6 APPLICATIONS OFINTEGRATION

6 APPLICATIONS OF INTEGRATION6.1 Areas Between Curves

1. A =x=4

x=0

(yT − yB) dx =4

0

(5x− x2)− x dx =4

0

(4x− x2) dx = 2x2 − 13x

3 4

0= 32− 64

3− (0) = 32

3

3. A =y=1

y=0

(xR − xL) dy =1

0

y − (y2 − 1) dy =1

0

(y1/2 − y2 + 1) dy = 23y3/2 − 1

3y3 + y

1

0

= 23− 1

3+ 1 − (0) = 4

3

5. A =2

−1(9− x2)− (x+ 1) dx

=2

−1(8− x− x2) dx

= 8x− x2

2− x3

3

2

−1

= 16− 2− 83− −8− 1

2+ 1

3

= 22− 3 + 12 =

392

7. The curves intersect when x = x2 ⇔ x2 − x = 0 ⇔x(x− 1) = 0 ⇔ x = 0 or 1.

A =1

0

(x− x2) dx = 12x2 − 1

3x3

1

0= 1

2− 1

3= 1

6

9. First find the points of intersection:√x+ 3 =

x+ 3

2⇒ √

x+ 32=

x+ 3

2

2

⇒ x+ 3 = 14 (x+ 3)

2 ⇒

4(x+ 3)− (x+ 3)2 = 0 ⇒ (x+ 3)[4− (x+ 3)] = 0 ⇒ (x+ 3)(1− x) = 0 ⇒ x = −3 or 1. So

A=1

−3

√x+ 3− x+ 3

2dx

= 23 (x+ 3)

3/2 − (x+ 3)2

4

1

−3

= 163 − 4 − (0− 0) = 4

3

213

214 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION

11. A =1

0

√x− x2 dx

= 23x3/2 − 1

3x3

1

0

= 23 − 1

3 =13

13. 12− x2 = x2 − 6 ⇔ 2x2 = 18 ⇔x2 = 9 ⇔ x = ±3, so

A=3

−3(12− x2)− (x2 − 6) dx

= 23

0

18− 2x2 dx [by symmetry]

= 2 18x− 23x3

3

0= 2 [(54− 18)− 0]

= 2(36) = 72

15. The curves intersect when 8 cosx = sec2 x ⇒ 8 cos3 x = 1 ⇒ cos3 x = 18 ⇒ cosx = 1

2 ⇒x = π

3for 0 < x < π

2. By symmetry,

A= 2π/3

0

(8 cosx− sec2 x) dx

= 2 8 sinx− tanx π/3

0

= 2 8 ·√32−√3 = 2 3

√3

= 6√3

17. 12x =

√x ⇒ 1

4x2 = x ⇒ x2 − 4x = 0 ⇒ x(x− 4) = 0 ⇒ x = 0 or 4, so

A=4

0

√x− 1

2x dx+

9

4

12x−√x dx = 2

3x3/2 − 1

4x2

4

0+ 1

4x2 − 2

3x3/2

9

4

= 163− 4 − 0 + 81

4− 18 − 4− 16

3= 81

4+ 32

3− 26 = 59

12

SECTION 6.1 AREAS BETWEEN CURVES ¤ 215

19. 2y2 = 4 + y2 ⇔ y2 = 4 ⇔ y = ±2, so

A=2

−2(4 + y2)− 2y2 dy

= 22

0

(4− y2) dy [by symmetry]

= 2 4y − 13y3

2

0= 2 8− 8

3= 32

3

21. The curves intersect when 1− y2 = y2 − 1 ⇔ 2 = 2y2 ⇔ y2 = 1 ⇔ y = ±1.

A =1

−1(1− y2)− (y2 − 1) dy

=1

−12(1− y2) dy

= 2 · 21

0

(1− y2) dy

= 4 y − 13y3

1

0= 4 1− 1

3= 8

3

23. Notice that cosx = sin 2x = 2 sinx cosx ⇔2 sinx cosx− cosx = 0 ⇔ cosx (2 sinx− 1) = 0 ⇔2 sinx = 1 or cosx = 0 ⇔ x = π

6 or π2 .

A=π/6

0

(cosx− sin 2x) dx+π/2

π/6

(sin 2x− cosx) dx

= sinx+ 12cos 2x

π/6

0+ − 1

2cos 2x− sinx π/2

π/6

= 12+ 1

2· 12− 0 + 1

2· 1 + 1

2− 1 − − 1

2· 12− 1

2= 1

2

25. From the graph, we see that the curves intersect at x = 0, x = π2 , and x = π. By symmetry,

A =π

0

cosx− 1− 2x

πdx = 2

π/2

0

cosx− 1− 2x

πdx = 2

π/2

0

cosx− 1 + 2x

πdx

= 2 sinx− x+ 1πx2

π/2

0= 2 1− π

2+ 1

π· π24

− 0 = 2 1− π2+ π

4= 2− π

2


27. Graph the three functions y = 1/x2, y = x, and y = 18x; then determine the points of intersection: (0, 0), (1, 1), and 2, 14 .

A=1

0

x− 1

8x dx+

2

1

1

x2− 1

8x dx

=1

078xdx+

2

1x−2 − 1

8x dx

=7

16x2

1

0

+ − 1x− 1

16x2

2

1

= 716+ − 1

2− 1

4− −1− 1

16= 3

4

29. An equation of the line through (0, 0) and (2, 1) is y = 12x; through (0, 0)

and (−1, 6) is y = −6x; through (2, 1) and (−1, 6) is y = − 53x+

133 .

A=0

−1− 53x+ 13

3− (−6x) dx+

2

0

− 53x+ 13

3− 1

2x dx

=0

−1133x+ 13

3dx+

2

0

− 136x+ 13

3dx

= 133

0

−1(x+ 1) dx+ 13

3

2

0

− 12x+ 1 dx

= 133

12x2 + x

0

−1 +133− 14x2 + x

2

0

= 1330− 1

2− 1 + 13

3[(−1 + 2)− 0] = 13

3· 12+ 13

3· 1 = 13

2

31. The curves intersect when sinx = cos 2x (on [0, π/2]) ⇔ sinx = 1− 2 sin2 x ⇔ 2 sin2 x+ sinx− 1 = 0 ⇔(2 sinx− 1)(sinx+ 1) = 0 ⇒ sinx = 1

2⇒ x = π

6.

A=π/2

0

|sinx− cos 2x| dx

=π/6

0

(cos 2x− sinx) dx+π/2

π/6

(sinx− cos 2x) dx

= 12 sin 2x+ cosx

π/6

0+ − cosx− 1

2 sin 2xπ/2

π/6

= 14

√3 + 1

2

√3 − (0 + 1) + (0− 0)− − 1

2

√3− 1

4

√3

= 32

√3− 1

33. Let f(x) = cos2 πx

4− sin2 πx

4and ∆x =

1− 04

.

The shaded area is given by

A =1

0f(x) dx ≈M4

= 14f 1

8+ f 3

8+ f 5

8+ f 7

8

≈ 0.6407


35. From the graph, we see that the curves intersect at x = 0 and x = a ≈ 0.896, with

x sin(x2) > x4 on (0, a). So the area A of the region bounded by the curves is

A=a

0

x sin(x2)− x4 dx = − 12 cos(x

2)− 15x

5 a

0

= − 12cos(a2)− 1

5a5 + 1

2≈ 0.037

37. From the graph, we see that the curves intersect at

x = a ≈ −1.11, x = b ≈ 1.25, and x = c ≈ 2.86, with

x3 − 3x+ 4 > 3x2 − 2x on (a, b) and 3x2 − 2x > x3 − 3x+ 4on (b, c). So the area of the region bounded by the curves is

A=b

a

(x3 − 3x+ 4)− (3x2 − 2x) dx+c

b

(3x2 − 2x)− (x3 − 3x+ 4) dx

=b

a

(x3 − 3x2 − x+ 4) dx+c

b

(−x3 + 3x2 + x− 4) dx

= 14x4 − x3 − 1

2x2 + 4x

b

a+ − 1

4x4 + x3 + 1

2x2 − 4x c

b≈ 8.38

39. As the figure illustrates, the curves y = x and y = x5 − 6x3 + 4xenclose a four-part region symmetric about the origin (since

x5 − 6x3 + 4x and x are odd functions of x). The curves intersect

at values of x where x5 − 6x3 + 4x = x; that is, where

x(x4 − 6x2 + 3) = 0. That happens at x = 0 and where

x2 =6±√36− 12

2= 3±√6; that is, at x = − 3 +

√6, − 3−√6, 0, 3−√6, and 3 +

√6. The exact area is

2

√3+√6

0

(x5 − 6x3 + 4x)− x dx = 2

√3+√6

0

x5 − 6x3 + 3x dx

= 2

√3−√6

0

(x5 − 6x3 + 3x) dx+ 2√3+√6

√3−√6

(−x5 + 6x3 − 3x) dx

CAS= 12

√6− 9


41. 1 second = 13600 hour, so 10 s = 1

360 h. With the given data, we can take n = 5 to use the Midpoint Rule.

