chapter 7 soulutions

7/29/2019 cHAPTER 7 SOULUTIONS

1/35

56. Method 1Compare graph ofy1

x2 lnx with

y2

NDERx3

3

ln x

x

9

3

. The graphs should be the same.Method 2Compare graph ofy

1 NINT(x2 lnx) with

y2

x3

3

ln x

x

9

3

. The graphs should be the same or differ

only by a vertical translation.

57. (a) 20,000 10,000(1.063)t

2 1.063t

ln 2 tln 1.063

tln

l

1

n

.0

2

63 11.345

It will take about 11.3 years.

(b) 20,000 10,000e0.063t

2 e0.063t

ln 2 0.063t

t0

l

.

n

06

2

3 11.002

It will take about 11.0 years.

58. (a) f(x) d

d

x x

0

u(t) dt u(x)

g(x) d

d

x x

3

u(t) dt u(x)

(b) C f(x) g(x)

x0

u(t) dt x3

u(t) dt

x

0

u(t) dt 3

x

u(t) dt

30

u(t) dt

59. (a) y 1 5

5

.8

6

9

.0

4

7

e

1

60.0205x

[20, 200] by [10, 60]

(b) The carrying capacity is about 56.0716 million people.(c) Use NDER twice to solvey 0. The solution is

x 86.52, representing (approximately) the year 1887.The population at this time was approximately

P(86.52) 28.0 million people.

60. (a) T 79.961(0.9273)t

[1, 33] by [5, 90]

(b) Solving T(t) 40 graphically, we obtain t 9.2 sec.The temperature will reach 40 after about 9.2 seconds.

(c) When the probe was removed, the temperature was

about T(0) 79.96C.

61. v

0

k

m coasting distance

(0.86)(

k

30.84) 0.97

k 27.343

s(t) v

0

k

m(1 e(k/m)t)

s(t) 0.97(1 e(27.343/30.84)t)

s(t) 0.97(1 e0.8866t)

A graph of the model is shown superimposed on a graph of

the data.

[0, 3] by [0,1]

Chapter 7

Applications of Definite Integrals

s Section 7.1 Integral as Net Change(pp. 363374)

Exploration 1 Revisiting Example 2

1. s(t) t2 (t8

1)2 dt t3

3

t

8

1 C

s(0) 0

3

3

0

8

1 C 9 C 1

Thus, s(t) t

3

3

t

8

1 1.

2. s(1) 1

3

3

1

8

1 1

1

3

6. This is the same as the

answer we found in Example 2a.

3. s(5) 5

3

3

5

8

1 1 44. This is the same answer we

found in Example 2b.

288 Section 7.1


2/35

Quick Review 7.1

1. On the interval, sin 2x 0 whenx

2, 0, or

2. Test one

point on each subinterval: forx 3

4

, sin 2x 1;

forx

4, sin 2x 1; forx

4, sin 2x 1; and for

x 3

4

, sin 2x 1. The function changes sign at

2, 0,

and

2. The graph is

2. x2 3x 2 (x 1)(x 2) 0 whenx 1 or 2. Test

one point on each subinterval: for x 0, x2 3x 2 2;

for x 3

2, x2 3x 2

1

4; and forx 3,

x2 3x 2 2. The function changes sign at 1 and 2.

The graph is

3. x2 2x 3 0 has no real solutions, since

b2 4ac (2)2 4(1)(3) 8 0. The function is

always positive. The graph is

4. 2x3 3x2 1 (x 1)2(2x 1) 0 whenx 1

2 or 1.

Test one point on each subinterval: for x 1,

2x3 3x2 1 4; forx 0, 2x3 3x2 1 1; and

x 3

2, 2x3 3x2 1 1. The function changes sign at

1

2. The graph is

5. On the interval, x cos 2x 0 whenx 0,

4,

3

4

, or

5

4

.

Test one point on each subinterval: for x

8,

x cos 2x

1

6

2; forx

2, x cos 2x

2; forx ,

x cos 2x ; and forx 4, x cos 2x 0.58. The

function changes sign at

4,

3

4

, and

5

4

. The graph is

6. xex 0 whenx 0. On the rest of the interval, xex is

always positive.

7. x2

x

1 0 whenx 0. Test one point on each subinterval:

forx 1,x2

x

1

1

2; forx 1,

x2

x

1

1

2. The

function changes sign at 0. The graph is

8. x

x

2

2

2

4 0 whenx 2 and is undefined when

x 2. Test one point on each subinterval: forx 5

2,

x

x

2

2

2

4

1

9

7; forx 1.9,

x

x

2

2

2

4 4.13; forx 0,

x

x2

2

4

2

1

2; forx 1.9,

x

x

2

2

4

2 4.13; and forx 5

2,

x

x

2

2

2

4

1

9

7. The function changes sign at 2, 2, 2

and 2. The graph is

9. sec (1 1 sin2x) cos (1

1

cos x) is undefined when

x 0.9633 kor 2.1783 k for any integer k. Test for

x 0: sec (1 1 sin20) 2.4030.

Test forx 1: sec (1 1 sin21) 32.7984. The

sign alternates over successive subintervals. The function

changes sign at 0.9633 kor 2.1783 k, where kis an

integer. The graph is

10. On the interval, sin 1x 0 whenx 31

or

2

1

. Test one

point on each subinterval: forx 0.1, sin 1x 0.54; forx 0.15, sin 1x 0.37; and forx 0.2,sin 1x 0.96. The graph changes sign at 3

1

, and

2

1

.

The graph is

+

0.1 0.213

12

f(x)

x

+ +

2.1783 0.9633 0.9633 2.1783

f(x)

x

+ ++

3 2 2 32 2

f(x)

x

+

5 0 30

f(x)

x

++

0 44

34

54

f(x)

x

+ +

2 2112

f(x)

x

+

4 2

f(x)

x

+ +

2 1 2 4

f(x)

x

++

3 20 2

2

f(x)

x

Section 7.1 289


3/35

Section 7.1 Exercises

1. (a) Right when v(t) 0, which is when cos t 0, i.e.,

when 0 t

2 or

3

2

t 2. Left when cos t 0,

i.e., when

2 t

3

2

. Stopped when cos t 0,

i.e., when t

2 or

3

2

.

(b) Displacement

20

5 cos t dt 5sin t 5[sin 2 sin 0] 0

(c) Distance 20

5 cos t dt

/20

5 cos t dt 3/2/2

5 cos t dt 23/2

5 cos t dt

5 10 5 20

2. (a) Right when v(t) 0, which is when sin 3t 0,

i.e., when 0 t

3

. Left when sin 3t 0, i.e., when

3 t

2. Stopped when sin 3t 0, i.e., when t 0

or

3.

(b) Displacement /20

6 sin 3t dt 613 cos 3t 2cos 32

cos 0 2

(c) Distance /20

6 sin 3t dt

/30

6 sin 3t dt /2/3

6 sin 3t dt 4 2 6

3. (a) Right when v(t)

49

9.8t

0, i.e., when 0

t

5.

Left when 49 9.8t 0, i.e., when 5 t 10.

Stopped when 49 9.8t 0, i.e., when t 5.

(b) Displacement 100

(49 9.8t) dt

49t 4.9t2 49[(10 10) 0] 0

(c) Distance 100

49 9.8t dt

50

(49 9.8t) dt 105

(49 9.8t) dt

122.5 122.5 245

4. (a) Right when

v(t) 6t2 18t 12 6(t 1)(t 2) 0,

i.e., when 0 t 1. Left when 6(t 1)(t 2) 0,

i.e., when 1 t 2. Stopped when

6(t 1)(t 2) 0, i.e., when x 1, or 2.

(b) Displacement 2

0

(6t2 18t 12) dt

2t3 9t2 12t [(16 36 24) 0] 4

(c) Distance 20

6t2 18t 12 dt

10

(6t2 18t 12) dt 21

(6t2 18t 12) dt

5 1 6

5. (a) Right when v(t) 0, which is when sin t 0 and

cos t 0, i.e., when 0 t

2 or

3

2

t 2. Left

when sin t 0 and cos t 0, i.e., when 2 t or

t 3

2

. Stopped when sin t 0 or cos t 0,

i.e., when t 0,

2, ,

3

2

, or 2.

(b) Displacement 20

5 sin2 tcos t dt 513 sin3 t

5[0 0] 0

(c) Distance 20

5 sin2 tcos t dt

/20

5 sin2 tcos t dt 3/2/2

5 sin2 tcos t dt

2

3/2

5 sin2 tcos t dt

5

3

1

3

0

5

3

2

3

0

6. (a) Right when v(t) 0, which is when 4 t 0, i.e.,

when 0 t 4. Left: never, since 4 t cannot be

negative. Stopped when 4 t 0, i.e., when t 4.

(b) Displacement 40

4 t dt 23(4 t)3/2

2

3[0 8]

1

3

6

(c) Distance 40

4 t dt 1

3

6

4

0

2

0

2

0

10

0

/2

0

2

0

290 Section 7.1


4/35

7. (a) Right when v(t) 0, which is when cos t 0,

i.e., when 0 t

2 or

3

2

t 2. Left when

cos t 0, i.e., when

2 t

3

2

. Stopped when

cos t 0, i.e., when t

2 or

3

2

.

(b) Displacement 20

esin tcos t dt esin t [e0 e0] 0

(c) Distance 20

esin tcos t dt /20

esin tcos t dt

3/2/2

esin tcos t dt 23/2

esin tcos t dt

(e 1) e 1e 1 1

e 2e 2e 4.7

8. (a) Right when v(t) 0, which is when 0 t 3. Left:

never, since v(t) is never negative. Stopped when t 0.

(b) Displacement 30

1

t

t2 dt 12 ln (1 t2)

1

2[ln (10) ln (1)]

ln

2

10

1.15

(c) Distance 30

1

t

t2 dt

ln

2

10 1.15

9. (a) v(t) a(t) dt t 2t3/2 C, and since v(0) 0,v(t) t 2t3/2. Then v(9) 9 2(27) 63 mph.

(b) First convert units:

t 2t3/2 mph 36

t

00

1

t

8

3

0

/2

0 mi/sec. Then

Distance 90

36

t

00

1

t

8

3

0

/2

0

dt

72t

0

2

0

4

t

5

5

0

/2

0 8

9

00

5

2

0

7

0 0 0.06525 mi

344.52 ft.

10. (a) Displacement 40

(t 2) sin t dt

sin t tcos t 2 cos t [(sin 4 4 cos 4 2 cos 4) 2] 1.44952 m

(b) Because the velocity is negative for 0 t 2, positive

for 2 t , and negative for t 4,

Distance 20

(t 2) sin t dt 2

(t 2) sin t dt

4

(t 2) sin t dt

[(2 sin 2) ( sin 2 2)

( 2 cos 4 sin 4 2)]

2 2 cos 4 2 sin 2 sin 4 2 1.91411 m.

11. (a) v(t) a(t) dt 32 dt 32t C1, where

C1

v(0) 90.

