − = 49.92 + 399.36 = 499.28 lb 9.

15 15 1 10.

9. The equation of the line through (0, 5) and (4, 3) is y = − x + 5. Using 2 11. symmetry,

Z 4 Z 4 12.

1 2 1 dx = 124.8 F = − x + 5x dx 13.

62.4x 2 − x + 5 2 2 0 0

4 14.

1 5 = 124.8 − x3 + x2 = 3660.8 lb 15.

6 2 0

0

x

1

F =

9.

10.

6.

7.

8.

9.

10.

384

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

11. The equation of the line through (8, 0) and (4, 12) is y = −3x + 24.

F =

Z

4

62.4x(12) dx +

0

= 374.4

Z

4

= 374.4x2 √

2x dx + 62.4

4 0

√

62.4x(−3x + 24) dx

Z

8

(−3x + 24x) dx

+ 62.4(−x3 + 12x2 )

8 4

25 − x2 . Using symmetry, Z 5 Z p

F = 60x 2 25 − x2 dx = 120 = −40(25 − x2 )3/2

i5 0

= 124.8

Z

5

5

x(25 − x )

2 1/2

0

4

−4

dx

p x 16 − x2 dx + 624

4

−4

y

5 x

= −40(0 − 125) = 5, 000 lb

Z

y

= 5, 990.4 + 7, 987.2 = 13, 977.6 lb

16 − x2 . Using symmetry, Z 4

p F = 62.4(x + 10) 2 16 − x2 dx −4

(4, 12) (8, 0) x

2

4

0

13. y =

8

4

0

12. y =

Z

–10 y p 2 16 − x2 dx.

4

x

The first integral has an odd integrand and is thus 0. The second integral represents the area of the circle and is thus π(4)2 . Therefore F = 624(16π) = 9984π lb. r 9 –10 14. y = 9 − x2 . Using symmetry, –2 4 ! r Z 2 y 9 2 F = ρ(x + 10) 2 9 − x dx 2 4 −2 x Z 2 r Z 2 r 9 2 9 2 = 2ρ x 9 − x dx + 10ρ 2 9 − x dx.

4 4 −2 −2 The first integral has an odd integrand and is thus 0. The second integral represents the area of the ellipse and is thus π(2)(3). Therefor F = 10ρ(6π) = 60πρ lb.

Z 5 y 15. F = Ftop + Fbottom + 4Fside = 62.4(3)(4) + 62.4(5)(4) + 4 62.4x(2) dx 3 3

2 5 = 748.8 + 1248 + 249.6x 3 = 1996.8 + 3993.6 = 5990.4 lb 5 x

16. Fbottom − Ftop = 62.4(5)(4) − 62.4(3)(4) = 499.2 lb. Since the volume of the block is 8 ft3 , the weight of the water displaced is 62.4(8) = 499.2 lb (illustrating Archimedes’ principle).

385

6.10. CENTERS OF MASS AND CENTROIDS √ √ 17. The length of the bottom of the pool√is 202 + 52 = 5 17.√Using similar √ triangles, h/x = 5/5 17 and h = x/ 17. Then d = 4 + x/ 17. Now

20 4 d

4

Z 5√17 h 5 x∗k x x √ √ F = lim 62.4 4 + 62.4 4 + (15)∆xk = 15 dx kP k→0 17 17 0 k=1 384 CHAPTER 6. APPLICATIONS O

5√17 Z 5√17 1 1 384 √ √ CHAPTER 6. APPLICATIONS OF THE INTEG 17. The length of the bottom of the pool√is 202 + 52 = 5 17.√Using simil 4 + √ x dx = 936 4x + √ x2 = 936 √ 17 2 17 triangles, h/x = 5/5 17 and h = x/ 17.

Then d = 4 + x/ 17. Now √ √ 0 0 2 + 52 = 5 17. Using similar 20 17. The length of the pool√is 20CHAPTER √ !

384 of the bottom 6. APPLICATIONS OF THE IN √ √ √ " 17. Then # d = 4 + x/ $ 17. # n h = x/ √ 5 17 triangles, h/x = 5/5 17 and Now " !

65 17 x∗k x 4 d √ F = lim 62.4 4 + √ 62.4 4 + √ (15)∆x 2= √ 15 d = 936 = 30, 420 17 lb. √ 202 + k !P 17. The of!→0 the bottom 5" = 50 17.√# Using similar517 h # of the pool $ 17 n length" 2 √is5 17 ∗√ ! xk=1 x n X

18.

19.

k 17 and h = x/ 17. Then d = 4 + x/ 17. Now√ =$ √ 5/5 √ F = lim triangles, 62.4 h/x 4+ 62.4" 4 + √ 15#% dx5 17 5 17 "(15)∆xk =# !P !→0 1 1 17" 17 √ 2 # k=1 √ # x 0dx = $ 5 n 4+ = 936 936 x 5 174x√+ " √ ∗√ 45° ! √ x "(15)∆xd = #%62.4 $ F5d =17 17√x 015 dx 0 # 4 + √ k 17 The slant height of the dam is 40 2 ft. Since x = d 2, we have ="lim x/ 2.

5 17 24 + 62.4 k 1 & ' 1 2 !P !→0 17 x 4x + 17 √ 65 4 + k=1 = 936 is 100∆x x √dx 936 √ √ 0x 17 = 40 Then, using the fact that the area of the rectangular element √ k √,17 2" lb.

17 0 # 30, 420 17 0 #%5 17 =$ 936 = 5 17 " & √ ' 1 1 2 4 +√√ x dx = 936 936 45° 4x + √ x2 ∗ Z 40√2 = 936 65= 17 n 17lb.

2 17 √ √ 0= 30, 420 17 0 √ X & ' 6240 xk 20 40 √ 2 slant height of the √ dam is 40 2 ft. Since x = d 2, we have d = x/ 65 17 √ x dx18.=The F = lim 62.4 √ 100∆xk = Then, that17the is 100∆xk 936 using the = fact 30, lb. area of √ 420 √the rectangular element √ kP k→0 2 2 45 0 18. The2slant height of the dam is 40 2 ft. Since x = d 2, we have d =√x/ 2. k=1 " # $ n √ √ √ √ 40 2 ∗ Then, using the fact that the area of! the rectangular element is 100∆x6240 40 x k, x 40 2 k 18. The slant heightFof=the lim dam is 4062.4 2 ft. √ Since x = d 2,=we have d = 2. √ x/x dx 100∆x √ k 3120 √ !P !→0 # area of the$2rectangular Then, using! the fact" that the element is 100∆x 2 k, 40 n 0 40 2 = 4, 992, 000 2 lb. = √ x2 x∗k k=1 √ 6240 20 √ x dx F = lim 62.4 n√ 100∆x %40 ∗ 2#k = " $√ 2 40 2 !