∆t = 1/360−05

= 11800

, so

distance Kelly − distance Chris =1/360

0vK dt− 1/360

0vC dt =

1/360

0(vK − vC) dt

≈M5 =1

1800[(vK − vC)(1) + (vK − vC)(3) + (vK − vC)(5)

+ (vK − vC)(7) + (vK − vC)(9)]

= 11800 [(22− 20) + (52− 46) + (71− 62) + (86− 75) + (98− 86)]

= 11800 (2 + 6 + 9 + 11 + 12) =

11800 (40) =

145 mile, or 117 13 feet

43. Let h(x) denote the height of the wing at x cm from the left end.

A ≈M5 =200− 05

[h(20) + h(60) + h(100) + h(140) + h(180)]

= 40(20.3 + 29.0 + 27.3 + 20.5 + 8.7) = 40(105.8) = 4232 cm2

45. We know that the area under curve A between t = 0 and t = x is x

0vA(t) dt = sA(x), where vA(t) is the velocity of car A

and sA is its displacement. Similarly, the area under curve B between t = 0 and t = x is x

0vB(t) dt = sB(x).

(a) After one minute, the area under curve A is greater than the area under curve B. So car A is ahead after one minute.

(b) The area of the shaded region has numerical value sA(1)− sB(1), which is the distance by which A is ahead of B after

1 minute.

(c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curve A from t = 0

to t = 2 is still greater than the corresponding area for curve B, so car A is still ahead.

(d) From the graph, it appears that the area between curves A and B for 0 ≤ t ≤ 1 (when car A is going faster), which

corresponds to the distance by which car A is ahead, seems to be about 3 squares. Therefore, the cars will be side by side

at the time x where the area between the curves for 1 ≤ t ≤ x (when car B is going faster) is the same as the area for

0 ≤ t ≤ 1. From the graph, it appears that this time is x ≈ 2.2. So the cars are side by side when t ≈ 2.2 minutes.

47. To graph this function, we must first express it as a combination of explicit

functions of y; namely, y = ±x√x+ 3. We can see from the graph that the loop

extends from x = −3 to x = 0, and that by symmetry, the area we seek is just

twice the area under the top half of the curve on this interval, the equation of the

top half being y = −x√x+ 3. So the area is A = 2 0

−3 −x√x+ 3 dx. We

substitute u = x+ 3, so du = dx and the limits change to 0 and 3, and we get

A= −2 3

0[(u− 3)√u ] du = −2 3

0(u3/2 − 3u1/2) du

= −2 25u5/2 − 2u3/2

3

0= −2 2

532√3 − 2 3√3 = 24

5

√3


49. By the symmetry of the problem, we consider only the first quadrant, where

y = x2 ⇒ x = y. We are looking for a number b such that

b

0

y dy =4

b

y dy ⇒ 23y3/2

b

0= 2

3y3/2

4

b⇒

b3/2 = 43/2 − b3/2 ⇒ 2b3/2 = 8 ⇒ b3/2 = 4 ⇒ b = 42/3 ≈ 2.52.

51. We first assume that c > 0, since c can be replaced by −c in both equations without changing the graphs, and if c = 0 the

curves do not enclose a region. We see from the graph that the enclosed area A lies between x = −c and x = c, and by

symmetry, it is equal to four times the area in the first quadrant. The enclosed area is

A = 4c

0(c2 − x2) dx = 4 c2x− 1

3x3

c

0= 4 c3 − 1

3c3 = 4 2

3c3 = 8

3c3

So A = 576 ⇔ 83c3 = 576 ⇔ c3 = 216 ⇔ c = 3

√216 = 6.

Note that c = −6 is another solution, since the graphs are the same.

53. A=2

1

1

x− 1

x2dx = lnx+

1

x

2

1

= ln 2 + 12− (ln 1 + 1)

= ln 2− 12≈ 0.19

55. The curves intersect when tanx = 2 sinx (on [−π/3, π/3]) ⇔ sinx = 2 sinx cosx ⇔

2 sinx cosx− sinx = 0 ⇔ sinx (2 cosx− 1) = 0 ⇔ sinx = 0 or cosx = 12⇔ x = 0 or x = ±π

3.

A=π/3

−π/3(2 sinx− tanx) dx

= 2π/3

0

(2 sinx− tanx) dx [by symmetry]

= 2 −2 cosx− ln |secx| π/3

0

= 2 [(−1− ln 2)− (−2− 0)]= 2(1− ln 2) = 2− 2 ln 2


6.2 Volumes

1. A cross-section is a disk with radius 2− 12x, so its area is A(x) = π 2− 1

2x

2.

V =2

1

A(x) dx =2

1

π 2− 12x

2dx

= π2

1

4− 2x+ 14x

2 dx

= π 4x− x2 + 112x3

2

1

= π 8− 4 + 812

− 4− 1 + 112

= π 1 + 712

= 1912π

3. A cross-section is a disk with radius 1/x, so its area is

A(x) = π(1/x)2.

V =2

1

A(x) dx =2

1

π1

x

2

dx

= π2

1

1

x2dx = π − 1

x

2

1

= π − 12− (−1) = π

2

5. A cross-section is a disk with radius 2 y, so its area is

A(y) = π 2 y2

.

V =9

0

A(y) dy =9

0

π 2 y2

dy = 4π9

0

y dy

= 4π 12y2

9

0= 2π(81) = 162π

7. A cross-section is a washer (annulus) with inner

radius x3 and outer radius x, so its area is

A(x) = π(x)2 − π(x3)2 = π(x2 − x6).

V =1

0

A(x) dx =1

0

π(x2 − x6) dx

= π 13x

3 − 17x

7 1

0= π 1

3 − 17= 4

21π

SECTION 6.2 VOLUMES ¤ 221

9. A cross-section is a washer with inner radius y2

and outer radius 2y, so its area is

A(y) = π(2y)2 − π(y2)2 = π(4y2 − y4).

V =2

0

A(y) dy = π2

0

(4y2 − y4) dy

= π 43y3 − 1

5y5

2

0= π 32

3− 32

5= 64

15π

11. A cross-section is a washer with inner radius 1−√x and outer radius 1− x, so its area is

A(x) = π(1− x)2 − π 1−√x

2

= π (1− 2x+ x2)− 1− 2√x+ x

= π −3x+ x2 + 2√x .

V =1

0

A(x) dx = π1

0

−3x+ x2 + 2√x dx

= π − 32x2 + 1

3x3 + 4

3x3/2

1

0= π − 3

2+ 5

3= π

6

13. A cross-section is a washer with inner radius (1 + secx)− 1 = secx and outer radius 3− 1 = 2, so its area isA(x) = π 22 − (secx)2 = π(4− sec2 x).

V =π/3

−π/3A(x) dx =

π/3

−π/3π(4− sec2 x) dx

= 2ππ/3

0

(4− sec2 x) dx [by symmetry]

= 2π 4x− tanx π/3

0= 2π 4π

3−√3 − 0

= 2π 4π3−√3

15. V =1

−1π(1− y2)2 dy = 2

1

0

π(1− y2)2 dy

= 2π1

0

(1− 2y2 + y4) dy

= 2π y − 23y

3 + 15y

5 1

0

= 2π · 815 =

1615π


17. y = x2 ⇒ x =√y for x ≥ 0. The outer radius is the distance from x = −1 to x = y and the inner radius is the

distance from x = −1 to x = y2.