Then v(3) 32(3) 90 6 ft/sec.

(b) s(t) v(t) dt 16t2 90t C2, where

C2

s(0)

0. Solve s(t)

0:

16t2 90t 2t(8t 45) 0

when t 0 or t 4

8

5 5.625 sec.

The projectile hits the ground at 5.625 sec.

(c) Since starting height ending height,

Displacement 0.

(d) Max. Height s5.6225

16

5.6

2

25

2

90

5.6

2

25

126.5625, andDistance 2(Max. Height) 253.125 ft.

12. Displacement c0

v(t) dt 4 5 24 23 cm

13. Total distance c0

v(t) dt 4 5 24 33 cm

14. At t a, s s(0) a0

v(t) dt 15 4 11.

At t b, s s(0) b0

v(t) dt 15 4 5 16.

At t c, s s(0)

c

0

v(t) dt 15 4 5 24 8.

15. At t a, where d

d

v

t is at a maximum (the graph is steepest

upward).

16. At t c, where d

d

v

t is at a maximum (the graph is steepest

upward).

17. Distance Area under curve 412 1 2 4(a) Final position Initial position Distance

2 4 6; ends atx 6.

(b) 4 meters

18. (a) Positive and negative velocities cancel: the sum of

signed areas is zero. Starts and ends atx 2.

(b) Distance Sum of positive areas 4(1 1) 4meters

4

0

9

0

3

0

2

0

Section 7.1 291


5/35

19. (a) Final position 2 70

v(t) dt

2 1

2(1)(2)

1

2(1)(2) 1(2)

1

2(2)(2)

1

2(2)(1)

5;

ends atx 5.

(b) 70

v(t) dt 12(1)(2)

1

2(1)(2) 1(2)

1

2(2)(2)

1

2(2)(1)

7 meters

20. (a) Final position 2 100

v(t) dt

2 1

2(2)(3)

1

2(1)(3) (3)(3)

1

2(1)(3)

1

2(3)(3)

2.5;

ends atx 2.5.

(b) Distance 100

v(t) dt

1

2(2 3)

1

2(1)(3) 3(3)

1

2(1)(3)

1

2(3)(3)

19.5 meters

21. 100

27.08 et/25 dt 27.0825et/25 27.08[25e0.4 25]

332.965 billion barrels

22. 2403.9 2.4 sin 1

2

t dt 3.9t28

.8 cos 1

2

t

93.6 28.8 28.8 93.6 kilowatt-hours23. (a) Solve 10,000(2 r) 0: r 2 miles.

(b) Width r, Length 2r: Area 2rr

(c) Population Population density Area

(d) 20

10,000(2 r)(2r) dr 20,0002

0

(2r r2) dr

20,000r2 13r3 20,0004 8

3 0

80,

3

000 83,776

24. (a) Width

r, Length

2

r: Area

2

r

r

(b) Volume per second

Inches per second Cross section area

8(10 r2)s

i

e

n

c

. (2r)rin

2 flow in

s

in

ec

3

(c) 30

8(10 r2)(2r) dr 163

0

(10r r3) dr

165r2 14r4 1645 84

1 0

396s

in

ec

3

1244.07 s

in

ec

3

25. (Answers may vary.)

Plot the speeds vs. time. Connect the points and find the

area under the line graph. The definite integral also gives

the area under the curve.

26. (a) Sum of numbers in Sales column 797.5 thousand

(b) Enter the table in a graphing calculator and use

QuadReg: B(x) 1.6x2 2.3x 5.0.

(c) 11

0 (1.6x2

2.3x 5.0) dx

13.6x3

2

2

.3x2 5.0x

904.02 thousand

(d) The answer in (a) corresponds to the area of left hand

rectangles. These rectangles lie under the curveB(x).

The answer in (c) corresponds to the area under the

curve. This area is greater than the area of rectangles.

27. (a) 10.50.5

(1.6x2 2.3x 5.0) dx

13.6x3

2

2

.3x2 5.0x 798.97 thousand

(b) The answer in (a) corresponds to the area of midpointrectangles. The curve now gives a better approximation

since part of each rectangle is above the curve and part

is below.

28. Treat 6 P.M. as 18 oclock:

b

2

n

a f(x0)

n1

i1

2f(xi) f(x

n)

1

2

8

(1

0)

8[120 2(110) 2(115) 2(115) 2(119)

2(120) 2(120) 2(115) 2(112) 2(110)

121]

1156.5

29. F(x) kx; 6 k(3), so k 2 and F(x) 2x.

(a) F(9) 2(9) 18N

(b) W 90

F(x) dx 90

2x dx x2 81 N cm30. F(x) kx; 10,000 k(1), so k 10,000.

(a) W d0

kx dx 12kx2 1

2kd2

1

2(10,000)(0.5)2

1250 inch-pounds

(b) For total distance: W 12(10,000)(1)2 5000

For second half of distance:

W 5000 1250 3750 inch-pounds

31.(1

2

2

(1

2)

0)[0.04 2(0.04) 2(0.05) 2(0.06) 2(0.05)

2(0.04) 2(0.04) 2(0.05) 2(0.04)

2(0.06) 2(0.06) 2(0.05) 0.05] 0.585

The overall rate, then, is0.

1

5

2

85 0.04875.

d

0

9

0

10.5

0.5

11

0

3

0

2

0

24

0

10

0

292 Section 7.1


6/35

32.(1

2

2

(1

2)

0)[3.6 2(4.0) 2(3.1) 2(2.8) 2(2.8) 2(3.2)

2(3.3) 2(3.1) 2(3.2) 2(3.4) 2(3.4)

2(3.9) 4.0] 40 thousandths or 0.040

33. (a) x M

M

y . Taking dm dA as m

kand letting

dA 0, k yields .

(b) y M

M

y . Taking dm dA as m

kand letting

dA 0, k yields .

34. By symmetry, x 0. Fory, use horizontal strips:

y

1

5

2

35. By symmetry, y 0. Forx, use vertical strips:

x

4

3

s Section 7.2Areas in the Plane(pp. 374382)

Exploration 1 A Family of Butterflies

1. For k 1:

0

[(2 sinx) sinx] dx 0

(2 2 sinx) dx

2x 2 cosx 2 4

For k 2:

/20

[(4 2 sin 2x) (2 sin 2x)] dx

/20

(4 4 sin 2x) dx

4x 2 cos 2x 2 42. It appears that the areas for k 3 will continue to be

2 4.

3. Ak /k

0[(2k ksin kx) ksin kx] dx

/k0

(2k 2ksin kx) dx

If we make the substitution u kx, then du k dx and the

u-limits become 0 to . Thus,

Ak

/k0

(2k 2ksin kx) dx

/k0

(2 2 sin kx)k dx

0

(2 2 sin u) du.

4. 2 4

5. Because the amplitudes of the sine curves are k, the kth

butterfly stands 2kunits tall. The vertical edges alone have

lengths (2k) that increase without bound, so the perimeters

are tending to infinity.

Quick Review 7.2

1. 0

sinx dx cosx [1 1] 2

2. 10

e2x dx 12e2x 12(e2 1) 3.195

3. /4

/4

sec2x dx tanx 1 (1) 2

4. 20

(4x x3) dx 2x2 14x4 (8 4) 0 4

5. 33

9 x2 dx 9

2

(This is half the area of a circle of

radius 3.)

2

0

/4

/4

1

0

0

/2

0

0

2

3x320

x220

20

x(2x) dx

2

0 2x dx

x dA

dAx dA

dAx dm

dm

225y5/24

0

223y3/24

0

40

y(2 y) dy

40

2 y dy

y dA

dAy dA

dAy dm

dm

y dm

dm

mky

mk

x dm

dm

mkxk

mk

Section 7.2 293


7/35

6. Solvex2 4x x 6.

x2 5x 6 0

(x 6)(x 1) 0

x 6 or x 1

y 6 6 12 or y 1 6 5

(6, 12) and (1, 5)

7. Solve ex x 1. From the graphs, it appears that ex is

always greater than or equal tox 1, so that if they are

ever equal, this is when ex (x 1) is at a minimum.

d

d

x[ex (x 1)] ex 1 is zero when ex 1, i.e., when

x 0. Test: e0 0 1 1. So the solution is (0, 1).

8. Inspection of the graphs shows two intersection points:

(0, 0), and (, 0). Check: 02 0 sin 0 0 and

2

2 sin 0.

9. Solvex2

2

x

1x3.

(0, 0) is a solution. Now divide byx.

x22

1x2

2 x4 x2

x4 x2 2 0

x2 1

2

1 8 2 or 1

Throw out the negative solution.

x 1

y x3 1

(0, 0), (1, 1) and (1, 1)

10. Use the intersect function on a graphing calculator:

[2, 2] by [2, 2]

(0.9286, 0.8008), (0, 0), and (0.9286, 0.8008)


1. 0

(1 cos2x) dx 12x 1

4 sin 2x 2

2. Use symmetry:

2/30

12 sec2 t 4 sin2 t dt /3

0

(sec2 t 8 sin2 t) dt

tan t 4t 2 sin 2t

3 43 3 0

4

3

3. 10

(y2 y3) dy 13y3

1

4y4 1

1

2

4. 10

[(12y2 12y3) (2y2 2y)] dy

10

(12y3 10y2 2y) dy

3y4

1

3

0y3 y2

3

1

3

0 1

4

3

5. Use the regions symmetry:

220

[2x2 (x4 2x2)] dx 220

(x4 4x2) dx

215x5 4

3x3

2352

3

3

2 0 11

2

5

8


210

(x2 2x4) dx 213x3 2

5x5 213

2

5 21

2

5

7. Integrate with respect toy:

10

(2 y y) dy 43y3/2

1

2y2

43 1

2 0 56

8. Integrate with respect toy:

1

0[(2 y) y] dy

2y 12y2 2

3y3/2 2 12

2

3 0 56

9. Integrate in two parts:

02

[(2x3 x2 5x) (x2 3x)] dx

20

[(x2 3x) (2x3 x2 5x)] dx

02

(2x3 8x) dx 20

(2x3 8x) dx

1

2

x4 4x2

1

2

x4 4x2

[0 (8 16)] [(8 16) 0] 16

2

0

0

2

1

0

1

0

1

0

2

0

1

0

1

0

/3

0

0

294 Section 7.2


8/35

10. Integrate in three parts:

12

[(x 2) (4 x2)] dx

21

[(4 x2) (x 2)] dx

3

2[(x 2) (4 x2)] dx

12

(x2 x 2) dx 21

(x2 x 2) dx

32

(x2 x 2) dx

13x3 1

2x2 2x 13x

3

1

2x

2 2x

13x3

1

2x

2 2x

13 1

2 2 83 2 4

83 2 4 1

3

1

2 2

9 92 6 8

3 2 4

4

6

9 8

1

6

11. Solvex2 2 2: x2 4, so the curves intersect at

x 2.