0 √ !P !→0 3120 2 2 xk 2 6240 0 20 Fk=1 = √ lim= √ 62.4 100∆x x √ = 4, 992, 000 2 lb.√ x dx k = 2 2 Z 40 0 n 2 k=12 0 √ i40 2 %40!P !→0 X √ √ √ 3120 √ $ 40 %40992, n= 4, = 2√ x 3120 (40 2 000 √2 lb. √ ! √ √ 62.4x∗k (100 2∆xk ) = F = lim 2x 6240 2x dx = 3120 19.

2 lim =0 √ x262.4x∗k (100 = 4, 992, 000k ) = 2 lb. 6240 2x dx = 3120 2x2 F = 2∆x kP k→0 0 0 !P !→0 0 2 $ n 0 0 k=1 40 (40 ! k=1 √ √ √ $ 402x dx√= 3120 2x2 √ √k ) √ √ n (40 19. F = lim 62.4x∗k!

(100 2∆x = 6240 45° 6240 2x dx = 3120 02x2 = 4, 992, 000∗ (100 2 lb 2∆x !P19. !→0F = lim 62.4x = 4, 992, 000 2 lb 0 k) = k k=1 !P !→0 0 0 √ k=1 √ = 4, 992, 000 2 lb x

6.10 1. x =

√

6.10

Centers of Mass and Centroids 1.

2(4) + 5(−2) 2 =− 2+5 7

6(−1/2) + 1(−3) + 3(8) 18 9 = = 2. x = 6+1+3 10 5 10(−5) + 5(2) + 8(6) + 7(−3) 13 3. x = =− 10 + 5 + 8 + 7 30 4. x =

√

= 4, 992, 000 2 lb 6.10 Centers of Mass and Centroids Centers of Mass and Centroids 6.10 1. Centers of Mass and Centroids 1.

2.

2.

2.

3.

3.

3.

4.

4.

4.

5.

5.

5.

6.

6.

7.

6.

7.

8.

7.

8.

9.

8.

9.

2810.

2(9) + (3/2)(−4) + (7/2)(−6) + (1/2)(−10) −14 9.

10. = =− 2 + 3/2 + 7/2 + 1/2 15/2 1511.

10.

11.

11.

5. x =

12.

10(−5) + 15(5) 25 = = 1 m (to the right of the12.center) 10 + 15 25

12.

x

x

386

CHAPTER 6. APPLICATIONS OF THE INTEGRAL y

2(−1) + 2(1) 6. The center of mass of m1 and m2 is x12 = = 0. Thus, we 2+2 consider masses m1 and m2 to be concentrated at the midpoint of the base of the triangle. Taking this point to be the √ origin, with √ the positive 4(0) + 1(2 3) 2 3 direction up through m3 , we have y = = .

4+1 5

7. m =

Z

5

0

5 (2x + 1) dx = (x2 + x) 0 = 30;

M0 =

Z

2

(−x2 + 2x) dx =

m=

Z

Z

0

5 0

x

=

=

575 6

4/3 =1 4/3

1

1/3

x=

3/7 4 = 3/4 7

1

4

(−x2 + 1) dx =

Z

3

Z

4

1 − x2 + 3x 2

|x − 3| dx = −(x − 3) dx + (x − 3) dx = 0 3

9 9 = + −4 + =5 2 2 Z 4 Z 3 Z 4 M0 = x|x − 3| dx = −(x2 − 3x) dx + (x2 − 3x) dx 0

x

4 kg

2 3 1 2 x + x 3 2

1 2 1 = − x3 + x 3 3 0 0 1

Z 1 Z 1 1 1 4 1 2 2 3 M0 = = ; x(−x + 1) dx = (−x + x) dx = − x + x 4 2 4 0 0 0 m=

m2

1 3 4/3 3 9. m = x dx = x = 4 4 0 0 1 Z 1 Z 1 3 3 M0 = xx1/3 dx = x4/3 dx = x7/3 = ; 7 7 0 0 0

11.

x(2x + 1) dx =

m1

2 4 1 = − x3 + x2 3 3 0 0 2

Z 2 Z 2 4 1 4 2 3 2 3 2 M0 = = ; x(−x + 2x) dx = (−x + 2x ) dx = − x + x 4 3 3 0 0 0 m=

Z

10.

5

0

575/6 115 x= = 30 36

8.

Z

m3

0

3

x =

3 0

+

1/4 3 = 2/3 8 1 2 x − 3x 2

4 3

3 4

1 3 1 3 3 2 27 64 27 19 = − x3 + x2 + x − x = −9 + + − 24 − 9 − = 3 2 3 2 2 3 2 3 0 3 19/3 19 x= = 5 15

387

6.10. CENTERS OF MASS AND CENTROIDS 12.

Z

m=

3

0

(1 + |x − 1|) dx =

Z

0

1

[1 − (x − 1)] dx +

Z

3

1

[1 + (x − 1)] dx

1 3

1 2 1 2 3 9 1 11 = (2 − x) dx + x dx = 2x − x + x = + − = 2 2 2 2 2 2 0 1 0 1 Z 3 Z 1 Z 3 x[1 + (x − 1)] dx x[1 − (x − 1)] dx + x(1 + |x − 1|) dx = M0 = Z

Z

1

0

Z

=

1

0

(2x − x2 ) dx +

28/3 56 x= = 11/2 33 13.

m=

Z

Z

2

Z

0

3

+

1 (2 − x) dx = x3 3

1

1

Z

2

1

1

x2 dx =

1

1 x2 − x3 3

1 3 x 3

0

3

1

=

1 28 2 + 9− = 3 3 3

1 2x − x2 2

2

ρ(x) dx = x dx + + 0 1 0 1

3 5 1 = = + 2− 3 2 6 1 2 Z 2 Z 1 Z 2 1 1 M0 = xρ(x) dx = x3 dx + (2x − x2 ) dx = x4 + x2 − x3 4 3 0 0 1 0 1

1 4 2 11 = + − = 4 3 3 12 11/12 11 x= = 5/6 10 0

14.

3

m=

Z

3

ρ(x) dx =

0

M0 =

Z

Z

2

2

x dx +

0

3

xρ(x) dx =

0

Z

Z

2

2

x2 dx +

0

3

Z

1 2 dx = x2 2 3

2x dx =

2

2

3

+ 2x]2 = 2 + 2 = 4

0

1 3 x 3

2 0

+ x2

3 2

=

8 23 +5= ; 3 3

x=

23/3 23 = 4 12

1 1 and ρ(x) = x2 .

2 2 10 Z 10 1 3 1 4 M0 = x dx = x = 1250 2 8 0 0

15. Since ρ(x) = kx2 and 12.5 = ρ(5) = 25k, k = m=

Z

0

10

1 2 1 x dx = x3 2 6

10 0

=

500 ; 3

1250 x= = 7.5 ft (from the left end) 500/3

16. Place the origin at the right end of the rod with the positive direction to the left. Then 3 Z 3 Z 3 1 9 4 ρ(x) = kx and 6 = ρ(x) dx = kx dx = kx2 = k. Solving for k we obtain k = .