V =1

0

π y − (−1)2

− y2 − (−1) 2dy = π

1

0

y + 12

− (y2 + 1)2 dy

= π1

0

y + 2 y + 1− y4 − 2y2 − 1 dy = π1

0

y + 2 y − y4 − 2y2 dy

= π 12y

2 + 43y

3/2 − 15y

5 − 23y

31

0= π 1

2 +43 − 1

5 − 23= 29

30π

19. R1 about OA (the line y = 0): V =1

0

A(x) dx =1

0

π(x3)2 dx = π1

0

x6 dx = π1

7x7

1

0

=π

7

21. R1 about AB (the line x = 1):

V =1

0

A(y) dy =1

0

π 1− 3 y2

dy = π1

0

(1− 2y1/3 + y2/3) dy = π y − 32y

4/3 + 35y

5/31

0

= π 1− 32 +

35= π

10

23. R2 about OA (the line y = 0):

V =1

0

A(x) dx =1

0

π(1)2 − π√x

2

dx = π1

0

(1− x) dx = π x− 12x2

1

0= π 1− 1

2= π

2


V =1

0

A(y) dy =1

0

π(1)2 − π(1− y2)2 dy = π1

0

1− (1− 2y2 + y4) dy = π1

0

(2y2 − y4) dy

= π 23y3 − 1

5y5

1

0= π 2

3− 1

5= 7

15π

27. R3 about OA (the line y = 0):

V =1

0

A(x) dx =1

0

π√x

2

− π(x3)2 dx = π1

0

(x− x6) dx = π 12x2 − 1

7x7

1

0= π 1

2− 1

7= 5

14π.

Note: Let R=R1 + R2 +R3. If we rotate R about any of the segments OA, OC, AB, or BC, we obtain a right circular

cylinder of height 1 and radius 1. Its volume is πr2h = π(1)2 · 1 = π. As a check for Exercises 19, 23, and 27, we can add the

answers, and that sum must equal π. Thus, π7+ π

2+ 5π

14= 2+ 7+ 5

14π = π.


V =1

0

A(y) dy =1

0

π(1− y2)2 − π 1− 3 y2

dy = π1

0

(1− 2y2 + y4)− (1− 2y1/3 + y2/3) dy

= π1

0

(−2y2 + y4 + 2y1/3 − y2/3) dy = π − 23y3 + 1

5y5 + 3

2y4/3 − 3

5y5/3

1

0= π − 2

3+ 1

5+ 3

2− 3

5= 13

30π

Note: See the note in Exercise 27. For Exercises 21, 25, and 29, we have π10+ 7π

15+ 13π

30= 3+ 14+13

30π = π.


31. V = ππ/4

0

(1− tan3 x)2 dx

33. V = ππ

0

(1− 0)2 − (1− sinx)2 dx

= ππ

0

12 − (1− sinx)2 dx

35. V = π

√8

−√8[3− (−2)]2 − y2 + 1− (−2)

2

dy

= π2√2

−2√252 − 1 + y2 + 2

2

dy

37. y = 2 + x2 cosx and y = x4 + x+ 1 intersect at

x = a ≈ −1.288 and x = b ≈ 0.884.

V = πb

a

[(2 + x2 cosx)2 − (x4 + x+ 1)2]dx ≈ 23.780

39. V= ππ

0

sin2 x− (−1) 2 − [0− (−1)]2 dx

CAS= 11

8π2


41. π π/2

0cos2 xdx describes the volume of the solid obtained by rotating the region R = (x, y) | 0 ≤ x ≤ π

2 , 0 ≤ y ≤ cosxof the xy-plane about the x-axis.

43. π1

0

(y4 − y8) dy = π1

0

(y2)2 − (y4)2 dy describes the volume of the solid obtained by rotating the region

R = (x, y) | 0 ≤ y ≤ 1, y4 ≤ x ≤ y2 of the xy-plane about the y-axis.

45. There are 10 subintervals over the 15-cm length, so we’ll use n = 10/2 = 5 for the Midpoint Rule.

V =15

0A(x) dx ≈M5 =

15−05[A(1.5) +A(4.5) +A(7.5) +A(10.5) +A(13.5)]

= 3(18 + 79 + 106 + 128 + 39) = 3 · 370 = 1110 cm3

47. (a) V =10

2π [f(x)]2 dx ≈ π 10− 24

[f(3)]2 + [f(5)]2 + [f(7)]2 + [f(9)]2

≈ 2π (1.5)2 + (2.2)2 + (3.8)2 + (3.1)2 ≈ 196 units3

(b) V =4

0π (outer radius)2 − (inner radius)2 dy

≈ π 4− 04

(9.9)2 − (2.2)2 + (9.7)2 − (3.0)2 + (9.3)2 − (5.6)2 + (8.7)2 − (6.5)2≈ 838 units3

49. We’ll form a right circular cone with height h and base radius r by

revolving the line y = rhx about the x-axis.

V = πh

0

r

hx

2

dx = πh

0

r2

h2x2dx = π

r2

h21

3x3

h

0

= πr2

h21

3h3 =

1

3πr2h

Another solution: Revolve x = − r

hy + r about the y-axis.

V = πh

0

− r

hy + r

2

dy∗= π

h

0

r2

h2y2 − 2r2

hy + r2 dy

= πr2

3h2y3 − r2

hy2 + r2y

h

0

= π 13r2h− r2h+ r2h = 1

3πr2h

∗ Or use substitution with u = r − r

hy and du = − r

hdy to get

π0

r

u2 −h

rdu = −π h

r

1

3u3

0

r

= −π h

r−13r3 =

1

3πr2h.


51. x2 + y2 = r2 ⇔ x2 = r2 − y2

V = πr

r−hr2 − y2 dy = π r2y − y3

3

r

r−h

= π r3 − r3

3− r2(r − h)− (r − h)3

3

= π 23r3 − 1

3(r − h) 3r2 − (r − h)2

= 13π 2r3 − (r − h) 3r2 − r2 − 2rh+ h2

= 13π 2r3 − (r − h) 2r2 + 2rh− h2

= 13π 2r3 − 2r3 − 2r2h+ rh2 + 2r2h+ 2rh2 − h3

= 13π 3rh2 − h3 = 1

3πh2(3r − h), or, equivalently, πh2 r − h

3

53. For a cross-section at height y, we see from similar triangles that α/2b/2

=h− y

h, so α = b 1− y

h.

Similarly, for cross-sections having 2b as their base and β replacing α, β = 2b 1− y

h. So

V =h

0

A(y) dy =h

0

b 1− y

h2b 1− y

hdy

=h

0

2b2 1− y

h

2

dy = 2b2h

0

1− 2y

h+

y2

h2dy

= 2b2 y − y2

h+

y3

3h2

h

0

= 2b2 h− h+ 13h

= 23b2h [ = 1

3Bh where B is the area of the base, as with any pyramid.]

55. A cross-section at height z is a triangle similar to the base, so we’ll multiply the legs of the base triangle, 3 and 4, by a

proportionality factor of (5− z)/5. Thus, the triangle at height z has area

A(z) =1

2· 3 5− z

5· 4 5− z

5= 6 1− z

5

2

, so

V =5

0

A(z) dz = 65

0

1− z

5

2

dz = 60

1

u2(−5 du) u = 1− z/5,du = − 1

5dz

= −30 13u3

0

1= −30 − 1

3= 10 cm3

57. If l is a leg of the isosceles right triangle and 2y is the hypotenuse,

then l2 + l2 = (2y)2 ⇒ 2l2 = 4y2 ⇒ l2 = 2y2.