22

[2 (x2 2)] dx 22

(4 x2) dx

4x 13x3 8 8

3 8 83

3

3

2 10

2

3

12. Solve 2x x2 3: x2 2x 3 (x 3)(x 1) 0,

so the curves intersect atx 1 andx 3.

31

(2x x2 3) dx x2 1

3x3 3x

(9 9 9) 1

1

3 3

3

3

2 10

2

3

13. Solve 7 2x2 x2 4: x2 1, so the curves intersect at

x 1.

11

[(7 2x2) (x2 4)] dx 11

(3x2 3) dx

311

(1 x2) dx

3 x 1

3x3

323 2

3 4

14.

[3, 3] by [1, 5]

The curves intersect atx 1 andx 2. Use the

regions symmetry:

210

[(x4 4x2 4)x2] dx 221

[x2 (x4 4x2 4)] dx

210

(x4 5x2 4) dx 221

(x4 5x2 4) dx

215x5 5

3x3 4x 215x5

5

3x3 4x

215 5

3 4 235

2

4

3

0 8 15

5

3 4 8

15.

32a, 3

2a by [a2, a2]

The curves intersect atx 0 andx a. Use the regions

symmetry:

2a0

x a2x2 dx 213(a2 x2)3/2 20 13a3

2

3a3

a

0

2

1

1

0

1

1

3

1

2

2

3

2

2

1

1

2

Section 7.2 295


9/35

16.

[2, 12] by [0, 3.5]

The curves intersect at three points:

x 1, x 4 andx 9.

Because of the absolute value sign, break the integral up atx 0 also:

01x5

6 x dx

4

0x5

6 x dx

94 xx 5

6 dx

23(x)3/2 23x3/2

2

3x3/2

0 11

1

0

2

3 35

2

1

3

6 0

18 118

0

9 13

6

3

5

2

1

3

3

0

1

1

6

5

1

6

5

3 1

2

3

17.

[5, 5] by [1, 14]

The curves intersect atx 0 andx 4. Because of the

absolute value sign, break the integral up

atx 2 also (where x2 4 turns the corner). Use thegraphs symmetry:

220x2

2

4 (4 x2) dx 24

2x2

2

4 (x2 4) dx 22

0

3

2

x2 dx 24

2x2

2

8 dx 2

x

2

3

2

x

6

3

8x

2[4] 233

2 32 43 16

6

3

4 21

1

3

18. Solvey2 y 2: y2 y 2 (y 2)(y 1) 0, so the

curves intersect aty 1 andy 2.

21

(y 2 y2) dy 12y2 2y 1

3y3

2 4 83 1

2 2

1

3

92 41

2

19. Solve forx: x y

4

2

1 andx 4

y 4.

Now solve y

4

2

1 4

y 4:

y

4

2

4

y 5 0,

y2 y 20 (y 5)(y 4) 0.

The curves intersect aty 4 andy 5.

544

y 4 y4

2

1 dy

5

4y4

2

4y 5 dy

1y

2

3

y

8

2

5y

112

2

5

2

8

5 25 13

6 2 20 248

3 30

3

8

20. Solve forx: x y2 andx 3 2y2. Now solve

y2 3 2y2: y2 1, so the curves intersect at y 1.

Use the regions symmetry:

210

(3 2y2 y2) dy 210

(3 3y2) dy

610

(1 y2) dy

6 y 1

3y3

61 13 0 4

21. Solve forx: x y2 andx 2 3y2.

Now solve y2 2 3y2: y2 1, so the curves intersect

at y 1. Use the regions symmetry:

210

(2 3y2 y2) dy 210

(2 2y2) dy

410

(1 y2) dy 4 y 1

3y3 41 13 0

8

3

1

0

1

0

5

4

2

1

4

2

2

0

9

4

1

2x2 6x

5

4

0

1

2x2 6x

5

0

1

1

2x2 6x

5

296 Section 7.2


10/35

22. Solve fory: y 4 4x2 andy x4 1.

Now solve 4 4x2 x4 1:

x4 4x2 5 (x2 1)(x2 5) 0.

The curves intersect at x = 1.


210

[(4 4x2) (x4 1)] dx

210

(x4 4x2 5) dx

215x5 4

3x3 5x

215 4

3 5 0

1

1

0

5

4 6

1

1

4

5

23. Solve forx: x

3

y

2

andx

y

4

2

.

Now solve 3 y2 y

4

2

: y2 4,

so the curves intersect aty 2.


220 3 y2 y4

2

dy 22

0 3 34y2 dy

23y y43

2(6 2) 0 8

24.

0

(2 sinx sin 2x) dx

2 cosx

1

2 cos 2x

2 12 2

1

2 4


2 /30

(8 cosx sec2x) dx 28 sinx tanx 2[(4 3 3) 0] 6 3

26.

[1.1, 1.1] by [0.1, 1.1]

The curves intersect atx 0 andx 1, but they do not

cross atx 0.

2 10 1 x2 cos 2

x dx

2 x 1

3x3

2 sin 2

x

21 13 2 0 43

4 0.0601

27.

[1.5, 1.5] by [1.5, 1.5]

The curves intersect atx 0 andx 1. Use the areas

symmetry:

210sin 2

xx dx 2

2 cos 2

x 12x2

212 2

4

0.273

28. Use the regions symmetry, and simplify before integrating:

2 /40

(sec2x tan2x) dx

2 /40

[sec2x (sec2x 1)] dx

2 /40

dx 2 x 2


2 /40

(tan2y tan2y) dy 4 /40

tan2y dy

4tany y 41 4 0

4 0.858

30. /20

3 siny cosy dy 323(cosy)3/2 30 23 2

31. Solve forx: x y3 andx y.

[1.5, 1.5] by [1.5, 1.5]


symmetry: 210

(y y3) dy 212y2 1

4y4 12

1

0

/2

0

/4

0

/4

0

1

0

1

0

/3

0

0

2

0

1

0

Section 7.2 297


11/35

32.

[0.5, 2.5] by [0.5, 1.5]

y x andy

x

12 intersect atx 1. Integrate in two parts:

10

x dx 21

x

12 dx 12x2

1

x

1

2 12 (1) 1

33. The curves intersect when sinx cosx, i.e., atx

4.

/40

(cosx sinx) dx sinx cosx 2 1 0.414

34.

[3, 3] by [2, 4]

(a) The curves intersect atx 2.


2 20

(3 x2 1) dx 2 20

(4 x2) dx

24x 13x3 28 83 0

3

3

2

(b) Solvey 3 x2 forx: x 3y. The

y-intercepts are 1 and 3.

31

2 3ydy 223(3 y)3/2

20 136 33

2

35. (a)

Ify x2 c, thenx c. So the points are

( c, c) and ( c, c).

(b) The two areas in Quadrant I, wherex y, are equal:

c0

y dy 4c

y dy

23y3/2 2

3y3/2

2

3c3/2

2

343/2

2

3c3/2

2c3/2

8

c3/2 4

c 42/3 24/3

(c) Divide the upper right section into a (4 c)-by- c

rectangle and a leftover portion:

c0

(c x2) dx (4 c) c 2 c

(4 x2) dx

cx 13x3

4 c c3/2 4x 13x

3

c3/2

1

3c3/2 4 c c3/2

8 83 4 c 1

3c3/2

2

3c3/2 4 c c3/2

1

3

6 4 c

1

3c3/2

4

3c3/2

1

3

6

c3/2 4

c 42/3 24/3

36.

[1, 5] by [1, 3]

The key intersection points are atx 0, x 1 andx 4.

Integrate in two parts:

101 x 4

x dx

4

1

2

x

4

x dx

x 2

3x3/2

x

8

2

4 x x82

1 23

1

8 (8 2) 4 18

1

3

1

4

1

1

0

2

c

c

0

4

c

c

0

y

5

x3

(c, c) (c, c)

y = 4(2, 4) (2, 4)

y = c

y =x2

3

1

2

0

/4

0

2

1

1

0

298 Section 7.2


12/35

37. First find the two areas.

For the triangle, 1

2(2a)(a2) a3

For the parabola, 2a0

(a2 x2) dx 2a2x 13x3 4

3a3

The ratio, then, is 3

4, which remains

constant as a

approaches zero.

38. ba

[2f(x) f(x)] dx ba

f(x) dx, which we already know

equals 4.

39. Neither; both integrals come out as zero because the

1-to-0 and 0-to-1 portions of the integrals cancel each

other.

40. Sometimes true, namely when dA [f(x) g(x)] dx is

always nonnegative. This happens whenf(x) g(x) over

the entire interval.

41.

[1.5, 1.5] by [1.5, 1.5]


symmetry:

210x2

2

x

1 x3 dx 2ln x2 1 14x4

2 ln 2 1

2

ln 4 12 0.886

42.

[1.5, 1.5] by [1.5, 1.5]

The curves intersect atx 0 andx 0.9286. Use NINT

to find 20.92860

(sinx x3) dx 0.4303.

43. First graphy cosx andy x2.

[1.5, 1.5] by [0.5, 1.5]

The curves intersect atx 0.8241. Use NINT to find

20.82410

(cosx x2) dx 1.0948. Multiplying both

functions by kwill not change thex-value of any

intersection point, so the area condition to be met is

2 20.82410

(kcosx kx2) dx

2 k 20.82410

(cosx x2) dx

2 k(1.0948)

k 1.8269.

44. (a) Solve fory:

a

x2

2 b

y2

2 1

y2 b21 ax2

2y b1 ax

2

2

(b) aab1 ax

2

2 b1 ax2

2 dx or2aa

b1 ax2

2 dx or 4a

0b1 ax

2

2 dx

(c) Answers may vary.

(d, e) 2aa

b1 ax2

2 dx 2b2x1 ax

2

2 a

2 sin1

a

x

2ba2 sin1 (1) a

2 sin1 (1)

ab

45. By hypothesis, f(x) g(x) is the same for each region,

wheref(x) and g(x) represent the upper and lower edges.

But then Area

b

a [f

(x

) g

(x

)]dx

will be the same foreach.

a

a

1

0

a3

4

3a3

a

0

Section 7.2 299


13/35

46. The curves are shown for m 1

2:

[1.5, 1.5] by [1,1]

In general, the intersection points are wherex2

x

1 mx,

which is wherex 0 or elsex m1 1. Then,because of symmetry, the area is

2 (1/m)10 x2

x

1 mx dx

212 ln (x2 1) 1

2mx2

ln m1 1 1

mm1

1 m ln (m) 1.

s Section 7.3Volumes(pp. 383394)

Exploration 1 Volume by Cylindrical Shells

1. Its height isf(xk) 3x

kx

k2.

2. Unrolling the cylinder, the circumference becomes one

dimension of a rectangle, and the height becomes the other.

The thickness x is the third dimension of a slab with

dimensions 2(xk 1) by 3x

kx

k2 by x. The volume is

obtained by multiplying the dimensions together.

3. The limit is the definite integral 3

0

2(x 1)(3x x2) dx.

4. 45

2

Exploration 2 Surface Area

1. ba

2y1 ddyx2

dxThe limit will exist iffandf are continuous on the interval

[a, b].

