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287

CHAPTER 6

Applications of the Definite Integral inGeometry, Science, and Engineering

EXERCISE SET 6.1

1. A =∫ 2

−1

(x2 + 1− x) dx = (x3/3 + x− x2/2)]2−1

= 9/2

2. A =∫ 4

0

(√x+ x/4) dx = (2x3/2/3 + x2/8)

]40

= 22/3

3. A =∫ 2

1

(y − 1/y2) dy = (y2/2 + 1/y)]21

= 1

4. A =∫ 2

0

(2− y2 + y) dy = (2y − y3/3 + y2/2)]20

= 10/3

5. (a) A =∫ 2

0

(2x− x2) dx = 4/3

(b) A =∫ 4

0

(√y − y/2) dy = 4/3

x

y

2

4

y=x 2

y=2x

(2,4)

6. Eliminate x to get y2 = 4(y + 4)/2, y2 − 2y − 8 = 0,(y − 4)(y + 2) = 0; y = −2, 4 with correspondingvalues of x = 1, 4.

(a) A =∫ 1

0

[2√x− (−2

√x)] dx+

∫ 4

1

[2√x− (2x− 4)] dx

=∫ 1

0

4√x dx+

∫ 4

1

(2√x− 2x+ 4) dx = 8/3 + 19/3 = 9

(b) A =∫ 4

−2

[(y/2 + 2)− y2/4] dy = 9

(4, 4)

(1, –2)

x

y

y2 = 4x

y = 2x – 4


288 Chapter 6

7. A =∫ 1

1/4

(√x− x2) dx = 49/192

14

(1, 1)

x

y

y = x2

y = √x

8. A =∫ 2

0

[0− (x3 − 4x)] dx

=∫ 2

0

(4x− x3) dx = 4

2x

y

y = x3 – 4x

9. A =∫ π/2

π/4

(0− cos 2x) dx

= −∫ π/2

π/4

cos 2x dx = 1/2

3 6

–1

1

x

y

y = cos 2x

10. Equate sec2 x and 2 to get sec2 x = 2,

1

2

x

y

y = sec2 x

(#, 2) (3, 2)

secx = ±√

2, x = ±π/4

A =∫ π/4

−π/4(2− sec2 x) dx = π − 2

11. A =∫ 3π/4

π/4

sin y dy =√

2

3

9

x

y

x = sin y

12. A =∫ 2

−1

[(x+ 2)− x2] dx = 9/2

(2, 4)

(–1, 1)x

y

y = x2

x = y – 2

13. A =∫ ln 2

0

(e2x − ex

)dx

=(

12e2x − ex

)]ln 2

0

= 1/2

2

4

x

y

ln 2

y = e2x

y = ex

14. A =∫ e

1

dy

y= ln y

]e1

= 1

1/e 1

1

e

x

y


Exercise Set 6.1 289

15. A =∫ 1

−1

(2

1 + x2− |x|

)dx

= 2∫ 1

0

(2

1 + x2− x)dx

=[4 tan−1 x− x2

]10

= π − 1

–1 1

1

2

x

y

16.1√

1− x2= 2, x = ±

√3

2, so

A =∫ √3/2

−√

3/2

(2− 1√

1− x2

)dx

=[2x− sin−1 x

]√3/2

−√

3/2

= 2√

3− 2π3

0.5

1

1.5

2

x

y

23– 2

3

1 – x21y =

y = 2

17. y = 2 + |x− 1| =

{3− x, x ≤ 11 + x, x ≥ 1

,

A =∫ 1

−5

[(−1

5x+ 7

)− (3− x)

]dx

+∫ 5

1

[(−1

5x+ 7

)− (1 + x)

]dx

=∫ 1

−5

(45x+ 4

)dx+

∫ 5

1

(6− 6

5x

)dx

= 72/5 + 48/5 = 24

(–5, 8)

(5, 6)y = 3 – x

y = 1 + x

y = – x + 715

x

y

18. A =∫ 2/5

0

(4x− x) dx

+∫ 1

2/5

(−x+ 2− x) dx

=∫ 2/5

0

3x dx+∫ 1

2/5

(2− 2x) dx = 3/5

(1, 1)

25

85( , )

x

y

y = 4x

y = x

y = –x + 2

19. A =∫ 1

0

(x3 − 4x2 + 3x) dx

+∫ 3

1

[−(x3 − 4x2 + 3x)] dx

= 5/12 + 32/12 = 37/12

4

–8

–1 4


290 Chapter 6

20. Equate y = x3 − 2x2 and y = 2x2 − 3xto get x3 − 4x2 + 3x = 0,x(x− 1)(x− 3) = 0;x = 0, 1, 3with corresponding values of y = 0,−1.9.