V =2

−2A(x) dx = 22

0A(x) dx = 2

2

012(l)(l) dx = 2

2

0y2 dx

= 22

014(36− 9x2) dx = 9

2

2

0(4− x2) dx

= 924x− 1

3x3

2

0= 9

28− 8

3= 24


59. The cross-section of the base corresponding to the coordinate x has length

y = 1− x. The corresponding square with side s has area

A(x) = s2 = (1− x)2 = 1− 2x+ x2. Therefore,

V =1

0

A(x) dx =1

0

(1− 2x+ x2) dx

= x− x2 + 13x3

1

0= 1− 1 + 1

3− 0 = 1

3

Or:1

0

(1− x)2 dx =0

1

u2(−du) [u = 1− x] = 13u3

1

0= 1

3

61. The cross-section of the base b corresponding to the coordinate x has length 1− x2. The height h also has length 1− x2,

so the corresponding isosceles triangle has area A(x) = 12bh = 1

2(1− x2)2. Therefore,

V =1

−112(1− x2)2 dx

= 2 · 12

1

0

(1− 2x2 + x4) dx [by symmetry]

= x− 23x3 + 1

5x5

1

0= 1− 2

3+ 1

5− 0 = 8

15

63. (a) The torus is obtained by rotating the circle (x−R)2 + y2 = r2 about

the y-axis. Solving for x, we see that the right half of the circle is given by

x = R+ r2 − y2 = f(y) and the left half by x = R− r2 − y2 = g(y). So

V = πr

−r [f(y)]2 − [g(y)]2 dy

= 2πr

0R2 + 2R r2 − y2 + r2 − y2 − R2 − 2R r2 − y2 + r2 − y2 dy

= 2πr

04R r2 − y2 dy = 8πR

r

0r2 − y2 dy

(b) Observe that the integral represents a quarter of the area of a circle with radius r, so

8πRr

0r2 − y2 dy = 8πR · 1

4πr2 = 2π2r2R.

65. (a) Volume(S1) = h

0A(z) dz = Volume(S2) since the cross-sectional area A(z) at height z is the same for both solids.

(b) By Cavalieri’s Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius r

and height h, that is, πr2h.


67. The volume is obtained by rotating the area common to two circles of radius r, as

shown. The volume of the right half is

Vright = πr/2

0y2 dx = π

r/2

0r2 − 1

2r + x2

dx

= π r2x− 13

12r + x

3 r/2

0= π 1

2r3 − 1

3r3 − 0− 1

24r3 = 5

24πr3

So by symmetry, the total volume is twice this, or 512πr

3.

Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can use the formula from

Exercise 51 with h = 12r: V = 2 · 13πh2(3r − h) = 2

3π12r

23r − 1

2r = 512πr

3.

69. Take the x-axis to be the axis of the cylindrical hole of radius r.

A quarter of the cross-section through y, perpendicular to the

y-axis, is the rectangle shown. Using the Pythagorean Theorem

twice, we see that the dimensions of this rectangle are

x = R2 − y2 and z = r2 − y2, so

14A(y) = xz = r2 − y2 R2 − y2, and

V =r

−r A(y) dy =r

−r 4 r2 − y2 R2 − y2 dy = 8r

0r2 − y2 R2 − y2 dy

71. (a) The radius of the barrel is the same at each end by symmetry, since the

function y = R− cx2 is even. Since the barrel is obtained by rotating

the graph of the function y about the x-axis, this radius is equal to the

value of y at x = 12h, which is R− c 1

2h2= R− d = r.

(b) The barrel is symmetric about the y-axis, so its volume is twice the volume of that part of the barrel for x > 0. Also, the

barrel is a volume of rotation, so

V = 2h/2

0

πy2 dx = 2πh/2

0

R− cx22dx = 2π R2x− 2

3Rcx3 + 1

5c2x5

h/2

0

= 2π 12R

2h− 112Rch

3 + 1160c

2h5

Trying to make this look more like the expression we want, we rewrite it as V = 13πh 2R2 + R2 − 1

2Rch2 + 3

80c2h4 .

But R2 − 12Rch2 + 3

80c2h4 = R− 1

4ch2

2 − 140c2h4 = (R− d)2 − 2

514ch2

2= r2 − 2

5d2.

Substituting this back into V , we see that V = 13πh 2R2 + r2 − 2

5d 2 , as required.


6.3 Volumes by Cylindrical Shells

1. If we were to use the “washer” method, we would first have to locate the

local maximum point (a, b) of y = x(x− 1)2 using the methods of

Chapter 4. Then we would have to solve the equation y = x(x− 1)2

for x in terms of y to obtain the functions x = g1(y) and x = g2(y)

shown in the first figure. This step would be difficult because it involves

the cubic formula. Finally we would find the volume using

V = πb

0[g1(y)]

2 − [g2(y)]2 dy.

Using shells, we find that a typical approximating shell has radius x, so its circumference is 2πx. Its height is y, that is,

x(x− 1)2. So the total volume is

V =1

0

2πx x(x− 1)2 dx = 2π1

0

x4 − 2x3 + x2 dx = 2πx5

5− 2x

4

4+

x3

3

1

0

=π

15

3. V =2

1

2πx · 1xdx = 2π

2

1

1 dx

= 2π x2

1= 2π(2− 1) = 2π

5. V =2

02πx(4− x2) dx = 2π

2

0(4x− x3) dx

= 2π 2x2 − 14x4

2

0= 2π(8− 4)

= 8π

7. The curves intersect when 4(x− 2)2 = x2 − 4x+ 7 ⇔ 4x2 − 16x+ 16 = x2 − 4x+ 7 ⇔3x2 − 12x+ 9 = 0 ⇔ 3(x2 − 4x+ 3) = 0 ⇔ 3(x− 1)(x− 3) = 0, so x = 1 or 3.

V = 2π3

1x x2 − 4x+ 7 − 4(x− 2)2 dx = 2π

3

1x(x2 − 4x+ 7− 4x2 + 16x− 16) dx

= 2π3

1x(−3x2 + 12x− 9) dx = 2π(−3) 3

1(x3 − 4x2 + 3x) dx = −6π 1

4x4 − 4

3x3 + 3

2x2

3

1

= −6π 814− 36 + 27

2− 1

4− 4

3+ 3

2= −6π 20− 36 + 12 + 4

3= −6π − 8

3= 16π

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 229

9. V =2

12πy(1 + y2) dy = 2π

2

1(y + y3) dy = 2π 1

2y2 + 1

4y4 2

1

= 2π (2 + 4)− 12+ 1

4= 2π 21

4= 21

2π

11. V = 2π8

0

y( 3 y − 0) dy

= 2π8

0

y4/3 dy = 2π 37y7/3

8

0

=6π

7(87/3) =

6π

7(27) =

768

7π

13. The height of the shell is 2− 1 + (y − 2)2 = 1− (y − 2)2 = 1− y2 − 4y + 4 = −y2 + 4y − 3.

V = 2π3

1y(−y2 + 4y − 3) dy

= 2π3

1(−y3 + 4y2 − 3y) dy

= 2π − 14y4 + 4

3y3 − 3

2y2

3

1

= 2π − 814+ 36− 27

2− − 1

4+ 4

3− 3

2

= 2π 83= 16

3π

15. The shell has radius 2− x, circumference 2π(2− x), and height x4.

V =1

02π(2− x)x4 dx

= 2π1

0(2x4 − x5) dx

= 2π 25x

5 − 16x

6 1

0

= 2π 25− 1

6− 0 = 2π 7

30= 7

15π


17. The shell has radius x− 1, circumference 2π(x− 1), and height (4x− x2)− 3 = −x2 + 4x− 3.

V=3

12π(x− 1)(−x2 + 4x− 3) dx

= 2π3

1(−x3 + 5x2 − 7x+ 3) dx

= 2π − 14x

4 + 53x

3 − 72x

2 + 3x3

1

= 2π − 814+ 45− 63

2+ 9 − − 1

4+ 5

3− 7

2+ 3

= 2π 43= 8

3π

19. The shell has radius 1− y, circumference 2π(1− y), and height 1− 3 y y = x3 ⇔ x = 3 y .

V=1

02π(1− y)(1− y1/3) dy

= 2π1

0(1− y − y1/3 + y4/3) dy

= 2π y − 12y

2 − 34y

4/3 + 37y

7/31

0

= 2π 1− 12− 3

4+ 3

7− 0

= 2π 528

= 514π

21. V =3π

2π2πx sinxdx 23. V =

1

02π[x− (−1)] sin π

2 x− x4 dx

25. V =π

02π(4− y)

√sin y dy

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 231

27. V =1

02πx

√1 + x3 dx. Let f(x) = x

√1 + x3.