2. y sinx, so d

d

y

x cosx and

ba

2y1 ddyx2

dx

0

2 sin x 1cos2xdx 14.424.

3. y x, so d

d

y

x

2

1

x and

40

2 x1 21

x

2

dx 36.177.

Quick Review 7.3

1. x2

2. s

x

2, so Area s2

x

2

2

.

3. 1

2r2 or

2

x2

4. 1

2

d

2

2

or

8

x2

5. b x and h

2

3x, so Area

1

2bh

4

3x2.

6. b hx, so Area 1

2bh

x

2

2

.

7. b h

x

2, so Area

1

2bh

x

4

2

.

8.

b x and h (2x)2 12x2

2

15x, so

Area 1

2bh

4

15x2.

9. This is a 3-4-5 right triangle. b 4x, h 3x, and

Area 1

2bh 6x2.

10. The hexagon contains six equilateral triangles with sides of

lengthx, so from Exercise 5, Area 6 43x232

3x2.


1. In each case, the width of the cross section is

w 2 1x2.

(a) A r2, where r w

2, soA(x) w2

2 (1 x2).

(b) A s2, where s w, soA(x) w2 4(1 x2).

(c) A s2, where s

w

2

, soA(x)

w

2

2 2(1 x2).

(d) A

4

3w2 (see Quick Review Exercise 5), so

A(x)

4

3(2 1x2)2 3(1 x2).

2x2x

x

(1/m)1

0

300 Section 7.3


14/35

2. In each case, the width of the cross section is w 2 x.

(a) A r2, where r w

2, so A(x) w2

2 x.

(b) A s2, where s w, soA(x) w2 4x.

(c) A s2, where s

w

2, soA(x)

w

2

2 2x.

(d) A

43w2 (see Quick Review Exercise 5), so

A(x)

4

3(2 x)2 3x.

3. A cross section has width w 2 x and area

A(x) s2

w

2

2 2x. The volume is

40

2x dx x2 16.

4. A cross section has width w (2x2)x2 2 2x2

and area A(x) r2

w

22

(1 x2

)2

. The volume is

11

(1 x2)2 dx 1

1

(x4 2x2 1) dx

15x5 2

3x3 x

1

1

6

5.

5. A cross section has width w 2 1x2 and area

A(x) s2 w2 4(1 x2). The volume is

11

4(1 x2) dx 411

(1 x2) dx 4 x 1

3x3 13

6.

6. A cross section has width w 2 1x2 and area

A(x) s2

w

2

2 2(1 x2). The volume is

11

2(1 x2) dx 211

(1 x2) dx 2 x 1

3x3 83.

7. A cross section has width w 2 sin x.

(a) A(x)

4

3w2 3 sinx, and

V 0

3 sinx dx

3

0

sinx dx

3cosx 2 3.

(b) A(x) s2 w2 4 sinx, and

V 0

4 sinx dx 40

sinx dx 4cosx 8.

8. A cross section has width w secx tanx.

(a) A(x) r2 w22

4(secx tanx)2, and

V /3/3

4(secx tanx)2 dx

4

/3

/3(sec2

x 2 secx tanx tan2

x) dx

4tanx 2 secx tanxx

2tanx secx 12x

2 3 2 6 3 2

6

3

6

2

.

(b) A(x) s2 w2 (secx tanx)2, and

V

/3

/3

(secx tanx)2 dx, which by same method as

in part (a) equals 4 3 2

3.

9. A cross section has width w 5y2 and area

r2 w22

5

4

y4. The volume is

20

5

4

y4 dy

4 y5 8.

10. A cross section has width w 2 1y2 and area

1

2s2

1

2w2 2(1 y2). The volume is

1

1

2(1 y2) dy 2y 13y3

83.11. (a) The volume is the same as if the square had moved

without twisting: VAh s2h.

(b) Still s2h: the lateral distribution of the square cross

sections doesnt affect the volume. Thats Cavalieris

Volume Theorem.

12. Since the diameter of the circular base of the solid extends

fromy 1

2

2 6 toy 12, for a diameter of 6 and a radius

of 3, the solid has the same cross sections as the right

circular cone. The volumes are equal by Cavalieris

Theorem.

13. The solid is a right circular cone of radius 1 and height 2.

V 1

3Bh

1

3(r2)h

1

3(12)2

2

3

14. The solid is a right circular cone of radius 3 and height 2.

V 1

3Bh

1

3(r2)h

1

3(32)2 6

1

1

2

0

/3

/3

/3

/3

0

0

1

1

1

1

1

1

4

0

Section 7.3 301


15/35

15. A cross section has radius r tan 4y and areaA(y) r2 tan2 4y. The volume is10

tan2 4y dy 4 tan 4yy

4 1

4 .

16. A cross section has radius r sinx cosx and area

A(x) r2 sin2x cos2x. The shaded region extends

fromx 0 to where sinx cosx drops back to 0, i.e., where

x

2. Now, since cos 2x 2 cos2x 1, we know

cos2x 1 c

2

os 2x and since cos 2x 1 2 sin2x, we

know sin2x1 c

2

os 2x. /2

0 sin2x cos2x dx

/2

0

1 c2os 2x

1 c2os 2x

dx

4 /2

0(1 cos2 2x) dx

4 /2

0sin2 2x dx

4 /2

0

1 c

2

os 4x dx

8 /2

0(1 cos 4x) dx

8 x

1

4 sin 4x 8

2 0 0 16

2

.

17.

[2, 4] by [1, 5]

A cross section has radius rx2 and area

A(x) r2 x4. The volume is

20

x4 dx 15x5 32

5

.

18.

[4, 6] by [1, 9]

A cross section has radius rx3 and area

A(x) r2 x6. The volume is

20

x6 dx 17x7 12

7

8.

19.

[6, 6] by [4, 4]

The solid is a sphere of radius r 3. The volume is

43r3 36.

20.

[0.5, 1.5] by [0.5, 0.5]

The parabola crosses the liney 0 when

xx2 x(1 x) 0, i.e., whenx 0 orx 1. A cross

section has radius rxx2 and area

A(x)

r

2

(x

x

2

)

2

(x

2

2x

3

x

4

).The volume is

10

(x2 2x3 x4) dx 13x3 1

2x4

1

5x5 3

0.

21.

[1, 2] by [1, 2]

Use cylindrical shells: A shell has radiusy and heighty.

The volume is

1

0

2(y)(y) dy 213y3

23.22.

[1, 3] by [1, 3]

Use washer cross sections: A washer has inner radius rx,

outer radiusR 2x, and areaA(x) (R2 r2) 3x2.

The volume is 10

3x2 dx 3

1

3

x3

.

1

0

1

0

1

0

2

0

2

0

/2

0

1

0

302 Section 7.3


16/35

23.

[2, 3] by [1, 6]

The curves intersect whenx2 1x 3, which is when

x2 x 2 (x 2)(x 1) 0, i.e., when

x 1 orx 2. Use washer cross sections: a washer hasinner radius rx2 1, outer radiusRx 3, and area

A(x) (R2 r2)

[(x 3)2 (x2 1)2]

(x4 x2 6x 8). The volume is

21

(x4 x2 6x 8) dx

15x5 1

3x3 3x2 8x

352

8

3 12 16 15

1

3 3 8 115

7.

24.

[2, 3] by [1, 5]

The curves intersect when 4 x2 2 x, which is when

x2 x 2 (x 2)(x 1) = 0, i.e., whenx1 or

x 2. Use washer cross sections: a washer has inner radius

r 2x, outer radiusR 4 x2, and area

A(x) (R2 r2)

[(4 x2)2 (2 x)2]

(12 4x 9x2 x4).

The volume is

2

1

(12 4x 9x2 x4) dx

12x 2x2 3x3 15x5

24 8 24 352 12 2 3 15

10

5

8.

25.

3,

3

by [0.5, 2]

Use washer cross sections: a washer has inner radius

r secx, outer radiusR 2, and areaA(x) (R2 r2) (2 sec2x).

The volume is

/4/4

(2 sec2x) dx 2x tanx

2 1

2 1

2 2.

26.

[1, 5] by [3, 1]

The curves intersect where x 2, which is wherex 4. Use washer cross sections: a washer has inner radius

r x, outer radiusR 2, and areaA(x) (R2 r2) (4 x).

The volume is 40

(4 x) dx 4x 12x2 827.

[0.5, 1.5] by [0.5, 2]

The curves intersect atx 0.7854. A cross section has

radius r 2 secx tanx and area

A(x) r2 ( 2 secx tanx)2. Use NINT to find

0.78540

( 2 secx tanx)2 dx 2.301.

28.

[1, 3] by [1, 3]

The curve and horizontal line intersect atx

2. A cross

section has radius 2 2 sinx and area

A(x) r2 4(1 sinx)2 4(1 2 sinx sin2x).

The volume is

/20

4(1 2 sinx sin2x) dx

432x 2 cosx 1

4sin 2x

434

2 (3 8)

29.

[1, 3] by [1.5, 1.5]

A cross section has radius r 5y2 and area

A(y) r2 5y4.

The volume is 11

5y4 dy y5 2.1

1

/2

0

4

0

/4

/4

2

1

2

1

Section 7.3 303


17/35

30.

[1, 4] by [1, 3]

A cross section has radius ry3/2 and area

A(y) r2 y3. The volume is

20

y3 dy 14y4 4.31.

[1.2, 3.5] by [1, 2.1]

Use washer cross sections. A washer has inner radius r 1,

outer radiusRy 1, and area

A(y) (R2 r2) [(y 1)2 1] (y2 2y). The

volume is 10

(y2 2y) dy 13y3 y2 4

3.

32.

[1.7, 3] by [1, 2.1]

Use cylindrical shells: a shell has radiusx and heightx. The

volume is 10

2(x)(x) dx 2

1

3

x3

2

3

.

33.

[2, 4] by [1, 5]

Use cylindrical shells: A shell has radiusx and heightx2.

The volume is 20

2(x)(x2) dx 214x4 8.34.

[0.5, 1.5] by [0.5, 1.5]

The curves intersect atx 0 andx 1. Use cylindrical

shells: a shell has radiusx and height xx. The volume

is 10

2(x)( xx) dx 225x5/2 1

3x3 21

5.

35.

[1, 5] by [1, 3]

The curved and horizontal line intersect at (4, 2).

(a) Use washer cross sections: a washer has inner radius

r x, outer radiusR 2, and area

A(x) (R2 r2) (4 x). The volume is

40(4x) dx 4x 12x2 8

(b) A cross section has radius ry2 and area

A(y) r2 y4.

The volume is 20

y4 dy 15y5 32

5

.

(c) A cross section has radius r 2 x and area

A(x) r2 (2 x)2 (4 4 xx).

The volume is

40

(4 4 xx) dx 4x 83x3/2 1

2x2 83

.

(d) Use washer cross sections: a washer has inner radius

r 4 y2, outer radiusR 4, and area

A(y) (R2 r2) [16 (4 y2)2]

(8y2 y4).