2 2 3 0 0 0 4 Then ρ(x) = x and ρ(3/2) = 2 kg/m.

3 17. m = 3 + 4 = 7;

x=

3(−2) + 4(1) 2 =− ; 7 7

y=

3(3) + 4(2) 17 = 7 7

388

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

18. m = 1 + 3 + 2 = 6;

x=

19. m = 4 + 8 + 10 = 22;

1(−4) + 3(2) + 2(5) = 2; 6

x=

y=

17 4(1) + 8(−5) + 10(7) = ; 22 11

1(1) + 3(2) + 2(−2) 1 = 6 2 y=

20 4(1) + 8(2) + 10(−6) =− 22 11

1(9) + (1/2)(−4) + 4(3/2) + (5/2)(−2) =1 8 1(3) + (1/2)(−6) + 4(−1) + (5/2)(10) 21 y= = 8 8 Z 2 2 21. A = (2x + 4) dx = (x2 + 4x) 0 = 12 20. m = 1 + 1/2 + 4 + 5/2 = 8;

0

My =

Z

2

x(2x + 4) dx =

0

0

2

(2x2 + 4x) dx =

2 3 x + 2x2 3

Z 1 2 (2x + 4) dx = (4x2 + 16x + 16) dx 2 0 0 2

1 3 112 x + 2x2 + 4x = =2 3 3 0 40/3 10 112/3 28 x= = ; y= = 12 9 12 9

1 Mx = 2

Z

Z

x=

2

2 0

=

8

40 3

2

4

2

3

1 2 15 1 22. A = (x + 1) dx = x +x = − − =8 2 2 2 −1 −1 3

Z 3 Z 3 1 3 1 2 x + x My = x(x + 1) dx = (x2 + x) dx = 3 2 −1 −1 −1 27 1 40 = − = 2 6 3 3 Z 1 3 32 32 1 Mx = = −0= (x + 1)2 dx = (x + 1)3 2 −1 6 3 3 −1 Z

3

40/3 5 32/3 4 = ; y= = 8 3 8 3 1 Z 1 1 1 23. A = x2 dx = x3 = 3 3 0 0 1 Z 1 Z 2 1 4 1 3 2 My = x dx = (2x + 4x) dx = x = 4 4 0 0 0 1 Z 1 1 1 5 1 1/4 3 Mx = x4 dx = x = ; x= = ; 2 0 10 10 1/3 4 0

4

3

x=

1

1

1/10 3 y= = 1/3 10

389

6.10. CENTERS OF MASS AND CENTROIDS 24.

A=

Z

2

(x + 2) dx = 2

−1

Z

1 3 x + 2x 3

2

−1 2

7 20 − − =9 = 3 3

6

1 2 27 9 (x + 2)2 =9− = 4 4 4 −1 −1 Z 2 Z 2 1 1 (x2 + 2)2 dx = (x4 + 4x2 + 4) dx Mx = 2 −1 2 −1

2

83 153 1 1 5 4 3 1 376 x + x + 4x − − = = = 2 5 3 2 15 15 10 −1 My =

2

x(x2 + 2) dx =

27/4 3 153/10 17 = ; y= = 9 4 9 10 3 Z 3 1 81 25. A = x3 dx = x4 = 4 4 0 0 3 3 Z 3 Z 243 2187 1 1 3 6 1 7 = My = x4 dx = x5 = ; Mx = x dx = x 5 5 2 14 14 0 0 0 0 243/5 12 2187/14 54 x= = ; y= = 81/4 5 81/4 7 x=

-1

1

2

20

10

3

26.

x = y 1/3 ; Z

A=

Z

8

y 1/3 dy =

0

8

3 4/3 y 4

384 = Mx = 7 0 0 8 Z 8 1 3 5/3 48 My = y 2/3 dy = y = 2 0 10 5 0 48/5 4 384/7 32 x= = ; y= = 12 5 12 7

27.

A=

Z

8

4

1

3 y 4/3 dy = y 7/3 7

x1/2 dx =

2 3/2 x 3

4 1

=

8

= 12

0

4

2

16 2 14 − = 3 3 3

4 2 5/2 64 2 62 3/2 My = x dx = x = − = 5 5 5 5 1 1 4 Z 4 1 1 1 15 62/5 93 Mx = x dx = x2 = 4 − = ; x= = ; 2 1 4 4 4 14/3 35 1 Z

8

2

4

y=

15/4 45 = 14/3 56

390

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

1

1 2 4 2 y − y3 = = − − 3 3 3 3 −1 −1

1 Z 1 Z 1 1 2 1 4 1 1 (y − y 3 ) dy = y(1 − y 2 ) dy = Mx = y − y = − =0 2 4 4 4 −1 −1 −1

1 Z 1 Z 1 1 1 1 1 [12 − (y 2 )2 ] dy = (1 − y 4 ) dy = y − y5 My = 2 −1 2 −1 2 5 −1

4/5 0 1 4 4 4 3 x= = − − = ; = ; y= =0 2 5 5 5 4/3 5 4/3

2 Z 2 7 9 1 2 1 3 10 2 x + 2x − x − − = 29. A = [(x + 2) − x ] dx = = 2 3 3 6 2 −1 −1 Z 2 Z 2 My = x(x + 2 − x2 ) dx = (x2 + 2x − x3 ) dx 28.

A=

Z

1

−1

(1 − y 2 ) dy =

−1

2 1 3 1 4 2 = = x +x − x 3 4 −1 Z 1 2 Mx = [(x + 2)2 − x4 ] dx = 2 −1

1

1

-1

4

-2

-1

9/4 1 36/5 8 = ; y= = 9/2 2 9/2 5 1

Z 1 1 1 2 3/2 1 3 1/2 2 x − x = 30. A = (x − x ) dx = 3 3 3 0 0

1 Z 1 Z 1 2 5/2 1 4 3 My = x(x1/2 − x2 ) dx = (x3/2 − x3 ) dx = x − x = 5 4 20 0 0 0 1

Z Z 1 1 1 3 1 1 1/2 2 1 1 2 5 2 2 4 x − x = Mx = [(x ) − (x ) ] dx = (x − x ) dx = 2 0 2 0 2 2 5 20 0 9 9 3/20 3/20 = ; y= = x= 1/3 20 1/3 20