Then the Midpoint Rule with n = 5 gives

1

0f(x) dx≈ 1−0

5[f(0.1) + f(0.3) + f(0.5) + f(0.7) + f(0.9)]

≈ 0.2(2.9290)

Multiplying by 2π gives V ≈ 3.68.

29. 3

02πx5 dx = 2π

3

0x(x4) dx. The solid is obtained by rotating the region 0 ≤ y ≤ x4, 0 ≤ x ≤ 3 about the y-axis using

cylindrical shells.

31. 1

02π(3− y)(1− y2) dy. The solid is obtained by rotating the region bounded by (i) x = 1− y2, x = 0, and y = 0 or

(ii) x = y2, x = 1, and y = 0 about the line y = 3 using cylindrical shells.

33. From the graph, the curves intersect at x = 0 and at x = a ≈ 1.32, with

x+ x2 − x4 > 0 on the interval (0, a). So the volume of the solid obtained

by rotating the region about the y-axis is

V = 2πa

0[x(x+ x2 − x4)] dx = 2π

a

0(x2 + x3 − x5) dx

= 2π 13x3 + 1

4x4 − 1

6x6

a

0≈ 4.05

35. V = 2ππ/2

0

π2 − x sin2 x− sin4 x dx

CAS= 1

32π3

37. Use shells:

V =4

22πx(−x2 + 6x− 8) dx = 2π 4

2(−x3 + 6x2 − 8x) dx

= 2π − 14x4 + 2x3 − 4x2 4

2

= 2π[(−64 + 128− 64)− (−4 + 16− 16)]= 2π(4) = 8π

39. Use shells:

V =4

12π[x− (−1)][5− (x2 − 5x+ 9)] dx

= 2π4

1(x+ 1)(−x2 + 5x− 4) dx

= 2π4

1(−x3 + 4x2 + x− 4) dx = 2π − 1

4x4 + 4

3x3 + 1

2x2 − 4x 4

1

= 2π −64 + 2563+ 8− 16 − − 1

4+ 4

3+ 1

2− 4

= 2π 634= 63

2 π


41. Use disks: x2 + (y − 1)2 = 1 ⇔ x = ± 1− (y − 1)2

V = π2

0

1− (y − 1)2 2

dy = π2

0

(2y − y2) dy

= π y2 − 13y3

2

0= π 4− 8

3= 4

3π

43. Use shells:

V = 2r

02πx

√r2 − x2 dx

= −2π r

0(r2 − x2)1/2(−2x) dx

= −2π · 23 (r

2 − x2)3/2r

0

= − 43π(0− r3) = 4

3πr3

45. V = 2πr

0

x −h

rx+ h dx = 2πh

r

0

−x2

r+ x dx

= 2πh −x3

3r+

x2

2

r

0

= 2πhr2

6=

πr2h

3

6.4 Work

1. W = Fd = mgd = (40)(9.8)(1.5) = 588 J

3. W =b

a

f(x) dx =9

0

10

(1 + x)2dx = 10

10

1

1

u2du [u = 1 + x, du = dx] = 10 − 1

u

10

1

= 10 − 110 + 1 = 9 ft-lb

5. The force function is given by F (x) (in newtons) and the work (in joules) is the area under the curve, given by8

0F (x) dx =

4

0F (x) dx+

8

4F (x) dx = 1

2 (4)(30) + (4)(30) = 180 J.

7. 10 = f(x) = kx = 13k [4 inches = 1

3foot], so k = 30 lb/ft and f(x) = 30x. Now 6 inches = 1

2foot, so

W =1/2

030xdx = 15x2

1/2

0= 15

4ft-lb.

9. (a) If 0.12

0kx dx = 2 J, then 2 = 1

2kx2 0.12

0= 1

2k(0.0144) = 0.0072k and k = 20.0072 =

25009 ≈ 277.78 N/m.

Thus, the work needed to stretch the spring from 35 cm to 40 cm is0.10

0.0525009

xdx = 12509

x21/10

1/20= 1250

91100− 1

400= 25

24≈ 1.04 J.

(b) f(x) = kx, so 30 = 25009 x and x = 270

2500 m = 10.8 cm

SECTION 6.4 WORK ¤ 233

11. The distance from 20 cm to 30 cm is 0.1 m, so with f(x) = kx, we get W1 =0.1

0kx dx = k 1

2x2 0.1

0= 1

200k.

Now W2 =0.2

0.1kx dx = k 1

2x2

0.2

0.1= k 4

200− 1

200= 3

200k. Thus, W2 = 3W1.

In Exercises 13 – 20, n is the number of subintervals of length ∆x, and x∗i is a sample point in the ith subinterval [xi−1, xi].

13. (a) The portion of the rope from x ft to (x+∆x) ft below the top of the building weighs 12∆x lb and must be lifted x∗i ft,

so its contribution to the total work is 12x∗i ∆x ft-lb. The total work is

W = limn→∞

n

i=1

12x∗i ∆x =

50

012xdx = 1

4x2

50

0= 2500

4= 625 ft-lb

Notice that the exact height of the building does not matter (as long as it is more than 50 ft).

(b) When half the rope is pulled to the top of the building, the work to lift the top half of the rope is

W1 =25

012xdx = 1

4x2

25

0= 625

4ft-lb. The bottom half of the rope is lifted 25 ft and the work needed to accomplish

that is W2 =50

2512 · 25dx = 25

2x

50

25= 625

2 ft-lb. The total work done in pulling half the rope to the top of the building

is W =W1 +W2 =6252+ 625

4= 3

4· 625 = 1875

4ft-lb.

15. The work needed to lift the cable is limn→∞

ni=1 2x

∗i ∆x =

500

02xdx = x2

500

0= 250,000 ft-lb. The work needed to lift

the coal is 800 lb · 500 ft = 400,000 ft-lb. Thus, the total work required is 250,000 + 400,000 = 650,000 ft-lb.

17. At a height of x meters (0 ≤ x ≤ 12), the mass of the rope is (0.8 kg/m)(12− x m) = (9.6− 0.8x) kg and the mass of the

water is 3612 kg/m (12− x m) = (36− 3x) kg. The mass of the bucket is 10 kg, so the total mass is

(9.6− 0.8x) + (36− 3x) + 10 = (55.6− 3.8x) kg, and hence, the total force is 9.8(55.6− 3.8x) N. The work needed to lift

the bucket ∆x m through the ith subinterval of [0, 12] is 9.8(55.6− 3.8x∗i )∆x, so the total work is

W = limn→∞

n

i=1

9.8(55.6− 3.8x∗i )∆x =12

0(9.8)(55.6− 3.8x) dx = 9.8 55.6x− 1.9x2

12

0= 9.8(393.6) ≈ 3857 J

19. A “slice” of water ∆x m thick and lying at a depth of x∗i m (where 0 ≤ x∗i ≤ 12

) has volume (2× 1×∆x) m3, a mass of

2000∆x kg, weighs about (9.8)(2000∆x) = 19,600∆x N, and thus requires about 19,600x∗i ∆x J of work for its removal.

So W = limn→∞

n

i=1

19,600x∗i ∆x =1/2

019,600xdx = 9800x2

1/2

0= 2450 J.