The volume is

20(8y2 y4) dy 83y3

1

5y5 221

4

5

36.

[1, 3] by [1, 3]

The slanted and vertical lines intersect at (1, 2)

(a) The solid is a right circular cone of radius 1 and

height 2. The volume is

1

3Bh

1

3(r2)h

1

3(12)2

2

3.

(b) Use cylindrical shells: a shell has radius 2x and

height 2x. The volume is

10

2(2 x)(2x) dx 41

0

(2xx2) dx

4 x2 1

3x3 83

.1

0

2

0

4

0

2

0

4

0

1

0

2

0

1

0

1

0

2

0

304 Section 7.3


18/35

37.

[2, 2] by [1, 2]

The curves intersect at (1, 1).

(a) A cross section has radius r 1x2

and area

A(x) r2 (1 x2)2 (1 2x2 x4).

The volume is

11

(1 2x2 x4) dx x 2

3x3

1

5x5 11

6

5

.

(b) Use cylindrical shells: a shell has radius 2y and

height 2 y. The volume is

10

2(2 y)(2 y) dy 41

0

(2 yy3/2) dy

443y3/2 2

5y5/2 51

6

5

.

(c) Use cylindrical shells: a shell has radiusy 1 and

height 2 y. The volume is

10

2(y 1)(2 y) dy 41

0

(y3/2 y) dy

425y5/2 2

3y3/2 61

4

5

.

38. (a) A cross section has radius r h1 bx and area

A(x) r2 h21 bx

2. The volume is

b

0h

2

1b

x

2

dx h2

b

31 b

x

3

3bh2

.

(b) Use cylindrical shells: a shell has radiusx and height

h1 bx. The volume is

b0

2(x)h1 bx dx 2h

b

0x xb

2

dx 2h12x2 3

x

b

3

3b2h.

39.

[2, 3] by [2, 3]

A shell has radiusx and heightx 2x 32x.

The volume is 20

2(x)32x dx x3 8.

40.

[2, 2] by [1, 3]

x2 2 x atx 1. A shell has radiusx and height

2 xx2. The volume is

10

2(x)(2 xx2) dx 2x2 1

3x3

1

4x4 56

.

41.

[1, 5] by [1, 3]

A shell has radiusx and height x. The volume is

40

2(x)( x) dx 225x5/2 12

5

8.

42.

[2, 2] by [2, 2]

The functions intersect where 2x 1 x, i.e., atx 1.

A shell has radiusx and height

x (2x 1) x 2x 1. The volume is

10

2(x)( x 2x 1) dx 21

0

(x3/2 2x2 x) dx

225x5/2 2

3x3

1

2x2

7

1

5.

43. A shell has height 12(y2 y3).

(a) A shell has radiusy. The volume is

10

2(y)12(y2 y3) dy 241

0

(y3 y4) dy

2414y4 1

5y5 65

.

(b) A shell has radius 1y. The volume is

1

0

2(1 y)12(y2 y3) dy

241

0

(y4 2y3 y2) dy

2415y5 1

2y4

1

3y3 45

.1

0

1

0

1

0

4

0

1

0

2

0

b

0

b

0

1

0

1

0

1

1

Section 7.3 305


19/35

43. continued

(c) A shell has radius 8

5y. The volume is

10

285y12(y2 y3) dy 24

1

0y4 15

3y3

8

5y2 dy

24

1

5

y5 1

2

3

0

y4

1

8

5

y3

2.

(d) A shell has radiusy 2

5. The volume is

10

2y 2512(y2 y3) dy 24

1

0y4 5

3y3

2

5y2 dx

2415y5 23

0y4

1

2

5y3 2.

44. A shell has height y

2

2

y44

y

2

2

y2 y44

.

(a) A shell has radiusy. The volume is

20

2(y)y2 y44

dy 214y4 21

4y6 83

.

(b) A shell has radius 2 y. The volume is

20

2(2y)y2 y44

dy 2

2

0 y4

5

y

2

4

y3 2y2 dy 22

1

4y6

1

1

0y5

1

4y4

2

3y3 85

.

(c) A shell has radius 5y. The volume is

20

2(5y)y2 y44

dy 2

2

0y4

5

5

4

y4y3 5y2 dy

221

4y6

1

4y5

1

4y4

5

3y3 8.

(d) A shell has radiusy 5

8. The volume is

20

2y 58y2 y

4

4

dy

2

2

0 y

4

5

5

3

y

2

4

y3

5

8

y2

dy 22

1

4y6

3

1

2y5

1

4y4

2

5

4y3 4.

45.

[1, 3] by [1.4, 9.1]

The functions intersect at (2, 8).

(a) Use washer cross sections: a washer has inner radius

rx3, outer radiusR 4x, and area

A(x) (R2 r2) (16x2 x6). The volume is

20

(16x2 x6) dx 136x3

1

7x7 512

2

1

.

(b) Use cylindrical shells: a shell has a radius 8y and

heighty1/3 4

y. The volume is

80

2(8 y)y1/3 4y dy

2

8

0

8y1/3 2yy4/3

y

4

2

dy

26y4/3 y2 37y7/3 11

2y3 832

2

1

.

46.

[0.5, 1.5] by [0.5, 1.5]

The functions intersect at (0, 0) and (1, 1).

(a) Use cylindrical shells: a shell has radiusx and height

2x

x

2

x

x

x

2

. The volume is

10

2(x)(xx2) dx 213x3 1

4x4 6.

(b) Use cylindrical shells: a shell has radius 1x and

height 2xx2 xx x2. The volume is

10

2(1 x)(xx2) dx 21

0

(x3 2x2 x) dx

214x4 2

3x3

1

2x2

6.

1

0

1

0

8

0

2

0

2

0

2

0

2

0

2

0

1

0

1

0

306 Section 7.3


20/35

47.

[0.5, 2.5] by [0.5, 2.5]

The intersection points are 14, 1, 1

4, 2, and (1, 1).

(a) A washer has inner radius r 14, outer radiusR

y12,

and area (R2 r2) y14 1

1

6. The volume is

21

y14 1

1

6 dy 3

1

y3

1

1

6y 14

1

8

.

(b) A shell has radiusx and height

1

x 1. The volume is

11/4

2(x)

1

x 1 dx 223x3/2

1

2x2 14

1

8

.

48. (a) For 0 x , x f(x) x(si

x

n x) sinx.

Forx 0, x f(x) 0 1 sin 0 sinx. So

x f(x) sinx for 0 x .

(b) Use cylindrical shells: a shell has radiusx and heighty.

The volume is 0

2xy dx, which from part (a) is

0

2 sinx dx 2cosx 4.

49. (a) A cross section has radius r 1

x

2 36x2 and area

A(x) r2 1

44(36x2 x4). The volume is

6

0

1

44(36x2 x4) dx

1

4412x

3 15x5 365

cm3.

(b) 365

cm3(8.5 g/cm3) 192.3 g

50. A cross section has radius r c sinx and area

A(x) r2 (c sinx)2 (c2 2c sinx sin2x).

The volume is

0

(c2 2c sinx sin2x) dx

c2x 2c cosx

1

2x 1

4sin 2x c2 2c 12 2c

2c2 4c

2

2

.

(a) Solve

d

d

c2c2 4c 2

2

022c 4 0

c 2 0

c

2

This value ofc gives a minimum for Vbecause

a

d

2

c

V

2 22 0.

Then the volume is 2

2

2 4

2 2

2

2

2

4

(b) Since the derivative with respect to c is not zero

anywhere else besides c

2, the maximum must occur

at c 0 or c 1. The volume for c 0 is

2

2

4.935,

and for c 1 it is(3

2

8) 2.238. c 0

maximizes the volume.

(c)

[0, 2] by [0, 6]

The volume gets large without limit. This makes sense,

since the curve is sweeping out space in an

ever-increasing radius.

51. (a) Using d C

, andA d22

4

C

2

yields the

following areas (in square inches, rounded to the

nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7,

10.7, 9.3, 6.4, 3.2.

(b) IfC(y) is the circumference as a function ofy, then the

area of a cross section is

A(y) C(y2)/

2

C

4

2

(y),

and the volume is 4

1

60

C2(y) dy.

(c) 4

1

60

A(y) dy 4

1

60

C2(y) dy

4

1

6 2

4

0[5.42 2(4.52 4.42

5.12 6.32 7.82 9.42 10.82 11.62

11.62 10.82 9.02) 6.32] 34.7 in.3

0

6

0

0

1

1/4

2

1

Section 7.3 307


21/35

52. (a) A cross section has radius r 2y and area

r2 2y. The volume is 50

2y dy y2 25.

(b) V(h) A(h) dh, so d

d

V

hA(h).

d

d

V

t

d

d

V

h

d

d

h

tA(h)

d

d

h

t,

so d

d

h

t

A(

1

h)

d

d

V

t

For h 4, the area is 2(4) 8,

so d

d

h

t

8

1

3un

se

it

c

s3 =

8

3

un

se

it

c

s3.

53. (a)

The remaining solid is that swept out by the shaded

region in revolution. Use cylindrical shells: a shell has

radiusx and height 2 r2x2. The volume is

..

2(x)(2 r2x2) dx

223(r2 x2)3/2

4

3

(

8)

32

3

.

(b) The answer is independent ofr.

54. Partition the appropriate interval in the axis of revolution

and measure the radius r(x) of the shadow region at these

points. Then use an approximation such as the trapezoidal

rule to estimate the integral ba

r2(x) dx.

55. Solve axx2 0: This is true atx 0 andx a. For

revolution about thex-axis, a cross section has radius

r axx2 and area

A(x) r2 (axx2)2 (a2x2 2ax3 x4).

The volume is

a

0

(a2x2 2ax3 x4) dx 13a2x3 1

2ax4

1

5x5

3

1

0a5.

For revolution about they-axis, a cylindrical shell has

radiusx and height axx2. The volume is

a0

2(x)(axx2) dx 213ax3 1

4x4 16a4.

Setting the two volumes equal,

3

1

0a5

1

6a4 yields

3

1

0a

1

6, so a 5.

56. The slant height s of a tiny horizontal slice can be

written as s x2y2 1(g(y))2y. So the

surface area is approximated by the Riemann sum

n

k1

2g(yk) 1(g(y))2y. The limit of that is the

integral.

57. g(y) ddxy

21

y

, and

20

2 y1 21

y

2

dy 2

0

4y1dy

6(4y 1)3/2

13

3

13.614

58. g(y) d

d

x

yy2, and

10

2y33

1(y2)2dy 231

6(1 y4)3/2

9(2 2 1) 0.638.

59. g(y) d

d

x

y

1

2y1/2 , and

31

2 y1/2 133/2

1 12y1/22

dy

23

1

y1/2 133/2

1 41y dy.Using NINT, this evaluates to 16.110

1

0

2

0

a

0

a

0

r

r24

r

r222

y

2

2

x

r

5

0

308 Section 7.3


22/35

60. g(y) d

d

x

y 2y

1

1, and

15/8

2 2y11 2y1

12

dy

21

5/8

2y dy

2 223y3/2

4

3

21 1

5

652 2.997.