A =∫ 1

0

[(x3 − 2x2)− (2x2 − 3x)] dx

+∫ 3

1

[(2x3 − 3x)− (x3 − 2x2)] dx

=∫ 1

0

(x3 − 4x2 + 3x) dx+∫ 3

1

(−x3 + 4x2 − 3x) dx

=512

+83

=3712

9

–2

–1 3

21. From the symmetry of the region

A = 2∫ 5π/4

π/4

(sinx− cosx) dx = 4√

2

1

–1

0 o

22. The region is symmetric about the origin so

A = 2∫ 2

0

|x3 − 4x| dx = 8

3.1

–3.1

–3 3

23. A =∫ 0

−1

(y3 − y) dy +∫ 1

0

−(y3 − y) dy

= 1/2

1

–1

–1 1

24. A =∫ 1

0

[y3 − 4y2 + 3y − (y2 − y)

]dy

+∫ 4

1

[y2 − y − (y3 − 4y2 + 3y)

]dy

= 7/12 + 45/4 = 71/6

4.1

0–2.2 12.1

25. The curves meet when x =√

ln 2, so

A =∫ √ln 2

0

(2x− xex2) dx =

(x2 − 1

2ex

2)]√ln 2

0

= ln 2− 12

0.5 1

0.5

1

1.5

2

2.5

x

y



26. The curves meet for x = e−2√

2/3, e2√

2/3 thus

A =∫ e2

√2/3

e−2√

2/3

(3x− 1x√

1− (lnx)2

)dx

=(3 lnx− sin−1(lnx)

) ]e2√2/3

e−2√

2/3

= 4√

2− 2 sin−1

(2√

23

)1 2 3

5

10

15

20

x

y

27. True. If f(x)− g(x) = c > 0 then f(x) > g(x) so Formula (1) implies that

A =∫ b

a

[f(x)− g(x)] dx =∫ b

a

c dx = c(b− a).

If g(x)− f(x) = c > 0 then g(x) > f(x) so

A =∫ b

a

[g(x)− f(x)] dx =∫ b

a

c dx = c(b− a).

28. False. Let f(x) = 2x, g(x) = 0, a = −2, and b = 1. Then∫ b

a

[f(x)− g(x)] dx =∫ 1

−2

2x dx = x2

]1−2

= −3,

but the area of A is∫ 0

−2

(−2x) dx+∫ 1

0

2x dx = −x2

]0−2

+x2

]10

= 4+1 = 5.x

y

1

-4

2

-2

29. True. Since f and g are distinct, there is some point c in [a, b] for which f(c) 6= g(c). Supposef(c) > g(c). (The case f(c) < g(c) is similar.) Let p = f(c)− g(c) > 0. Since f − g is continuous,there is an interval [d, e] containing c such that f(x)− g(x) > p/2 for all x in [d, e]. So∫ e

d

[f(x)− g(x)] dx ≥ p

2(e− d) > 0. Hence

0 =∫ b

a

[f(x)− g(x)] dx =∫ d

a

[f(x)− g(x)] dx+∫ e

d

[f(x)− g(x)] dx+∫ b

e

[f(x)− g(x)] dx

>

∫ d

a

[f(x)− g(x)] dx+∫ e

b

[f(x)− g(x)] dx,

so at least one of∫ d

a

[f(x)− g(x)] dx and∫ e

b

[f(x)− g(x)] dx is negative. Therefore f(t)− g(t) < 0

for some point t in one of the intervals [a, d] and [b, e]. So the graph of f is above the graph of gat x = c and below it at x = t; by the Intermediate-Value Theorem, the curves cross somewherebetween c and t.(Note: It is not necessarily true that the curves cross at a point. For example, let

f(x) =

x if x < 0;0 if 0 ≤ x ≤ 1;

x− 1 if x > 1,

and g(x) = 0. Then∫ 2

−1

[f(x) − g(x)] dx = 0, and the curves cross between -1 and 2, but there’s

no single point at which they cross; they coincide for x in [0, 1].)