21. A rectangular “slice” of water ∆x m thick and lying x m above the bottom has width x m and volume 8x∆x m3. It weighs

about (9.8× 1000)(8x∆x) N, and must be lifted (5− x) m by the pump, so the work needed is about

(9.8× 103)(5− x)(8x∆x) J. The total work required is

W ≈ 3

0(9.8× 103)(5− x)8xdx = (9.8× 103) 3

0(40x− 8x2) dx = (9.8× 103) 20x2 − 8

3x3

3

0

= (9.8× 103)(180− 72) = (9.8× 103)(108) = 1058.4× 103 ≈ 1.06× 106 J


23. Let x measure depth (in feet) below the spout at the top of the tank. A horizontal

disk-shaped “slice” of water ∆x ft thick and lying at coordinate x has radius38(16− x) ft ( ) and volume πr2∆x = π · 9

64(16− x)2∆x ft3. It weighs

about (62.5) 9π64(16− x)2∆x lb and must be lifted x ft by the pump, so the

work needed to pump it out is about (62.5)x 9π64(16− x)2∆x ft-lb. The total

work required is

W ≈ 8

0(62.5)x 9π

64 (16− x)2 dx = (62.5) 9π648

0x(256− 32x+ x2) dx

= (62.5) 9π64

8

0(256x− 32x2 + x3) dx = (62.5) 9π

64128x2 − 32

3x3 + 1

4x4

8

0

= (62.5)9π

64

11, 264

3= 33,000π ≈ 1.04× 105 ft-lb

( ) From similar triangles,d

8− x=3

8.

So r = 3 + d = 3 + 38 (8− x)

=3(8)

8+3

8(8− x)

= 38 (16− x)

25. If only 4.7× 105 J of work is done, then only the water above a certain level (call

it h) will be pumped out. So we use the same formula as in Exercise 21, except that

the work is fixed, and we are trying to find the lower limit of integration:

4.7× 105 ≈ 3

h(9.8× 103)(5− x)8xdx = 9.8× 103 20x2 − 8

3x3 3

h⇔

4.79.8× 102 ≈ 48 = 20 · 32 − 8

3· 33 − 20h2 − 8

3h3 ⇔

2h3 − 15h2 + 45 = 0. To find the solution of this equation, we plot 2h3 − 15h2 + 45 between h = 0 and h = 3.

We see that the equation is satisfied for h ≈ 2.0. So the depth of water remaining in the tank is about 2.0 m.

27. V = πr2x, so V is a function of x and P can also be regarded as a function of x. If V1 = πr2x1 and V2 = πr2x2, then

W =x2

x1

F (x) dx =x2

x1

πr2P (V (x)) dx =x2

x1

P (V (x)) dV (x) [Let V (x) = πr2x, so dV (x) = πr2 dx.]

=V2

V1

P (V ) dV by the Substitution Rule.

29. W =b

a

F (r) dr =b

a

Gm1m2

r2dr = Gm1m2

−1r

b

a

= Gm1m21

a− 1

b

6.5 Average Value of a Function

1. fave =1

b− a

b

af(x) dx = 1

4− 04

0(4x− x2) dx = 1

42x2 − 1

3x3 4

0= 1

432− 64

3− 0 = 1

4323= 8

3

3. gave =1

b− a

b

ag(x) dx = 1

8− 18

1

3√xdx = 1

734x

4/38

1= 3

28 (16− 1) = 4528

5. fave = 15− 0

5

0t√1 + t2 dt = 1

5

26

1

√u 1

2du [u = 1 + t2, du = 2t dt]

= 110

26

1u1/2 du = 1

10 · 23 u3/226

1= 1

15 (263/2 − 1)

SECTION 6.5 AVERAGE VALUE OF A FUNCTION ¤ 235

7. have =1

π− 0π

0cos4 x sinxdx = 1

π

−11

u4(−du) [u = cosx, du = − sinxdx]

= 1π

1

−1 u4 du = 1

π· 2 1

0u4 du [by Theorem 5.5.6(a)] = 2

π15u5

1

0= 2

5π

9. (a) fave =1

5 − 2

5

2

(x− 3)2 dx = 1

3

1

3(x− 3)3

5

2

= 1923 − (−1)3 = 1

9(8 + 1) = 1

(c)

(b) f(c) = fave ⇔ (c− 3)2 = 1 ⇔c− 3 = ±1 ⇔ c = 2 or 4

11. (a) fave =1

π − 0

π

0

(2 sinx− sin 2x) dx

= 1π−2 cosx+ 1

2 cos 2xπ

0

= 1π

2 + 12− −2 + 1

2= 4

π

(c)

(b) f(c) = fave ⇔ 2 sin c− sin 2c = 4π

⇔c1 ≈ 1.238 or c2 ≈ 2.808

13. f is continuous on [1, 3], so by the Mean Value Theorem for Integrals there exists a number c in [1, 3] such that

3

1f(x) dx = f(c)(3− 1) ⇒ 8 = 2f(c); that is, there is a number c such that f(c) = 8

2= 4.

15. fave =1

50− 2050

20

f(x)dx ≈ 130M3 =

1

30· 50− 20

3[f(25) + f(35) + f(45)] = 1

3(38 + 29 + 48) = 115

3= 38 1

3

17. Let t = 0 and t = 12 correspond to 9 AM and 9 PM, respectively.

Tave =1

12− 0

12

050 + 14 sin 1

12πt dt = 1

1250t− 14 · 12

πcos 1

12πt

12

0

= 11250 · 12 + 14 · 12

π+ 14 · 12

π= 50 + 28

π◦F ≈ 59◦F

19. ρave =1

8

8

0

12√x+ 1

dx =3

2

8

0

(x+ 1)−1/2 dx = 3√x+ 1

8

0= 9− 3 = 6 kg/m

21. Vave =15

5

0V (t) dt = 1

5

5

054π

1− cos 25πt dt = 1

4π

5

01− cos 2

5πt dt

= 14π

t− 52πsin 2

5πt

5

0= 1

4π[(5− 0)− 0] = 5

4π≈ 0.4 L

23. Let F (x) = x

af(t) dt for x in [a, b]. Then F is continuous on [a, b] and differentiable on (a, b), so by the Mean Value

Theorem there is a number c in (a, b) such that F (b)− F (a) = F 0(c)(b− a). But F 0(x) = f(x) by the Fundamental

Theorem of Calculus. Therefore, b

af(t) dt− 0 = f(c)(b− a).


6 Review

1. (a) See Section 6.1, Figure 2 and Equations 6.1.1 and 6.1.2.

(b) Instead of using “top minus bottom” and integrating from left to right, we use “right minus left” and integrate from bottom

to top. See Figures 11 and 12 in Section 6.1.

2. The numerical value of the area represents the number of meters by which Sue is ahead of Kathy after 1 minute.

3. (a) See the discussion in Section 6.2, near Figures 2 and 3, ending in the Definition of Volume.

(b) See the discussion between Examples 5 and 6 in Section 6.2. If the cross-section is a disk, find the radius in terms of x or y

and use A = π(radius)2. If the cross-section is a washer, find the inner radius rin and outer radius rout and use

A = π r2out − π r2in .

4. (a) V = 2πrh∆r = (circumference)(height)(thickness)

(b) For a typical shell, find the circumference and height in terms of x or y and calculate

V =b

a(circumference)(height)(dx or dy), where a and b are the limits on x or y.

(c) Sometimes slicing produces washers or disks whose radii are difficult (or impossible) to find explicitly. On other

occasions, the cylindrical shell method leads to an easier integral than slicing does.

5. 6

0f(x) dx represents the amount of work done. Its units are newton-meters, or joules.

6. (a) The average value of a function f on an interval [a, b] is fave =1

b− a

b

a

f(x) dx.

(b) The Mean Value Theorem for Integrals says that there is a number c at which the value of f is exactly equal to the average

value of the function, that is, f(c) = fave. For a geometric interpretation of the Mean Value Theorem for Integrals, see

Figure 2 in Section 6.5 and the discussion that accompanies it.

1. The curves intersect when x2 = 4x− x2 ⇔ 2x2 − 4x = 0 ⇔2x(x− 2) = 0 ⇔ x = 0 or 2.

A=2

0(4x− x2)− x2 dx =

2

0(4x− 2x2) dx

= 2x2 − 23x3

2

0= 8− 16

3− 0 = 8

3

3. If x ≥ 0, then |x | = x, and the graphs intersect when x = 1− 2x2 ⇔ 2x2 + x− 1 = 0 ⇔ (2x− 1)(x+ 1) = 0 ⇔x = 1

2 or −1, but −1 < 0. By symmetry, we can double the area from x = 0 to x = 12 .