61. f(x) d

d

y

x 2x, and

20

2x2 1(2x)2 dx 20

2x2 14x2dx evaluates,

using NINT, to 53.226.

62. f(x)

d

d

y

x 3 2x, and

30

2(3xx2) 1(32x)2 dx evaluates, using NINT,

to 44.877.

63. f(x) d

d

y

x

1

2x

x

x2, and

1.50.5

2 2xx21 1

2x

x

x22

dx 21.5

0.5

1 dx

2 x 2 6.283

64. f(x) d

d

y

x

2 x

1

1, and

51

2 x11 2x1

12

dx

25

1 x 5

4 dx

223x 5

4

3/2

4

3

2

4

5

3/2

9

4

3/2

49

3

51.313

65. Hemisphere cross sectional area: ( R2h2)2 A1.

Right circular cylinder with cone removed cross sectional

area: R2 h2 A2

SinceA1A

2, the two volumes are equal by Cavalieris

theorem. Thus, volume of hemisphere

volume of cylinder volume of cone

R3 1

3R3

2

3R3.

66. Use washer cross sections: a washer has inner radius

r b a2y2, outer radiusR b a2y2, and

area (R2 r2)

[(b a2y2)2 (b a2y2)2]

4b a2y2. The volume is

a

a

4b a2y2 dy 4ba

a

a2y2 dy

4b2a2

22a2b

67. (a) Put the bottom of the bowl at (0, a). The area of a

horizontal cross section is ( a2y2)2 (a2 y2).

The volume for height h is

haa

(a2 y2) dy a2y 13y3 h2(3

3

a h).

(b) For h 4, y 1 and the area of a cross section is

(52 12) 24. The rate of rise is

d

d

h

t

A

1

d

d

V

t

24

1

(0.2)12

1

0 m/sec.

68. (a) A cross section has radius r a2x2 and area

A(x) r2 ( a2x2)2 (a2 x2). The

volume is

a

a

(a2 x2) dx

a2x

1

3

x3

a3 13a3 a3

1

3a3

4

3a3.

(b) A cross section has radiusx r1 hy and area

A(y) x2 r21 hy

2 r21 2h

y

h

y2

2. Thevolume is

h0

r21 2hy

h

y2

2 dy r2 y yh2

3

y

h

3

2

1

3r2h.

h

0

a

a

ha

a

5

1

1.5

0.5

1

5/8

Section 7.3 309


23/35

s Section 7.4 Lengths of Curves(pp. 395401)

Quick Review 7.4

1. 12xx2 (1x)2, which, sincex 1, equals1 x orx 1.

2.

1x

x

42

1 2

x

2

, which, sincex 2, equals1

2

x or

2

2

x.

3. 1(tan x)2 (secx)2, which, since 0 x

2,

equals secx.

4. 1 4x 1x2

12 116x2 x12 14(x2

x

24

)2 which,

sincex 0, equalsx2

4

x

4.

5. 1cos2x 2 cos2x, which, since

2 x

2,

equals 2 cosx.

6. f(x) has a corner atx 4.

7. d

d

x(5x2/3)

3

1

3

0

x is undefined atx 0.f(x) has a cusp

there.

8. d

d

x(

5

x3)5(x

1

3)4/5 is undefined forx 3.

f(x) has a vertical tangent there.

9. x24x4 x 2 has a corner atx 2.

10. d

d

x1

3

sin x3(sc

i

o

n

s

x

x

)2/3 is undefined forx k,

where kis any integer.f(x) has vertical tangents at these

values ofx.


1. (a) y 2x, so

Length 21

1(2x)2 dx 21

14x2dx.

(b)

[

1, 2] by [

1, 5](c) Length 6.126

2. (a) y sec2x, so Length 0/3

1sec4x dx.

(b)

3, 0 by [3, 1](c) Length 2.057

3. (a) x cosy, so Length 0

1cos2ydy.

(b)

[1, 2] by [1, 4]

(c) Length 3.820

4. (a) x y(1 y2)1/2, so

Length 1/21/211

y

2

y2 dy.(b)

[1, 2] by [1, 1]

(c) Length 1.047

5. (a) y2 2y 2x 1, so

y2 2y 1 (y 1)2 2x 2, and

y 2x2 1. Theny 2x

1

2, and

Length 711 2x1 2 dx.

(b)

[1, 7] by [2, 4]

(c) Length 9.2946. (a) y cosxx sinx cosxx sinx, so

Length 0

1x2sin2x dx.

(b)

[0, ] by [1, 4]

(c) Length 4.698

7. (a) y tanx, so Length /60

1tan2x dx.

(b) y tanx dx ln (secx)

0, 6 by [0.1, 0.2](c) Length 0.549

310 Section 7.4


24/35

8. (a) x sec2y1, so Length /4/3

secy dy.

(b) x sec2y1 tany,

sox

[2.4, 2.4] by 2,

2

(c) Length 2.198

9. (a) y secx tanx, so

Length /3/3

1 sec2xtan2x dx.

(b)

3,

3 by [1, 3]

(c) Length 3.139

10. y (ex

2

ex), so Length 3

31 ex

2e

x

2

dx.(b)

[3, 3] by [2, 12]

(c) Length 20.036

11. y 1

2(x2 2)1/2(2x) x x22, so the length is

30

1 (xx22)2 dx 30

x42x21dx

30

(x2 1) dx 13x3 x 12.

12. y 3

2 x, so the length is

401 32x

2

dx 4

0 1 94x dx

28

71 9

4

x

3/2

80

2

1

7

0 8.

13. x y2 4

1

y2, so the length is 3

1 1 y2 41y22

dy 3

1 y2 41y22

dy 13y3 41

y 56

3.

14. x y3 4

1

y3, so the length is

21 1 y3 41y3

2

dy 2

1 y3 41y32

dy

14y4 81y2

1

3

2

2

3.

15. x y

2

2

2

1

y2, so the length is

21 1 y2

2

21y22

dy 2

1 y22

21

y22

dy

16y3 21

y 11

7

2.

16. y

x2

2x

1

(4x

4

4)2

(x

1)

2

4(x

1

1)2

so the length is

20 1 (x 1)24(x

1 1)2

2

dx

20 (x 1)24(x

1 1)2

2

dx

13(x 1)3 4(x1

1) 56

3.

2

0

2

1

2

1

3

1

4

0

3

0

3 y 0

0 y

4

ln (cosy)

ln (cosy)

Section 7.4 311


25/35

17. x sec4y1, so the length is

/4/4

1 (sec4y1) dy /4/4

sec2y dy

tany 2.

18. y 3x41, so the length is

1

2

1 (3x41) dx 1

2

3x2 dx

313x3 7

3

3.

19. (a) ddy

x

2corresponds to

4

1

x here, so take

d

d

y

x as

2

1

x. Then

y x C, and, since (1, 1) lies on the curve, C 0.

Soy x.

(b) Only one. We know the derivative of the function and

the value of the function at one value ofx.

20. (a)

d

d

x

y

2

corresponds to

y

1

4

here, so take

d

d

x

y

as

y

1

2

.

Thenx 1

y Cand, since (0, 1) lies on the curve,

C 1. Soy 1

1

x.

(b) Only one. We know the derivative of the function and

the value of the function at one value ofx.

21. y cos2x, so the length is

/40

1 cos2x dx /40

2 cos2xdx

2sinx 1.

22. y (1 x2/3)1/2x1/3, so the length is

81 2/4 1 (1 x2/3)x2/3 dx

81 24 x2/3 dx

81 2/4

x1/3 dx

832x2/3

832 3

212 6.

23. Find the length of the curvey sin 3

2

0x for 0 x 20.

y 3

2

0 cos

3

2

0x, so the length is

200 1 32

0 cos

3

2

0x

2

dx, which evaluates, using NINT,

to 21.07 inches.

24. The area is 300 times the length of the arch.

y 2 sin

50

x, so the length is

25251

2

2

sin250x dx, which evaluates, using NINT,

to 73.185. Multiply that by 300, then by $1.75 to obtain

the cost (rounded to the nearest dollar): $38,422.

25. For track 1: y1

0 atx 10 5 22.3607, and

y1

10

0

0

.2x

0.2x2. NINT fails to evaluate

10 510 5

1 (y1

)2 dx because of the undefined slope at

the limits, so use the tracks symmetry, and back away

from the upper limit a little, and find

2

22.36

0

1

(

y1

)2

dx 52.548. Then, pretending the lastlittle stretch at each end is a straight line, add

2 100 0.2(22.36)2 0.156 to get the total length of

track 1 as 52.704. Using a similar strategy, find the

length of the right halfof track 2 to be 32.274. Now

enter Y1

52.704 and

Y2

32.274 + NINT1 15

00

.2

t

0.2t22

, t, x, 0 andgraph in a [30, 0] by [0, 60] window to see the effect of

thex-coordinate of the lane-2 starting position on the length

of lane 2. (Be patient!) Solve graphically to find the

intersection atx 19.909, which leads to starting point

coordinates (19.909, 8.410).

26. f(x) 1

3x2/3

2

3x1/3, but NINT fails on

20

1 (f(x))2 dx because of the undefined slope at

x 0. So, instead solve forx in terms ofy using the

quadratic formula. (x1/3)2 x1/3 y 0, and

x1/3 1

2

1 4y. Using the positive values,

x 1

8( 1 4y 1)3. Then,

x 3

8( 1 4y 1)2

1

2

4y, and

21/3 2

2/3

0

1 (x)2dy 3.6142.

1

2/4

/4

0

1

2

/4

/4

312 Section 7.4


26/35

27. f(x) , so the

length is 11/21

2

dx which evaluates,


28. There is a corner atx 0:

[2, 2] by [1, 5]

Break the function into two smooth segments:

y and

y .

The length is 12

1 (y)2dy

02

1 (3x25)2dx 10

1 (3x25)2dx,

which evaluates, using NINT for each part, to 13.132.

29. y (1 x)2, 0 x 1

[0.5, 2.5] by [0.5, 1.5]

y

x x

1, but NINT may fail usingy over the entire

interval becausey is undefined atx 0. So, split the curve

into two equal segments by solving x y 1 with

y x: x 1

4. The total length is 21

1/41 xx 12

dx,which evaluates, using NINT, to 1.623.

30.

[0, 16] by [0, 2]

y 1

4x3/4, but NINT may fail usingy over the entire

interval, becausey is undefined atx 0. So, use

x y4, 0 y 2: x 4y3 and the length is

20

1 (4y3)2dy, which evaluates,


31. Because the limit of the sum xkas the norm of the

partition goes to zero will always be the length (b a) of

the interval (a, b).

32. No; the curve can be indefinitely long. Consider, for

example, the curve 1

3

sin

1

x

+ 0.5 for 0 x 1.