1 Z 1 1 3 4/3 1 4 1 1/3 3 31. A = (x − x ) dx = x − x = 4 4 2 0 0

1 Z 1 Z 1 3 7/3 1 5 8 My = x(x1/3 − x3 ) dx = (x4/3 − x4 ) dx = x − x = 7 5 35 0 0 0

1 Z Z 1 1 1 1/3 2 1 1 3 1 7 8 3 2 2/3 6 5/3 Mx = [(x ) − (x ) ] dx = (x − x ) dx = x − x = 2 0 2 0 2 5 7 35 0 8/35 16 8/35 16 x= = ; y= = 1/2 35 1/2 35 x=

1

2

8 5 9 − = 3 12 4

2

1 224 1 1 1 5

8 36 3 (x + 2) − x = − = 2 3 5 2 15 15 5 −1

1

1

391

6.10. CENTERS OF MASS AND CENTROIDS

0 4 1 3 16 32. A = (4 − x ) dx = 4x − x = 3 3 −2 −2

0 Z 0 Z 0 1 4 2 3 2 My = x(4 − x ) dx = (4x − x ) dx = 2x − x = −4 4 -2 −2 −2 −2

0 Z 0 Z 0 1 1 8 1 1 128 (4 − x2 )2 dx = (16 − 8x2 + x4 ) dx = 16x − x3 + x5 Mx = = 2 −2 2 −2 2 3 5 15 −2 Z

x=

33.

0

2

−4 3 =− ; 16/3 4

y = x−3 ;

A=

128/15 8 = 16/3 5

y=

Z

3

1

x−3 dx = −

34.

A=

Z

3

2

1

=−

= My =

Z

[(−4x + 9) − (x − 2x + 1)] dx = 2

−4

1 x−2 dx = − x

3

1 1 − 18 2

=

1 2 My = =− −1 = 3 3 1 1 3

Z 3 1 1 1 121 1 Mx = = − − = x−6 dx = − 2 1 10x5 1 2430 10 1215 2/3 3 121/1215 121 x= = ; y= = 4/9 2 4/9 540 Z

3

1 2x2

−4

2

−4

1

2

(8 − 2x − x2 ) dx

20

28 80 − − = 36 3 3 −4 Z 2 2 x(8 − 2x − x ) dx = (8x − 2x2 − x3 ) dx

1 8x − x2 − x3 3 2

2

Z

4 9

=

−4

2

20 128 1 2 = − = −36 = 4x2 − x3 − x4 -4 -2 3 4 3 3 −4 Z 2 Z 2 1 1 Mx = [(−4x + 9)2 − (x2 − 2x + 1)2 ] dx = [16x2 − 72x + 81 − (x − 1)4 ] dx 2 −4 2 −4

2

1 907 1 1849 1692 1 16 3 2 5

x − 36x + 81x − (x − 1) = − − = = 2 3 5 2 15 3 5 −4

1692/5 47 = 36 5

2

Z 2 1 3 14 4 2 35. A = (y − 1 + 2) dy = y +y = − − =6 3 3 3 −1 −1

2 Z 2 Z 2 1 4 1 2 Mx = y(y 2 + 1) dy = (y 3 + y) dy = y + y 4 2 −1 −1 −1 3 21 =6− = 4 4 x=

−36 = −1; 36

2

y=

4 2

-3

3 -2

392

CHAPTER 6. APPLICATIONS OF THE INTEGRAL My =

1 2

1 = 2

Z

2

−1

[(y 2 − 1)2 − (−2)2 ] dy =

1 5 2 3 y − y − 3y 5 3

2

−1

1 = 2

1 2

Z

2

−1

(y 4 − 2y 2 − 3) dy

74 52 − − 15 15

=−

21 5

−21/5 7 21/4 7 x= =− ; y= = 6 10 6 8

4 Z 4 1 3 40 36. A = (x2 − 4x + 6) dx = x − 2x2 + 6x = 3 3 0 0 Z 4 Z 4 My = x(x2 − 4x + 6) dx = (x3 − 4x2 + 6x) dx 0

= Mx = = = x= 37.

A= My =

Mx = = x=

4 1 4 4 3 80 x − x + 3x2 = 4 3 3 0 Z 4 Z 1 1 4 2 2 (x − 4x + 6) dx = [(x − 2)2 + 2]2 dx 2 0 2 0

4 Z

1 4 4 1 1 5 3 4 2 (x − 2) + (x − 2) + 4x

[(x − 2) + 4(x − 2) + 4] dx = 2 0 2 5 3 0

1 496 256 376 − − = 2 15 15 15 80/3 376/15 47 = 2; y = = 40/3 40/3 25

Z

1

−1 1

Z

[(4 − 4x ) − (1 − x )] dx = 2

2

x(3 − 3x2 ) dx =

Z

1

−1

Z

1

−1

4

1 (3 − 3x2 ) dx = (3x − x3 ) −1 = 4

(3x − 3x3 ) dx

4

-2

2

1 3 3 3 2 3 4 = − =0 x − x 2 4 4 4 −1 Z 1 Z 1 1 1 [(4 − 4x2 )2 − (1 − x2 )2 ] dx = [16(1 − x2 )2 − (1 − x2 )2 ] dx 2 −1 2 −1

1

Z 2 3 1 5 15 8 8 15 1 15 2 4 x− x + x = − − =8 (1 − 2x + x ) dx = 2 −1 2 3 5 2 15 15 −1 0 8 = 0; y = = 2 4 4

38. x = 1 − y 2 ; x = −(1 + y) Z 2 Z 2 A= [(1 − y ) + (1 + y)] dy = −1

=

3

0

−1

=

6

1 1 2y + y 2 − y 3 2 3

2

−1

2

2

−1

(2 + y − y 2 ) dy

10 7 9 = − − = 3 6 2

-2

393

6.10. CENTERS OF MASS AND CENTROIDS

2 1 1 8 5 9 y2 + y3 − y4 = − = 3 4 3 12 4 −1 −1 −1

2 Z 2 Z 2 1 1 1 5 1 [(1 − y 2 )2 − (1 + y)2 ] dy = (y 4 − 3y 2 − 2y) dy = My = y − y3 − y2 2 −1 2 −1 2 5 −1

1 28 1 27 = − − − =− 2 5 5 10 −27/10 3 9/4 1 x= =− ; y= = 9/2 5 9/2 2

Mx =

Z

2

y(2 + y − y 2 ) dy =

Z

2

(2y + y 2 − y 3 ) dy =

39. By symmetry, x = 0.

Z π/2 Z A=2 (1 + cos x − 1) dx = 2 0

2

π/2

π/2

cos x dx = 2 sin x]0

=2

Z –π / 2 1 π/2 [(1 + cos x) − 1 ] dx = (cos2 x + 2 cos x) dx 2 −π/2 −π/2

π/2 Z π/2 1 1 1 1 1 1 cos 2x + + 2 cos x dx = sin 2x + x + 2 sin x = 2 −π/2 2 2 2 4 2 −π/2

i h 1 π π π+8 = +2 − − −2 = 2 4 4 4 π+8 (π + 8)/4 = y= 2 8

1 Mx = 2

Z

0

π/2

2

40. By symmetry, x = π/2.

Z π Z A= (4 sin x + sin x) dx = 0

π/2

2

0

4

π

π

5 sin x dx = −5 cos x]0

= −5(−1 − 1) = 10 Z Z 15 π 2 1 π 2 2 (16 sin x − sin x) dx = sin x dx Mx = 2 0 2 0

π Z π 15π 1 1 15 1 1 15 = − cos 2x dx = x − sin 2x = 2 0 2 2 2 2 4 4 0 15π/4 3π y= = 10 8

2

!