A= 21/2

0(1− 2x2)− x dx = 2

1/2

0(−2x2 − x+ 1) dx

= 2 − 23x

3 − 12x

2 + x1/2

0= 2 − 1

12 − 18 +

12− 0

= 2 724

= 712

CHAPTER 6 REVIEW ¤ 237

5. A =2

0

sinπx

2− (x2 − 2x) dx

= − 2πcos

πx

2− 1

3x3 + x2

2

0

= 2π− 8

3+ 4 − − 2

π− 0 + 0 = 4

3+ 4

π

7. Using washers with inner radius x2 and outer radius 2x, we have

V = π2

0(2x)2 − (x2)2 dx = π

2

0(4x2 − x4) dx

= π 43x

3 − 15x

5 2

0= π 32

3 − 325

= 32π · 215= 64

15π

9. V = π3

−3 (9− y2)− (−1) 2 − [0− (−1)]2 dy

= 2π3

0(10− y2)2 − 1 dy = 2π

3

0(100− 20y2 + y4 − 1) dy

= 2π3

0(99− 20y2 + y4) dy = 2π 99y − 20

3y3 + 1

5y5

3

0

= 2π 297− 180 + 2435

= 16565

π

11. The graph of x2 − y2 = a2 is a hyperbola with right and left branches.

Solving for y gives us y2 = x2 − a2 ⇒ y = ±√x2 − a2.

We’ll use shells and the height of each shell is√x2 − a2 − −√x2 − a2 = 2

√x2 − a2.

The volume is V =a+h

a2πx · 2√x2 − a2 dx. To evaluate, let u = x2 − a2,

so du = 2xdx and xdx = 12 du. When x = a, u = 0, and when x = a+ h,

u = (a+ h)2 − a2 = a2 + 2ah+ h2 − a2 = 2ah+ h2.

Thus, V = 4π2ah+h2

0

√u

1

2du = 2π

2

3u3/2

2ah+h2

0

=4

3π 2ah+ h2

3/2.

13. A shell has radius π2− x, circumference 2π π

2− x , and height cos2 x− 1

4.

y = cos2 x intersects y = 14

when cos2 x = 14⇔

cosx = ± 12

[ |x| ≤ π/2] ⇔ x = ±π3

.

V =π/3

−π/32π

π

2− x cos2 x− 1

4dx


15. (a) A cross-section is a washer with inner radius x2 and outer radius x.

V =1

0π (x)2 − (x2)2 dx =

1

0π(x2 − x4) dx = π 1

3x3 − 1

5x5

1

0= π 1

3− 1

5= 2

15π

(b) A cross-section is a washer with inner radius y and outer radius y.

V =1

0π y

2

− y2 dy =1

0π(y − y2) dy = π 1

2y2 − 1

3y3 1

0= π 1

2 − 13= π

6

(c) A cross-section is a washer with inner radius 2− x and outer radius 2− x2.

V =1

0π (2− x2)2 − (2− x)2 dx =

1

0π(x4 − 5x2 + 4x) dx = π 1

5x5 − 5

3x3 + 2x2

1

0= π 1

5− 5

3+ 2 = 8

15π

17. (a) Using the Midpoint Rule on [0, 1] with f(x) = tan(x2) and n = 4, we estimate

A =1

0tan(x2) dx ≈ 1

4tan 1

8

2+ tan 3

8

2+ tan 5

8

2+ tan 7

8

2 ≈ 14 (1.53) ≈ 0.38

(b) Using the Midpoint Rule on [0, 1] with f(x) = π tan2(x2) (for disks) and n = 4, we estimate

V =1

0f(x) dx ≈ 1

4π tan2 1

8

2+ tan2 3

8

2+ tan2 5

8

2+ tan2 7

8

2 ≈ π4(1.114) ≈ 0.87

19. π/2

02πx cosxdx =

π/2

0(2πx) cosxdx

The solid is obtained by rotating the region R = (x, y) | 0 ≤ x ≤ π2, 0 ≤ y ≤ cosx about the y-axis.

21. π

0π(2− sinx)2 dx

The solid is obtained by rotating the region R = {(x, y) | 0 ≤ x ≤ π, 0 ≤ y ≤ 2− sinx} about the x-axis.

23. Take the base to be the disk x2 + y2 ≤ 9. Then V =3

−3A(x) dx, where A(x0) is the area of the isosceles right triangle

whose hypotenuse lies along the line x = x0 in the xy-plane. The length of the hypotenuse is 2√9− x2 and the length of

each leg is√2√9− x2. A(x) = 1

2

√2√9− x2

2= 9− x2, so

V = 23

0A(x) dx = 2

3

0(9− x2) dx = 2 9x− 1

3x3

3

0= 2(27− 9) = 36

25. Equilateral triangles with sides measuring 14x meters have height 1

4x sin 60◦ =

√38x. Therefore,

A(x) = 12· 14x ·

√38x =

√3

64x2. V =

20

0A(x) dx =

√3

64

20

0x2 dx =

√3

6413x3

20

0= 8000

√3

64 · 3 = 125√3

3m3.

27. f(x) = kx ⇒ 30 N = k(15− 12) cm ⇒ k = 10 N/cm = 1000 N/m. 20 cm− 12 cm = 0.08 m ⇒

W =0.08

0kx dx = 1000

0.08

0xdx = 500 x2

0.08

0= 500(0.08)2 = 3.2 N·m = 3.2 J.

CHAPTER 6 REVIEW ¤ 239

29. (a) The parabola has equation y = ax2 with vertex at the origin and passing through

(4, 4). 4 = a · 42 ⇒ a = 14⇒ y = 1

4x2 ⇒ x2 = 4y ⇒

x = 2 y. Each circular disk has radius 2 y and is moved 4− y ft.

W =4

0π 2 y

2

62.5(4− y) dy = 250π4

0y(4− y) dy

= 250π 2y2 − 13y

3 4

0= 250π 32− 64

3= 8000π

3 ≈ 8378 ft-lb

(b) In part (a) we knew the final water level (0) but not the amount of work done. Here

we use the same equation, except with the work fixed, and the lower limit of

integration (that is, the final water level — call it h) unknown: W = 4000 ⇔

250π 2y2 − 13y3

4

h= 4000 ⇔ 16

π= 32− 64

3− 2h2 − 1

3h3 ⇔

h3 − 6h2 + 32− 48π= 0. We graph the function f(h) = h3 − 6h2 + 32− 48

π

on the interval [0, 4] to see where it is 0. From the graph, f(h) = 0 for h ≈ 2.1.

So the depth of water remaining is about 2.1 ft.

31. limh→0

fave = limh→0

1

(x+ h)− x

x+h

x

f(t) dt = limh→0

F (x+ h)− F (x)

h, where F (x) = x

af(t) dt. But we recognize this

limit as being F 0(x) by the definition of a derivative. Therefore, limh→0

fave = F 0(x) = f(x) by FTC1.

PROBLEMS PLUS1. (a) The area under the graph of f from 0 to t is equal to t

0f(x) dx, so the requirement is that t

0f(x) dx = t3 for all t. We

differentiate both sides of this equation with respect to t (with the help of FTC1) to get f(t) = 3t2. This function is

positive and continuous, as required.

(b) The volume generated from x = 0 to x = b is b

0π[f(x)]2 dx. Hence, we are given that b2 = b

0π[f(x)]2 dx for all

b > 0. Differentiating both sides of this equation with respect to b using the Fundamental Theorem of Calculus gives

2b = π[f(b)]2 ⇒ f(b) = 2b/π, since f is positive. Therefore, f(x) = 2x/π.

3. Let a and b be the x-coordinates of the points where the line intersects the

curve. From the figure, R1 = R2 ⇒

a

0c− 8x− 27x3 dx=

b

a8x− 27x3 − c dx

cx− 4x2 + 274x4

a

0= 4x2 − 27

4x4 − cx

b

a

ac− 4a2 + 274 a

4 = 4b2 − 274 b

4 − bc − 4a2 − 274 a

4 − ac

0 = 4b2 − 274 b

4 − bc = 4b2 − 274 b

4 − b 8b− 27b3

= 4b2 − 274b4 − 8b2 + 27b4 = 81

4b4 − 4b2

= b2 814b2 − 4

So for b > 0, b2 = 1681 ⇒ b = 4

9 . Thus, c = 8b− 27b3 = 8 49− 27 64

729= 32

9 − 6427 =

3227 .