33. (a) The fin is the hypotenuse of a right triangle with leg

lengths xk

and d

d

x

fxxk1xkf(xk1)xk.

(b) limnn

k1

(xk

)2(f(xk

1)x

k)2

limnn

k1

xk 1 (f(x

k

1))2

ba

1 (f(x))2 dx

34. Yes. Any curve of the formy x c, where c is a

constant, has constant slope 1, so that

a0

1 (y)2dx a0

2 dx a 2.

s Section 7.5Applications from Scienceand Statistics(pp. 401411)

Quick Review 7.5

1. (a) 10

ex dx ex 1 1e(b) 0.632

2. (a) 10

ex dx ex e 1(b) 1.718

3. (a) /2/4

sinx dx cosx 22

(b) 0.707

4. (a) 30

(x2 2) dx 13x3 2x 15(b) 15

5. (a)

2

1x3

x

2

1dx 13 ln (x

3 1)

1

3[ln 9 ln 2]

1

3ln 92

(b) 0.501

6. 70

2(x 2) sinx dx

7. 70

(1 x2)(2x) dx

2

1

3

0

/2

/4

1

0

1

0

2 x 0

0 x 1

3x2 5

3x2 5

2 x 0

0 x 1

x3 5x

x3 5x

4x2 8x 1

(4x2 1)2

4x2 8x 1

(4x 1)2(4x2 1) (8x2 8x)

(4x2 1)2

Section 7.5 313


27/35

8. 70

cos2x dx

9. 70

2y

2(10 y) dy

10. 70

4

3 sin2x dx


1. 50

xex/3 dx 3ex/3(3 x) e25

4/3 9 4.4670 J

2. 30

x sin 4x dx

4

4 sin 4

x x cos 4

x

4

2

2

3

2

3.8473 J

3. 30

x 9 x2 dx 13(9 x2)3/2 9 J

4.

10

0 (esinx

cosx 2) dx esinx

2x esin 10 19 19.5804 J

5. When the bucket isx m off the ground, the water weighs

F(x) 490202

0

x 4901 2

x

0 490 24.5x N.

Then

W 200

(490 24.5x) dx 490x 12.25x2 4900 J.6. When the bucket isx m off the ground, the water weighs

F(x) 49020 204x/5 4901 2

x

5 490 19.6x N.

Then

W 200

(490 19.6x) dx 490x 9.8x2 5880 J.7. When the bag isx ft off the ground, the sand weighs

F(x) 14418 18x/2 1441 3

x

6 144 4x lb.

Then

W 180

(144 4x) dx 144x 2x2 1944 ft-lb8. (a) F ks, so 800 k(14 10) and k 200 lb/in.

(b) F(x) 200x, and 20

200x dx

100x2

400 in.-lb.

(c) F 200s, so s 1

2

6

0

0

0

0 8 in.

9. (a) F ks, so 21,714 k(8 5) and k 7238 lb/in.

(b) F(x) 7238x. W 1/20

7238x dx 3619x2 904.75 905 in.-lb, and W 1

1/2

7238x dx

3619x2 2714.25 2714 in.-lb.

10. (a) F ks, so 150 k11

6 and k 2400 lb/in. Then for

s 1

8, F 240018 300 lb.

(b) 1/80

2400x dx 1200x2 18.75 in.-lb

11. When the end of the rope is x m from its starting point, the

(50 x) m of rope still to go weigh

F(x) (0.624)(50 x) N. The total work is

500

(0.624)(50 x) dx 0.62450x 12x2 780 J.

12. (a) Work (p2, V2)(p

1, V

1)

F(x) dx (p2, V2)(p

1, V

1)

pA dx (p2, V2)(p

1, V

1)

p dV

(b) p1V

11.4

(50)(243)1.4 109,350, so p 10

V

91

,3.4

50 and

Work (p2, V2)

(p1, V

1)

10V9

1,3.450

dV

109,3502.5V0.4 37,968.75 in.-lb

13. (a) From the equationx2 y2 32, it follows that a thin

horizontal rectangle has area 2 9 y2 y, wherey is

distance from the top, and pressure 62.4y. The total

force is approximatelyn

k1

(62.4yk

)(2 9 yk

2) y

n

k1

124.8yk 9 y

k2y.

(b) 30

124.8y 9 y2 dy 41.6(9 y2)3/2 1123.2 lb

14. (a) From the equation 3

x2

2 8

y2

2 1, it follows that a thin

horizontal rectangle has area 61 6y42

y, wherey isdistance from the top, and pressure 62.4y. The total

force is approximately

n

k1

(62.4yk)61

y

6k42

y n

k1

374.4yk1

y

6k42

y.

(b) 80

374.4y1 6y42

dy 7987.21 6y

4

2

3/2

7987.2 lb

8

0

3

0

V 32

V 243

50

0

1/8

0

1

1/2

1/2

0

2

0

18

0

20

0

20

0

10

0

3

0

3

0

5

0

314 Section 7.5


28/35

15. (a) From the equationx 3

8y, it follows that a thin

horizontal rectangle has area 3

4yy, wherey is the

distance from the top of the triangle, the pressure is

62.4(y 3). The total force is approximately

n

k

1

62.4(yk 3)34yk y

n

k

1

46.8(yk2 3y

k) y.

(b) 83

46.8(y22 3y) dy 15.6y3 70.2y2

3494.4 (210.6) 3705 lb

16. (a) From the equationy x

2

2

, it follows that a thin

horizontal rectangle has area 2 2yy, wherey is

distance from the bottom, and pressure 62.4(4y).

The total force is approximately

n

k1

62.4(4 yk)(2 2y

k)y

n

k1124.8 2(4 yk

yk3/2

)y.

(b) 40

124.8 2(4 yy3/2) dy

124.8 283y3/2 2

5y5/2

1064.96 2 1506.1 lb

17. (a) Work to raise a thin slice 62.4(10 12)(y)y.

Total work 200

62.4(120)y dy 62.460y2 1,497,600 ft-lb

(b) (1,497,600 ft-lb) (250 ft-lb/sec) 5990.4 sec

100 min

(c) Work to empty half the tank 100

62.4(120)y dy

62.460y2 374,400 ft-lb, and374,400 250 1497.6 sec 25 min

(d) The weight per ft3 of water is a simple multiplicative

factor in the answers. So divide by 62.4 and multiply

by the appropriate weight-density

For 62.26:

1,497,600662

2

.

.

2

4

6 1,494,240 ft-lb and

5990.4662

2

.

.

2

4

6 5976.96 sec 100 min.

For 62.5:

1,497,600662

2

.

.

5

4 1,500,000 ft-lb and

5990.4662

2

.

.

5

4 6000 sec 100 min.

18. The work needed to raise a thin disk is (10)2(51.2)yy,

wherey is height up from the bottom. The total work is

300

100(51.2)y dy 512012y2 7,238,229 ft-lb

19. Work to pump through the valve is (2)2(62.4)(y 15)y

for a thin disk and

60

4(62.4)(y 15) dy 249.612y2 15y 84,687.3 ft-lb

for the whole tank. Work to pump over the rim is

(2)2(62.4)(6 15)y for a thin disk and

60

4(62.4)(21) dy 4(62.4)(21)(6) 98,801.8 ft-lb for

the whole tank. Through a hose attached to a valve in the

bottom is faster, because it takes more time for a pump with

a given power output to do more work.

20. The work is the same as if the straw were initially an inch

long and just touched the surface, and lengthened as the

liquid level dropped. For a thin disk, the volume is

y1417.5

2y and the work to raise it is

y1417.5

2

49(8y)y. The total work is

7

0

y

14

17.5

2

4

9

(8

y) dy, which using NINT evaluates

to 91.3244 in.-oz.

21. The work is that already calculated (to pump the oil to the

rim) plus the work needed to raise the entire amount 3 ft

higher. The latter comes to

13r2h(57)(3) 57(4)2(8) 22,921.06 ft-lb, and thetotal is 22,921.06 30,561.41 53,482.5 ft-lb.

22. The weight density is a simple multiplicative factor: Divide

by 57 and multiply by 64.5.

30,561.416547.5 34,582.65 ft-lb.

6

0

30

0

10

0

20

0

4

0

8

3

Section 7.5 315


29/35

23. The work to raise a thin disk is

r2(56)h ( 102y2)2(56)(10 2 y)y

56(12 y)(100 y2)y. The total work is

100

56(12 y)(100 y2) dy, which evaluates using NINT

to 967,611 ft-lb. This will come to

(967,611)($0.005) $4838, so yes, theres enough moneyto hire the firm.

24. Pipe radius 1

6 ft;

Work to fill pipe 3600

162(62.4)y dy 112,320 ft-lb.

Work to fill tank 385360

(10)2(62.4)y dy

58,110,000ft-lb.

Total work 58,222,320 ft-lb, which will take

58,222,320 1650 110,855 sec 31 hr.

25. (a) The pressure at depthy is 62.4y, and the area of a thin

horizontal strip is 2y. The depth of water is 1

6

1 ft, so

the total force on an end is

11/60

(62.4y)(2 dy) 209.73 lb.

(b) On the sides, which are twice as long as the ends, the

initial total force is doubled to 419.47 lb. When the

tank is upended, the depth is doubled to 1

3

1 ft, and the

force on a side becomes 11/30

(62.4y)(2 dy) 838.93 lb,

which means that the fluid force doubles.

26. 3.75 in. 1

5

6 ft, and 7.75 in.

3

4

1

8 ft.

Force on a side p dA 31/480

(64.5y)156 dy 4.2 lb.27. (a) 0.5 (50%), since half of a normal distribution lies

below the mean.

(b) Use NINT to find

65

63f(x) dx, where

f(x) 3.2

1

2e(x63.4)

2/(2 3.2

2). The result is

0.24 (24%).

(c) 6 ft 72 in. Pick 82 in. as a conveniently high upper

limit and with NINT, find 8272

f(x) dx. The result is

0.0036 (0.36%).

(d) 0 if we assume a continuous distribution. Between

59.5 in. and 60.5 in., the proportion is

60.559.5

f(x) dx 0.071 (7.1%)

28. Usef(x) 100

1

2e(x498)

2/(2 1002)

(a)

500

400f(x) dx 0.34 (34%)

(b) Take 1000 as a conveniently high upper limit:

1000700

f(x) dx 0.217, which means about

0.217(300) 6.5 people

29. Integration is a good approximation to the area (which

represents the probability), since the area is a kind of

Riemann sum.

30. The proportion of lightbulbs that last between 100 and 800

hours.

31.

35,780,000

6,370,000 100

r02MG

dr 1000MG

1

r

, which for

M 5.975 1024, G 6.6726 1011 evaluates to

5.1446 1010 J.

32. (a) The distance goes from 2 m to 1 m. The work by an

external force equals the work done by repulsion in

moving the electrons from a 1-m distance to a 2-m

distance:

Work 2

1

23 r120

29

dr

23 10291r 1.15 1028 J

(b) Again, find the work done by the fixed electrons in

pushing the third one away. The total work is the sum

of the work by each fixed electron. The changes in

distance are 4 m to 6 m and 2 m to 4 m, respectively.