41. (a) The circumference of a circle with radius y is 2πy. Thus, V = 2πyA. (b) By the same reasoning, V = 2πxA when the region R is revolved around the y-axis.

394

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

2 1 3 8 (x + 1 − 1) dx = x 42. A = = 3 3 0 0 Z 2 Z 2 1 1 [(x2 + 1)2 − 12 ] dx = (x4 + 2x2 ) dx Mx = 2 0 2 0

2 1 1 5 2 3 88/15 11 88 y= x + x ; = = = 2 5 3 15 8/3 5 0

11 8 176π By the theorem of Pappus, V = 2π = . To verify this 5 3 15 result we compute the volume using the washer method.

Z

2

5

2

V =π

Z

0

2

[(x + 1) − 1 ] dx = π 2

2

2

Z

0

2

(x + 2x) dx = π 4

1 5 2 3 x + x 5 3

2

2

=

0

176π .

15

43. We identify A = πa2 . The centroid of the region R is b units from L, so V = 2πb(πa2 ) = 2π 2 a2 b.

44. Since the graph is symmetric around x = 3, we would expect the center of mass to occur at the center of the rod.

6 Z 6 1 3 2 2 = 42 m= (6x − x + 1) dx = 3x − x + x 3 0 0

6 Z 6 1 4 1 2 2 3 3 M0 = (6x − x + x) dx = 2x − x + x = 126 4 2 0 0 126 =3 x= 42

10

5

-3

3

6

9

45. Thinking geometrically, the centroid of a triangle would appear to be the intersection of its three medians (a median is a line segment from one of the triangle’s vertices to the midpoint of the opposing side). Doing some research on the centroid of a triangle shows this to be true, and in fact this intersection is the mean of the coordinates of the triangle’s vertices.

46. Decomposing the region R into three 1 × 1 squares whose centers are (1/2, 3/2), (3/2, 3/2), and (3/2, 1/2), we get the centroid of R = ([1/2 + 3/2 + 3/2]/3, [3/2 + 3/2 + 1/2]/3) = ([7/2]/3, [7/2]/3) = (7/6, 7/6).

Chapter 6 in Review A. True/False 1. False. x-axis.

Z

b

f (x) dx may be positive even though a portion of the graph of f lies below the

a

2. False. On [0, 3] a portion of the graph of f (x) = x − 1 lies below the x-axis.

3. True

395

CHAPTER 6 IN REVIEW 4. True 5. True 6. True 7. True 8. True 9. True

10. False. Liquid pressure depends on the density and depth of the liquid, not the area covered.

Z t2 |v(t)| dt.

11. False. The distance moved is given by t1

12. False. The formula for s(t) depends only on acceleration, initial velocity, and initial position.

The mass and shape of the object are immaterial.

B. Fill in the Blanks 1. Newton-meter of joule 2. 400 mi-lb or 2, 112, 000 ft-lb 3. (100 cos 60◦ )(50) = 2, 500 ft-lb 4. 80 = k(1/2), k = 160, F = 160x. When F = 100, x = 5/8 = 0.625 and the spring will measure 1.625 m.

5. 6 6. 62.4 7. smooth 8. vimpact = −gT and vave =

1 − gT 2 2

f (a) x. Thus, the integral is a

Z

1 T −0

Z

T

(−gt) dt =

0

1 T

=

−gT vimpact . Thus, vave = .

2 2

C. Exercises 1. − 2.

Z

Z

a

f (x) dx

0

b

a

f (x) dx −

Z

b

c

f (x) dx +

Z

d

3. The line in Figure 6.R.3 is y = 4.

Z

a

b

[c − f (x)] dx

f (x) dx

c

0

a

f (a) f (x) − x dx. a

396

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

5. − 6.

Z

Z

b

2f (x) dx +

a

b

7.

c

2f (x) dx

b

g(x) dx +

a

Z

Z

Z

c

f (x) dx

b

c

d

[f (y) − a] dy CHAPTER 6 IN REVIEW Z b Z c Z d ! c dx 8. [f (x) − g(x)] dx + [g(x) − f (x)] dx!+b [f (x) − g(x)] a b c 5. − 2f (x) dx + 2f (x) dx a b 2b

Z 2b Z 0 0 ! 1b ! c 1 x x a2 a2 b− dx = − x2 + bx − x2 dx + + 2b2 − b2 = + b2 9. A = − = 6. g(x) dx + f (x) dx 2 2 4 4 4 4 0 a a 0 b

[a − f (y)] dy +

Z

c

a

10.

11.

12.

13.

14.

b

2

√

2 y dy = y 3/2 3

b2

!

b

! d 2 37. 3 [a − f (y)] dy + [f (y) − a] dy A= = (b − a ) 3 b c a2 a2 ! ! c ! d

R b 2 R2 1 8.2 2dx + [f (x) − g(x)] [g(x) − f (x)] dx + [f (x) − g(x)] dx [g(x)] − [f (x)] dx x[g(x) − f (x)] dx b c x = R0 2 ; y = 2 0 R 2a [g(x) − f (x)] dx [g(x) − $0 % &$2b ! f0 (x)] dx ! 2b " 0 0 x 1 2 x# 1 2 a2 9.

A = − dx = − dx + b − x + bx − x = + Z 2 2 4 4 4

a 2 0 a 0 2 2 V =π [g(x)] − [f (x)] dx $b2 ! b2 0 2 2 3/2 √ = (b3 − a3 ) 10. A = y dy = y Z 2 3 3 a2 a2 V = 2π x[g(x) − f (x)] dx ) '2 ' ( 1 2 2 2 0 x[g(x) − f (x)] dx dx 2 0 [g(x)] − [f (x)] 0 11. x = ' 2 ; y= '2 Z 2 [g(x) − f (x)] dx [g(x) − f (x)] dx

0 0 V =π [g(x) + 1]2 − [f (x) + 1]2 dx ! 2 ( ) 0 12. V = π [g(x)]2 − [f (x)]2 dx

Z

c

0

13. V = 2π 14. V = π

!

!