5. (a) V = πh2(r − h/3) = 13πh2(3r − h). See the solution to Exercise 6.2.51.

(b) The smaller segment has height h = 1 − x and so by part (a) its volume is

V = 13π(1− x)2 [3(1)− (1− x)] = 1

3π(x− 1)2(x+ 2). This volume must be 1

3of the total volume of the sphere,

which is 43π(1)3. So 1

3π(x− 1)2(x+ 2) = 1

343π ⇒ (x2 − 2x+ 1)(x+ 2) = 4

3⇒ x3 − 3x+ 2 = 4

3⇒

3x3 − 9x+ 2 = 0. Using Newton’s method with f(x) = 3x3 − 9x+ 2, f 0(x) = 9x2 − 9, we get

xn+1 = xn − 3x3n − 9xn + 29x2n − 9 . Taking x1 = 0, we get x2 ≈ 0.2222, and x3 ≈ 0.2261 ≈ x4, so, correct to four decimal

places, x ≈ 0.2261.

(c) With r = 0.5 and s = 0.75, the equation x3 − 3rx2 + 4r3s = 0 becomes x3 − 3(0.5)x2 + 4(0.5)3(0.75) = 0 ⇒x3 − 3

2x2 + 4 1

834= 0 ⇒ 8x3 − 12x2 + 3 = 0. We use Newton’s method with f(x) = 8x3 − 12x2 + 3,

f 0(x) = 24x2 − 24x, so xn+1 = xn − 8x3n − 12x2n + 324x2n − 24xn . Take x1 = 0.5. Then x2 ≈ 0.6667, and x3 ≈ 0.6736 ≈ x4.

So to four decimal places the depth is 0.6736 m.

241

242 ¤ CHAPTER 6 PROBLEMS PLUS

(d) (i) From part (a) with r = 5 in., the volume of water in the bowl is

V = 13πh2(3r − h) = 1

3πh2(15− h) = 5πh2 − 1

3πh3. We are given that dV

dt= 0.2 in3/s and we want to find dh

dt

when h = 3. Now dV

dt= 10πh

dh

dt− πh2

dh

dt, so dh

dt=

0.2

π(10h− h2). When h = 3, we have

dh

dt=

0.2

π(10 · 3− 32) =1

105π≈ 0.003 in/s.

(ii) From part (a), the volume of water required to fill the bowl from the instant that the water is 4 in. deep is

V = 12 · 43π(5)3 − 1

3π(4)2(15− 4) = 2

3 · 125π− 163 · 11π = 74

3 π. To find the time required to fill the bowl we divide

this volume by the rate: Time = 74π/30.2

= 370π3≈ 387 s ≈ 6.5 min.

7. We are given that the rate of change of the volume of water is dV

dt= −kA(x), where k is some positive constant and A(x) is

the area of the surface when the water has depth x. Now we are concerned with the rate of change of the depth of the water

with respect to time, that is, dxdt

. But by the Chain Rule, dVdt

=dV

dx

dx

dt, so the first equation can be written

dV

dx

dx

dt= −kA(x) ( ). Also, we know that the total volume of water up to a depth x is V (x) = x

0A(s) ds, where A(s) is

the area of a cross-section of the water at a depth s. Differentiating this equation with respect to x, we get dV/dx = A(x).

Substituting this into equation , we get A(x)(dx/dt) = −kA(x) ⇒ dx/dt = −k, a constant.

9. We must find expressions for the areas A and B, and then set them equal and see what this says about the curve C. If

P = a, 2a2 , then area A is just a

0(2x2 − x2) dx =

a

0x2 dx = 1

3a3. To find area B, we use y as the variable of

integration. So we find the equation of the middle curve as a function of y: y = 2x2 ⇔ x = y/2, since we are

concerned with the first quadrant only. We can express area B as

2a2

0

y/2− C(y) dy =4

3(y/2)3/2

2a2

0

−2a2

0

C(y) dy =4

3a3 −

2a2

0

C(y) dy

where C(y) is the function with graph C. Setting A = B, we get 13a3 = 4

3a3 − 2a2

0C(y) dy ⇔ 2a2

0C(y) dy = a3.

Now we differentiate this equation with respect to a using the Chain Rule and the Fundamental Theorem:

C(2a2)(4a) = 3a2 ⇒ C(y) = 34

y/2, where y = 2a2. Now we can solve for y: x = 34

y/2 ⇒

x2 = 916(y/2) ⇒ y = 32

9x2.

11. (a) Stacking disks along the y-axis gives us V =h

0π [f(y)]2 dy.

(b) Using the Chain Rule, dVdt

=dV

dh· dhdt= π [f(h)]2

dh

dt.

(c) kA√h = π[f(h)]2

dh

dt. Set dh

dt= C: π[f(h)]2C = kA

√h ⇒ [f(h)]2 =

kA

πC

√h ⇒ f(h) =

kA

πCh1/4; that

is, f(y) = kA

πCy1/4. The advantage of having dh

dt= C is that the markings on the container are equally spaced.

CHAPTER 6 PROBLEMS PLUS ¤ 243

13. The cubic polynomial passes through the origin, so let its equation be

y = px3 + qx2 + rx. The curves intersect when px3 + qx2 + rx = x2 ⇔px3 + (q − 1)x2 + rx = 0. Call the left side f(x). Since f(a) = f(b) = 0,

another form of f is

f(x) = px(x− a)(x− b) = px[x2 − (a+ b)x+ ab]

= p[x3 − (a+ b)x2 + abx]

Since the two areas are equal, we must have a

0f(x) dx = − b

af(x) dx ⇒

[F (x)]a0 = [F (x)]ab ⇒ F (a)− F (0) = F (a)− F (b) ⇒ F (0) = F (b), where F is an antiderivative of f .

Now F (x) = f(x) dx = p[x3 − (a+ b)x2 + abx] dx = p 14x4 − 1

3(a+ b)x3 + 1

2abx2 + C, so

F (0) = F (b) ⇒ C = p 14b4 − 1

3(a+ b)b3 + 1

2ab3 + C ⇒ 0 = p 1

4b4 − 1

3(a+ b)b3 + 1

2ab3 ⇒

0 = 3b− 4(a+ b) + 6a [multiply by 12/(pb3), b 6= 0] ⇒ 0 = 3b− 4a− 4b+ 6a ⇒ b = 2a.

Hence, b is twice the value of a.

15. We assume that P lies in the region of positive x. Since y = x3 is an odd

function, this assumption will not affect the result of the calculation. Let

P = a, a3 . The slope of the tangent to the curve y = x3 at P is 3a2, and so

the equation of the tangent is y − a3 = 3a2(x− a) ⇔ y = 3a2x− 2a3.

We solve this simultaneously with y = x3 to find the other point of intersection:

x3 = 3a2x− 2a3 ⇔ (x− a)2(x+ 2a) = 0. So Q = −2a,−8a3 is

the other point of intersection. The equation of the tangent at Q is

y − (−8a3) = 12a2[x− (−2a)] ⇔ y = 12a2x+ 16a3. By symmetry,

this tangent will intersect the curve again at x = −2(−2a) = 4a. The curve lies above the first tangent, and

below the second, so we are looking for a relationship between A = a

−2a x3 − (3a2x− 2a3) dx and

B =4a

−2a (12a2x+ 16a3)− x3 dx. We calculate A = 14x4 − 3

2a2x2 + 2a3x

a

−2a =34a4 − (−6a4) = 27

4a4, and

B = 6a2x2 + 16a3x− 14x4

4a

−2a = 96a4 − (−12a4) = 108a4. We see that B = 16A = 24A. This is because our

calculation of area B was essentially the same as that of area A, with a replaced by −2a, so if we replace a with −2a in our

expression for A, we get 274 (−2a)4 = 108a4 = B.

6 APPLICATIONS OFINTEGRATION

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