30. True. Let

h(x) ={f(x)− g(x) if f(x) ≥ g(x);

0 if f(x) < g(x),


292 Chapter 6

and

k(x) ={

0 if f(x) ≥ g(x);g(x)− f(x) if f(x) < g(x).

Let B =∫ b

a

h(x) dx and C =∫ b

a

k(x) dx. If the curves cross, then f(x) > g(x) on some interval

and f(x) < g(x) on some other interval, so B > 0 and C > 0. Note that h(x)+k(x) = |f(x)−g(x)|and h(x)− k(x) = f(x)− g(x), so

A =∫ b

a

|f(x)− g(x)| dx =∫ b

a

h(x) dx+∫ b

a

k(x) dx = B + C.

But

A =

∣∣∣∣∣∫ b

a

[f(x)− g(x)] dx

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

[h(x)− k(x)] dx

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

h(x) dx−∫ b

a

k(x) dx

∣∣∣∣∣= |B − C| < max(B,C) < B + C.

Our assumption that the graphs cross leads to a contradiction, so the graphs don’t cross.

31. The area is given by∫ k

0

(1/√

1− x2 − x) dx = sin−1 k − k2/2 = 1; solve for k to get

k ≈ 0.997301.

32. The curves intersect at x = a = 0 and x = b = 0.838422 so the area is∫ b

a

(sin 2x− sin−1 x) dx ≈ 0.174192.

33. Solve 3−2x = x6+2x5−3x4+x2 to find the real roots x = −3, 1; from a plot it is seen that the line

is above the polynomial when −3 < x < 1, so A =∫ 1

−3

(3−2x−(x6+2x5−3x4+x2)) dx = 9152/105

34. Solve x5 − 2x3 − 3x = x3 to find the roots x = 0,±12

√6 + 2

√21. Thus, by symmetry,

A = 2∫ √(6+2

√21)/2

0

(x3 − (x5 − 2x3 − 3x)) dx =274

+74

√21

35.∫ k

0

2√y dy =

∫ 9

k

2√y dy

∫ k

0

y1/2 dy =∫ 9

k

y1/2 dy

23k3/2 =

23

(27− k3/2)

k3/2 = 27/2

k = (27/2)2/3 = 9/ 3√

4

y = 9

y = k

x

y



36.∫ k

0

x2 dx =∫ 2

k

x2 dx

13k3 =

13

(8− k3)

k3 = 4

k = 3√

42

x

y

x = √y

x = k

37. (a) A =∫ 2

0

(2x− x2) dx = 4/3

(b) y = mx intersects y = 2x− x2 where mx = 2x− x2, x2 + (m− 2)x = 0, x(x+m− 2) = 0 sox = 0 or x = 2−m. The area below the curve and above the line is∫ 2−m

0

(2x−x2−mx) dx =∫ 2−m

0

[(2−m)x−x2] dx =[

12

(2−m)x2 − 13x3

]2−m0

=16

(2−m)3

so (2−m)3/6 = (1/2)(4/3) = 2/3, (2−m)3 = 4,m = 2− 3√

4.

38. The line through (0, 0) and (5π/6, 1/2) is y =3

5πx;

A =∫ 5π/6

0

(sinx− 3

5πx

)dx =

√3

2− 5

24π + 1

c

1 12

c56( ),

x

yy = sin x

39. The curves intersect at x = 0 and, by Newton’s Method, at x ≈ 2.595739080 = b, so

A ≈∫ b

0

(sinx− 0.2x) dx = −[cosx+ 0.1x2

]b0

≈ 1.180898334

40. By Newton’s Method, the points of intersection are at x ≈ ±0.824132312, so with

b = 0.824132312 we have A ≈ 2∫ b

0

(cosx− x2) dx = 2(sinx− x3/3)]b0≈ 1.094753609

41. By Newton’s Method the points of intersection are x = x1 ≈ 0.4814008713 and

x = x2 ≈ 2.363938870, and A ≈∫ x2

x1

(lnxx− (x− 2)

)dx ≈ 1.189708441.