0

2

0

2

x[g(x) − f (x)] dx g(x) + 1

(

) [g(x) + 1]2 − [f (x) + 1]2 dx

TODO figure ! 2 15. V = 2π (2 − x)[g(x) − f (x)] dx

–1

0

TODO figure ! 2 16. V = [g(x) − f (x)]2 dx 0

17. Solving sin x = sin 2x, we get sin x = 2 sin x cos x, cos x = A=

!

0

%

π/3

(sin 2x − sin x) dx +

!

π

π/3

π 1 , and x = 2 3

(sin x − sin 2x) dx

&$π/3 % &$ 1 1 = − cos 2x + cos x + − cos x + cos 2x 2 2 0 % & 1 1 1 1 1 1 5 = + − − −1+1+ + + = 4 2 2 2 2 4 2 TODO figure

'2

x[g(x) − f (x)] dx

11. x = '0 2

; y= [g(x) − f (x)] dx 0 ! 2 ( ) 12. V = π [g(x)]2 − [f (x)]2 dx

1 2

'2( 0

) [g(x)]2 − [f (x)]2 dx

7.

CHAPTER 6 IN REVIEW

15. V = 2π

Z

0

2

(2 − x)[g(x) − f (x)] dx

13. V = 2π !

14. V = π

0

15. V = 2π

[g(x) − f (x)] dx 0

8.

!

16. V =

0

b

a

A = +− c dy [a − 9. f! (y)]

7.

8.!

!

2 a12.

'2

π/3

V =π

[g(x) 10.013.

A V=−=f2(

12. V = π

!

2

(

a

[g(x)] '2

2

0 14.

V =π

11. !x2 = '0 !

0

2

x[g(x) TODO

0

(

= 2+ 14. V = π 15. V [g(x)

2

!

π

12. 0 V = π–1

TODO TODO figure ! 2 !

CHAPTER 6.=x)A 15. V = 2π16.V (2 V=− 13.

2π

! 2 1 V we = 2π − 2x)[g(x) (x)]xdx 17. Solving sin x = 15. sin 2x, get sin(2 x= sin x cos−x,f cos = , 2 2= 16.

V [g(x) − f (x)] dx [g(x) − f (x)] dx 2 0

!

!

x[g(x) −!f

11. x = ' 2 0

(2 − x)[g(x) − f (x)] dx

!

b

0

13. V = 2π 2

[a −

b

[f (x) 10.− g(x)] A = dx

' x [f (x 9. A = − a2 dx +0 11. a x = ' ! b2 √ 10. A = 9. Ay= dy − =

) [g(x) + 1]2 − [f (x) + 1]2 dx

0

0

!

a

397

TODO figure ! 2 x 2 396 2 16. V =CHAPTER [g(x) −6. f (x)] dx APPLICATIONS OF THE INTEGRAL

396 2

x[g(x) − f (x)] dx

0 2(

0

Z

2

c

b

'2

0

!

!

0

TODO figure ! 17.

2 πV Solving 14. πf ( 16. Vx= and = [g(x) on=−(0, 0

3

17. Solving sin x = sin

π x) dx + A =x = 1 , and (sin 17. Solving sin x = sin 2x, we get sin x = 2 sin x cos 1 x, cos on (0, π). Thus,(sin x − sin 2x) dx π x2x=−3sin 2 0 π/3 17. Solving sin x = sin 2x, we get sin x = 2 sin x cos x, cos x = , and x = 396 CA ! π/3 % &$π/3 % &$π 2! π 3 1 1 on (0, π). Thus, A= (sin 2x − sin x) dx + (sin − sin 2x) !+2 − cos x + cos 15. = x− cos 2xdx + cos x 2x V = 2π 0 π/3 2 22 0= ! [g(x) − f (x)] π/36.

396 CHAPTER 16.

V dx Z π/3 Z π " #$π/3 " % #$ &π 01 1 1 TODO 1 1 1 1 1 5 TODO cos 2x 2x − 1 + 1 + + + = A= (sin 2x − sin x) dx = + − (sin x+ − cos sinx2x) dx+ −=cos x++ −cos− !

2 4 22 239617.

4 2x,2we get sin x = 0 π/3 Solving2sin x2= sin 0 π/3 2 CHc # " 17.

Solving sin x = sin 2x, we get sin x = 2 sin x cos x, 16.

V = π/3 1 1

TODO figure 1 π1 1 1 5 ! − = + − − 1 1 + 1 + + + != 1 0 π/3 ! 2 4+ 2− cos x2+ cos 2x 2 2on (0, 4 π). = − cos 2x + cos x 22 Thus, A =2 dx (sin 2x − si 2 2 2 16.

V = [g(x) − f (x)] 0 16.π/3V = [g(x)!− f (x)] dx0 TODO figure

!0 π 17. Solving π/3 " 0 1 1 1 ! ln 2 1 1 1 5 1 (sin x − s = + − − − 1 +x 1 +−x + + x = −x %ln 2 A= (sin 2x − dx2x,= +we−get 1 1 cos + 2cos Solving sinsin x =x)sin sin2x x= (e − e 2 ) dx2= (e4 + e 2 ) 0 =17. − 1 − 1sin = x =0 17.

2 + Solving 4 18.2 (a) A2= 2x cos x, co π/3 sin 2x, we get sin x = 2 sin 2 2 0 " #$π/3! 1π/3"1 " 1 # ! ln 2 ! ln 2 Z ln 2 on (0, π). Thus, 1 x 2 −x 2 2x −2x

1 1 −2x − ln 2 − sin 2 (b) V = π [(e ) − (e ) ] dx = π (e − e ) dx = − cos 2x + cos x A == 04++ (sin 2− cos x2+ 18. (a) A = (ex − e−x ) dx =0 (ex + e−x ) 0 = 2 + 0− 1 − 1 = 2 ! ! '" # $ 2 2 0 π/3 0 &ln 2 " π π π 1 9π " figure # 1 TODO = = (e2x + e−2x ) 4+ − (1 + 1) = A = (sin 2x − sin x) dx + x+ −1cos sin 1 1 1 = −1 (sin cos12x 2 2 4 8 0 = 0 + − − !−ln12 + 1 +π/32+ + = x −x x 4 2 2 2 2 4 " (e #+ TODO figure(s?) " 18. (a) A = #$ (eπ/3 − 1e ")1dx = 1 1 1#$ 2 √ 0 = + − −x + − " ! ! √2 ln 2 2 2 cos 2x + cos ! = ln− x ln 2 4+ 2−%ln cos 2 √ 1 22 x −(b) 2 −x4 x x 2 −x −x 22 19. (a) x = y 2 ; A(x) = (2y)2 = (2 x)2 = 4x; V18. = 4 (a) xA dx== 4 x(e = e dx ) V) = π = (e0[(e+ )e − (e ) ]= dx2=+ 2 0 # 0 Z ln 2 Z ln 2 0 " TODO figure 0 0 1 !