Work 64

23

r

12

029 dr 4

2

23

r

12

029 dr

23 1029

1

r

1

r

7.6667 1029 J.

4

2

6

4

2

1

35,780,000

6,370,000

316 Section 7.5


30/35

33. F mddvt mvddvx, soW x2

x1

F(x) dx

x2x

1

mvddvx dx v2

v

1

mv dv

1

2mv

22

1

2mv

12

34. Work Change in kinetic energy 1

2mv2.

m32

2

ft

o

/s

z

ec2

32

1/

f

8

t/s

l

e

b

c2

2

1

56 slug, so

Work 1

22

1

56(160)2 50 ft-lb.

35. 0.3125 lb 3

0

2

.31

ft

2

/s

5

ec

lb2 0.009765625 slug, and

90 mph 905218

m

0

i

ft36

1

00

hr

sec 132 ft/sec, so

Work change in kinetic energy 12(0.009765625)(132)2

85.1 ft-lb.

36. 1.6 oz 1.6 oz116lobz/(32 ft/sec2) 0.003125 slug, soWork

1

2(0.003125)(280)2 122.5 ft-lb.

37. 2 oz 2 oz116lobz/(32 ft/sec2) 2156 slug, and124 mph 124 mph521

8

m

0

i

ft36

1

00

hr

sec 181.867 ft/sec,

so Work 1

22

1

56(181.867)2 64.6 ft-lb.

38. 14.5 oz 14.5 oz11

6

l

o

b

z/(32 ft/sec2) 0.02832 slug, so

Work 1

2(0.02832)(88)2 109.7 ft-lb.

39. 6.5 oz 6.5 oz11

6

l

o

b

z/(32 ft/sec2) 0.01270 slug, so

Work 1

2(0.01270)(132)2 110.6 ft-lb.

40. 2 oz 1

8 lb

2

1

56 slug. Compression energy of spring

1

2ks2

1

2(18)14

2 0.5625 ft-lb, and final height is

given by mgh 0.5625 ft-lb, so h

(1/

0

2

.

5

5

6

6

)

2

(

5

32)

4.5 ft.

s Chapter 7 Review Exercises

(pp. 413415)

1. 50

v(t) dt 50

(t2 0.2t3) dt

13t3 0.05t4 10.417 ft

2.

7

0 c(t) dt 7

0 (4 0.001t4

) dt

4t 0.0002t5 31.361 gal

3. 1000

B(x) dx 1000

(21 e0.03x) dx

21x 33.333e0.03x 1464

4. 20

(x) dx 20

(11 4x) dx 11x 2x2 14 g

5.

24

0 E(t) dt

24

0 3002

cos

1

2

t

dt

3002t 12 sin 12t 14,4006.

[1, 3] by [1, 2]

The curves intersect atx 1. The area is

1

0

x dx 2

1

x12 dx

12x2

1x

1

2 12 1 1.

7.

[4, 4] by [4, 4]

The curves intersect atx 2 andx 1. The area is

1

2

[3 x2 (x 1)] dx 1

2

(x2 x 2) dx

13x3 12x2 2x

13 12 2 83 2 4

9

2.

1

2

2

1

1

0

24

0

2

0

100

0

7

0

5

0

Chapter 7 Review 317


31/35

8. x y 1 impliesy (1 x)2 1 2 xx.

[0.5, 2] by [0.5, 1]

The area is

1

0(1 2 xx) dx

x 43x3/2 1

2x2

1

6.

9. x 2y2 impliesy 2x.

[1, 19] by [1, 4]

The curves intersect atx 18. The area is

180 3 2

x dx 3x 432x

3/2

18,or 3

02y2 dy 23y3 18.

10. 4xy2 4 impliesx 1

4y2 1, and 4xy 16 implies

x 1

4y 4.

[6, 6] by [6, 6]

The curves intersect at (3, 4) and (5.25, 5). The area is

54

1

4y 4 14y2 1 dy

54

1

4y2

1

4y 5 dy

112y3 18y2 5y

4

2

2

4

5 338 2483 30.375.

11.

[0.1, 1] by [0.1, 1]

The area is /40

(x sinx) dx 12x2 cosx

32

2

2

2 1

0.0155.

12.

2, 3

2

by [3,3]

The area is

0

(2 sinx sin 2x) dx 2 cosx 12 cos 2x 4.13.

[5, 5] by [5, 5]

The curves intersect atx 2.1281. The area is

2.12812.1281

(4x2 cosx) dx,

which using NINT evaluates to 8.9023.

14.

[4, 4] by [4, 4]

The curves intersect atx 0.8256. The area is

0.82560.8256

(3 x sec2x) dx,


15. Solve 1 cosx 2 cosx for thex-values at the two

ends of the region: x 2

3, i.e.,

5

3

or

7

3

. Use the

symmetry of the area:

27/32

[(1 cosx) (2 cosx)] dx

27/32

(2 cosx 1) dx

22 sinxx 2 3

2

3 1.370.

16. 5/3

/3

[(2 cosx) (1 cosx)] dx

5/3/3

(1 2 cosx) dx

x 2 sinx 2 3

4

3 7.653

5/3

/3

7/3

2

0

/4

0

5

4

3

0

18

0

1

0

318 Chapter 7 Review


32/35

17. Solvex3 x x2

x

1 to find the intersection points at

x 0 andx 21/4. Then use the areas symmetry:

the area is

221/4

0 x2 x

1 (x3 x) dx

2

12 ln (x2 1) 14x

4 12x2

ln ( 2 1) 2 1 1.2956.

18. Use the intersect function on a graphing calculator to

determine that the curves intersect atx 1.8933.

The area is

1.89331.893331x

2

x2

1

0

3 dx,


19. Use thex- andy-axis symmetries of the area:

40

x sinx dx 4sinx x cosx 4.20. A cross section has radius r 3x4 and area

A(x) r2 9x8.

V 11

9x8 dx x9 2.21.

[5, 5] by [5, 5]

The graphs intersect at (0, 0) and (4, 4).

(a) Use cylindrical shells. A shell has radiusy and height

y y

4

2

. The total volume is

40

2(y) y y42

dy 24

0 y2 y

4

3

dy 213y3 116y4

32

3

.

(b) Use cylindrical shells. A shell has radiusx and height

2 xx. The total volume is

40

2(x)(2 xx) dx 24

0

(2x3/2 x2) dx

245x5/2 13x3

12

1

8

5

.

(c) Use cylindrical shells. A shell has radius 4 x and

height 2 xx. The total volume is

40

2(4 x)(2 xx) dx

24

0

(8 x 4x 2x3/2 x2) dx

2136x3/2 2x2 45x5/2 13x3 645.

(d) Use cylindrical shells. A shell has radius 4 y and

heighty y

4

2

. The total volume is

40

2(4 y)y y42

dy 2

4

0 4y 2y2 y

4

3

dy 22y2 23y3 116y4 323.

22. (a) Use disks. The volume is

2

0 ( 2y)2 dy

2

0 2y dy y2 4.

(b) k

02y dy y2 k2

(c) Since V k2, d

d

V

t 2k

d

d

k

t.

When k 1, d

d

k

t

2

1

k

d

d

V

t 2

1

(2)

1,

so the depth is increasing at the rate of

1 unit per

second.

23. The football is a solid of revolution about thex-axis. A

cross section has radius 121 142x12

and arear2 121 142x1

2

. The volume is, given the symmetry,211/2

0

121 142x12

dx 2411/2

0 1 14

2

x

1

2

dx 24x 121

2

13x3 24121 161 88 276 in3.

24. The width of a cross section is 2 sinx, and the area is

1

2r2

1

2 sin2x. The volume is

0

1

2 sin2x dx

212x 14sin 2x 4

2

.

0

11/2

0

k

0

2

0

4

0

4

0

4

0

4

0

1

1

0

21/4

0

Chapter 7 Review 319


33/35

25.

[1, 2] by [1, 2]

Use washer cross sections. A washer has inner radius r 1,

outer radiusR ex/2, and area (R2 r2) (ex 1).

The volume is

ln 30

(ex 1) dx ex x (3 ln 3 1)

(2 ln 3).

26. Use cylindrical shells. Taking the hole to be vertical, a shell

has radiusx and height 2 22x2. The volume of the

piece cut out is

3

0

2(x)(2 22x2) dx 2 3

0

2x 4x2 dx

223(4 x2)3/2

4

3(1 8)

28

3

29.3215 ft3.

27. The curve crosses thex-axis atx 3.y 2x, so the

length is 33

1(2x)2 dx 33

14x2dx, which

using NINT evaluates to 19.4942.

28.

[2, 2] by [2, 2]

The curves intersect atx 0 andx 1. Use the graphs

x- andy-axis symmetry:

d

d

x(x3 x) 3x2 1, and the total perimeter is

4

1

0

1(3x21)2dx, which using NINT evaluates to

5.2454.

29.

[1, 3] by [3, 3]

y 3x2 6x equals zero whenx 0 or 2. The maximum

is atx 0, the minimum at x = 2. The distance between

them along the curve is 20

1(3x26x)2 dx, which

using NINT evaluates to 4.5920. The time taken is about

4.5

2

920 2.296 sec.

30. If (b) were true, then the curvey ksinx would have to

get vanishingly short as kapproached zero. Since in fact the

curves length approaches 2 instead, (b) is false and (a) is

true.

31. F(x) x41, so

5

2 1(F(x))2dx

5

2 x4dx

52

x2 dx

13x3 39.32. (a) (100 N)(40 m) = 4000 J

(b) When the end has traveled a distancey, the weight of

the remaining portion is (40 y)(0.8) 32 0.8y.

The total work to lift the rope is

40

0

(32 0.8y) dy

32y 0.4y2

640 J.

(c) 4000 640 4640 J

33. The weight of the water at elevationx (starting fromx 0)

is (800)(8)47540750x/2 192584750 12x. The total workis 4750

0

1

9

2

5

84750 12x dx 192584750x 14x2

22,800,000 ft-lb.

34. F ks, so k Fs

08.03 80

30 N/m. Then

Work 0.30

80

3

0x dx 806

0x2 12 J.

To stretch the additional meter,

Work 1.30.3

80

3

0x dx 806

0x2 213.3 J.

35. The work is positive going uphill, since the force pushes in

the direction of travel. The work is negative going downhill.

1.3

0.3

0.3

0

4750

0

40

0

5

2

3

0

ln 3

0

320 Chapter 7 Review


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36. The radius of a horizontal cross section is 82y2, where

y is distance below the rim. The area is (64 y2), the

weight is 0.04(64 y2) y, and the work to lift it over

the rim is 0.04(64 y2)(y) y. The total work is

82

0.04y(64 y2) dy 0.048

2

(64y y3) dy

0.0432y2 14y4 36 113.097 in.-lb.

37. The width of a thin horizontal strip is

chapter 7 soulutions

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