1 &!ln12 1 '" $1√2 1 ! √2 2 πln12 2x ln 2 − − ln + 2 π = 1 =2 !2 + − + 1 +−2x + (b) V = π [(ex )2 − (e−x (e22x − e−2x ) dx 2 ) ] dx = π 2 4 (b) x = y ; A(x) = πr 0= πy = πx; V = π x dx = π x 4 =2π x 2 2 = −x(e2 x+ e −x 2π)) dx 2==(e 0 18. (a) A = (e − e e− (b) V = π [(e ) − (e ) ] dx = (e24x2x+ − 2 2 0 0

0 i 0 ln 2 0 0 ! π π 1 9π ! ln 2 '" ln 2 TODO&figure(s?) #2 %ln = (e2x + TODO e−2x ) figures = 4+ − (1 + 1) = x −x ln 2 x 2 −x 1−x 2 πx[(e+ π 2x (b) V) = π = (e )e − (e 18. (a) A = (e − e dx ) 2=+π 2 2 4 8 −2x 0 2 20. x = 2y − y = = 0 (e + e ) 4 + 0 )−]=dx (1 √ +2 0 2 2 '" ! 2 ! 2 2! 19. (a) x2 = 4 0 y ; 2 & A(x) = (2y) = (2 x) 2 !ln ln π 2x ln 2ln 2 2 π −2x V = 2π (3 − y)(2y − y 2 ) dy = 2π (y 3 − 5y 2 + 6y) dy = = (e + e ) 4+ x 2 −x 2 2x TODO figure(s?) (b) V = π [(e ) − (e 2 ) ] dx = π 0 (e2 − e 0 0 " #$2 2 2 2 0 TODO (b) x figure(s?) = y ; A(x) 1 4 5 3 16 '" = πr0#= πy = πx &ln 2 = 2π y − y + 3y 2 = π π2 2x π √ 2 1 2 −2x 4 3 3 0 19. (a) x = y (e ; A(x) = )(2y) = (2 x) + eTODO = 4 += 24x;− (1 √V+=1 figures 2 4 = (2 x)2 = 0 y2 ; 2 19. (a) x = A(x) = (2y) TODO figure 20. x ( = 2y − y 2 ! √ ( 5 TODO figure(s?) ! 2 =2 !

21. The equation of the line is y = − x + 10. Then y # = −5/4 and 21 + (y # )2 = 1 + 225/16 2 πy = πx; 2 2 V = 2 π (b) x = y ; A(x)(b) πr 4 = y= ;(3 − A(x) = πr πy= =2ππx; V ==x 2π y)(2y −y= ) dy 1√ 00 √ 41. The surface area is 20 " #$2 V = 4 4 19. TODO (a) x =figures y 2 ; A(x)TODO = (2y) = (2 x)2 = 4x; figures 16 √ #√ " #)7 = 2π 1 y 4 − 5 y 3 + 3y 2 ! 7" = π 2 5 41 41π 5 22 20. x = 2y − y 4 3 3√ 0 !

S = 2π − x + 10 dx20. = x = 2y−−xy + 10x 2 ! !

2 2 4 4 2 8 2 0 2figure 2 ! !

2V = π 0TODO (b) x = y ; A(x) = πr = πy = πx; 2 2 V = 2π (3 − y)(2y − y ) dy = 2π √ √ " # 2 3 41π 5 315π 41 2 − y)(2y − y 0) dy = 2π V≈ =396.03 2π ft(3 5y 25 0+0 of the line(y is#$ y− " = − · 49 + 70 = . 21. The equation 2= − x + 4 2 8 16 TODO figures 0 1 4 5 3 0 2 16 = 2π #$ y 2− y + 3y = π 1√ " 4 3 3 41.

The surface area is 1 5 43 16 0 2 20. x ==2y2π − y 2 y 4 − TODO y + 3yfigure = π 3 3! 2 ! 7 " ! 24 0 525 3 S =line 2π(y ++6 10 V = 2π figure (3 −21. y)(2y y 2 ) dy = 5y TODO The−equation of 2π the is y−=− −4 xx+ 1 0 4 0 0 √ 1 √ " #$surface 5 area 2 41. The is " 5 1 4 of 5the43 line 21. The equation is x + 41π 10. Then y# = 2 y = −16= = 2π y − y + 3y = 4 π 2 " − 8 · 49 +# ! 7 4 3 3 1√ 0 5

398

CHAPTER 6. APPLICATIONS OF THE INTEGRAL

√ 19. (a) x = y 2 ; A(x) = (2y)2 = (2 x)2 = 4x

√2 Z √2 1 2 x dx = 4 V =4 x =4 2 0 0

1

(x, y)

2y

y

2y

2 -1

(b) x = y 2 ; A(x) = πr2 = πy 2 = πx √2 Z √2 1 2 x dx = π x V =π =π 2 0 0

y

3

20. x = 2y − y 2 Z 2 Z 2 V = 2π (3 − y)(2y − y 2 ) dy = 2π (y 3 − 5y 2 + 6y) dy 0

= 2π

1 4 5 3 y − y + 3y 2 4 3

2

x

0

=

0

16 π 3

2

5 21. The equation of the line is y = − x + 10. Then y 0 = −5/4 and 4 p p 1√ 1 + (y 0 )2 = 1 + 25/16 = 41. The surface area is 4 √ √ #7

Z 7 5 41 41π 5 2 dx = S = 2π − x + 10 − x + 10x 4 4 2 8 0 0 √ √

5 315π 41 41π = − · 49 + 70 = ≈ 396.03 ft2 .

2 8 16 22. fave = 23. fave = 24. fave

1 4 − (−3)

Z

1 4−1

(x3/2 + x1/2 ) dx =

Z

4

4

−3

f (x) dx =

10

y

7 8

x

1 (21) = 3 7

1

1 3

2 5/2 2 3/2 x + x 5 3

4 1

=

272 16 − 15 15

1

= 50 −

1 3

=

256 45

3 3 1 1 (2x − 1) dx = (x2 − x) = 2. Setting f (c) = 2c − 1 = 2 we obtain c = 3/2. = 3−0 0 3 0 Z

25. 50 = k(1/2), k = 100, F = 100x;

Z

W =

1

1/2

26. Using 6 in = and k = 80.

1 ft, we need to solve 10 = 2

Z

0

1/2

100x dx = 50x2

1/2

kx dx for x. We have

k 2 x 2

25 = 37.5 joules 2

1/2 0

= 10, k/8 = 10,

399

CHAPTER 6 IN REVIEW

27. W =

Z

10

62.4(10)2 (x + 5) dx = 6240

0

1 2 x + 5x 2

10

= 624, 000 ft-lb 10

0

y

5 x

1 28. The weight of the bucket after it has been lifted x feet is 32 − x pounds.

4

5 Z 5 1 1 2 W = 32 − x dx = 32x − x = 156.875 ft-lb 4 8 0 0

29. At a rate of loss of 1/4 pound per second, it will take 120 seconds to lose the entire 30 pounds.

In 120 seconds, the bucket will be raised 120 feet.

120 Z 120 1 2 1 = 2040 ft-lb W = 32 − x dx = 32x − x 4 8 0 0 1 30. The weight of the rope after the bucket has been lifted x feet is (5 − x). Thus, the weight

8 1 1 261 3 of the system after the bucket has been lifted x feet is 32 − x + (5 − x) = − x.

4 8 8 8

5 Z 5 261 3 2535 261 3 W = − x dx = x − x2 = = 158.4375 ft-lb 8 8 8 16 16 0 0 ( x+3 0≤x≤2 8 31. y = 5 2≤x≤8 Z 8 y W = 62.4(πy 2 )(x + 15) dx 0

=

Z

2

62.4π(x + 3) (x + 15) dx +

0

= 62.4π

2

Z

Z

8

62.4π(25)(x + 15) dx

2

2

(x + 21x + 99x + 135) dx + 1560π 3

2

0

= 62.4π

5 y

1 4 99 x + 7x3 + x2 + 135x 4 2

2 0

Z

8

(x + 15) dx

2

+ 1560π

1 2 x + 15x 2

8 2

–15

= 62.4π(528) + 1560π(152 − 32) = 220, 147.2π ≈ 691, 612.83 ft-lb Z 32. (a) a = −5.5, v(0) = 44, s(0) = 0; v(t) = −5.5 dt = −5.5t + c; 44 = v(0) = c Z s(t) = (−5.5t + 44) dt = −2.75t2 + 44t + c; 0 = s(0) = c; s(t) = −2.75t2 + 44t

The maximum height is attained when v(t) = −5.5t + 44 = 0, or at t = 8 seconds. The maximum height is s(8) = 176 feet. On the earth, a = −32, v(t) = −32t + 44, and s(t) = −16t2 + 44t. The maximum height is attained when v(t) = −32t + 44 = 0, or at t = 1.375 seconds. The maximum height is s(1.375) = 30.25 feet.

400

CHAPTER 6. APPLICATIONS OF THE INTEGRAL (b) When the rock hits the astronaut, 6 = s(t) = −2.75t2 + 44t or 2.75t2 − 44t + 6 = 0.

Solving for t, we obtain t ≈ 1.04 (on the way up) and t ≈ 15.86 (on the way down). The velocity is v(15.86) ≈ −43.24 ft/s.

CHAPTER 6 IN REVIEW

3 33. y 0 = (x − 1)1/2 (b) When the rock hits the astronau 2 5 Z 5 Z 5r Solving for t, we obtain t ≈ 1.04 √ 9 1 1 1 3/2 3/2 3/2 is v(15.86) ≈ −43.24 ft/s 1 + (x − 1) dx = 9x − 5 dx = s= (9x − 5) (40 − 4 velocity = ) 4 2 1 27 27 1 1 3

(x − 1)1/2 2 ! 5" !

9 1 5√ s = 1 + (x − 9 2 1) dx = (b) When the rock hits the astronaut, 6 = s(t) = −2.75t + 44t or22.75t2 − 4 1 1

403/2 − 8 ≈ 9.07 = 27

33. y ! =

CHAPTER 6 IN REVIEW

Solving t, we obtain t ≈ 1.04 up) and t ≈ 15.86 (on the wa 34. Place the origin at the left end of the rod. Then the density offorthe rod is ρ(x) = (on ax the + b.way 3/2Since −8 velocity is v(15.86) ≈ −43.24 ft/s. = 40 ≈ 9.07 ρ(3) = 11 and ρ(6) = 17, we solve the system 3a + b = 11, 6a + b = 17 to obtain ρ(x) = 2x 27 + 5.

Z 6 3 !

1/2 6 33. y = (x − 1) 34. Place the origin at the left end of the r 2 (2x + 5) dx = x2 + 5x 0 = 66 m= #5 ! " !

0

M0 =

Z

0

6

x(2x + 5) dx =

Z

6

(2x2 + 5x) dx =

0

5

9

1

5

√ρ(3) = 11 and 1 ρ(6) = 17, we solve 1 the

3/2 ! 6= 1 + (x − 1) dx = 9x − 5 dx (9x − 5)3/2$ = (40 %&6

6 4 2 1 2 3 5 12 234 m =39 (2x27+ 5) dx = x1 2 +275x 0 = x + 40 x 3/2 − 8 = 234; x = = 0 ! 6 ! 6 =2 ≈ 9.07 3 66 11 27 0

s=

M0 = x(2x + 5) dx = (2x + 5x p 34. Place the origin at the left end of the rod. Then is ρ(x) = 0 the density of the rod 0 35. y = 16 − x2 )3a + b = 11, 6a + b = 17 to obtain ρ(3) = 11 and ρ(6) = 17, we solve the system !

2 6 4 Z 4 35. $ 2 %&6 y = 16 − x p m 800 = (2x + 5) dx = 51, x +200 5x 0 = 66 !

800 4 ) (16 − x2 )3/2 = − !0 (0 − 64) = ! ≈ 17, 066.67 N (# 2 F = 800x 16 − x2 dx = − 800 F = '800x 16 − x6 dx = − (16 63 3 3 6 0 2 3 5 2 0 3 23 2 2

M0 =

x(2x + 5) dx =

5 3(−1) + 8(1) m.

36. Taking the origin at the center of the bar, x = 35. y = )16 − x=2 3 + 8!

11 0

(2x + 5x) dx =0

0

3

x + x 2

= 234;

0

x=

6

36. Taking the origin at the center of the

#4 ) 800 Consider 800 51, 200 2 3/2 first the 2 dx = − 37.(16 F = 800x 16 − x − x ) =1− system (0 − 64)of = 1 kg and ≈ 0 weights. Taking the origin at 3 3 + 3(1) 3 0 0 1(0) 3 4

37. Consider first the system of 1 kg and 3 kg 1(0) + 3(1) 3 the left end, we obtain x1 = = .

1+3 4

Now consider the system left end of the upper bar.

12 6 4(0) + 6(2) = = .

4+6 10 5

obtain x1 =

=

.

1+ 3 3(−1) + 8(1) 5 4 36. Taking the origin at the center of theNow bar, consider x= = m.

3 the + 8 system11with the left

1

bar.

Taking the Taking origin at left,atweth 37. Consider first the system of 1 kg and 3 kg weights. thethe origin 3 1(0) + 3(1) 3 obtain x1 = = .

TODO figures 1+3 4 consider the system with the left with the left two weightsNow concentrated at the 0 two weights 2concentrated at the left en 4(0) + 6(2) 12 6 Taking origin at = = .

Taking the origin at thebar. left, we the obtain x2the=left, we obtain x2 = 4+6 10 5 TODO figures

4

6