42. By Newton’s Method the points of intersection are x = ±x1 where x1 ≈ 0.6492556537, thus

A ≈ 2∫ x1

0

(2

1 + x2− 3 + 2 cosx

)dx ≈ 0.826247888

43. The x-coordinates of the points of intersection are a ≈ −0.423028 and b ≈ 1.725171; the area is∫ b

a

(2 sinx− x2 + 1) dx ≈ 2.542696.

44. Let (a, k), where π/2 < a < π, be the coordinates of the point of intersection of y = k withy = sinx. Thus k = sin a and if the shaded areas are equal,∫ a

0

(k − sinx) dx =∫ a

0

(sin a− sinx) dx = a sin a+ cos a− 1 = 0

Solve for a to get a ≈ 2.331122, so k = sin a ≈ 0.724611.


294 Chapter 6

45.∫ 60

0

[v2(t) − v1(t)] dt = s2(60) − s2(0) − [s1(60) − s1(0)], but they are even at time t = 60, so

s2(60) = s1(60). Consequently the integral gives the difference s1(0) − s2(0) of their startingpoints in meters.

46. Since a1(0) = a2(0) = 0, A =∫ T

0

[a2(t)−a1(t)] dt = v2(T )−v1(T ) is the difference in the velocities

of the two cars at time T .

47. (a) It gives the area of the region that is between f and g when f(x) > g(x) minus the area ofthe region between f and g when f(x) < g(x), for a ≤ x ≤ b.

(b) It gives the area of the region that is between f and g for a ≤ x ≤ b.

48. (b) limn→+∞

∫ 1

0

(x1/n − x) dx = limn→+∞

[n

n+ 1x(n+1)/n − x2

2

]10

= limn→+∞

(n

n+ 1− 1

2

)= 1/2

49. Solve x1/2 + y1/2 = a1/2 for y to get

y = (a1/2 − x1/2)2 = a− 2a1/2x1/2 + x

A =∫ a

0

(a− 2a1/2x1/2 + x) dx = a2/6

a

ax

y

50. Solve for y to get y = (b/a)√a2 − x2 for the upper half of the ellipse; make use of symmetry to

get A = 4∫ a

0

b

a

√a2 − x2 dx =

4ba

∫ a

0

√a2 − x2 dx =

4ba· 1

4πa2 = πab.

51. First find all solutions of the equation f(x) = g(x) in the interval [a, b]; call them c1, · · · , cn. Letc0 = a and cn+1 = b. For i = 0, 1, · · · , n, f(x) − g(x) has constant sign on [ci, ci+1], so the area

bounded by x = ci and x = ci+1 is either∫ ci+1

ci

[f(x)−g(x)] dx or∫ ci+1

ci

[g(x)−f(x)] dx. Compute

each of these n+ 1 areas and add them to get the area bounded by x = a and x = b.

52. Let f(x) be the length of the intersection of R with the vertical line with x-coordinate x. Dividethe interval [a, b] into n subintervals, and use those to divide R into n strips. If the width of thek’th strip is ∆xk, approximate the area of the strip by f(x∗k)∆xk, where x∗k is a point in the k’thsubinterval. Add the approximate areas to approximate the entire area of R by the Riemann sumn∑k=1

f(x∗k)∆xk. Take the limit as n→ +∞ and the widths of the subintervals all approach zero, to

obtain the area of R,∫ b

a

f(x) dx.

Since f(x) is also the length of the intersection of S with the vertical line with x-coordinate x, wesimilarly find that the area of S equals the same integral, so R and S have the same area.

EXERCISE SET 6.2

1. V = π

∫ 3

−1

(3− x) dx = 8π 2. V = π

∫ 1

0

[(2− x2)2 − x2] dx

= π

∫ 1

0

(4− 5x2 + x4) dx

= 38π/15



3. V = π

∫ 2

0

14

(3− y)2 dy = 13π/6 4. V = π

∫ 2

1/2

(4− 1/y2) dy = 9π/2

5. V = π

∫ π/2

π/4

cosx dx = (1−√

2/2)π

3 6

–1

1

x

yy = √cos x

6. V = π

∫ 1

0

[(x2)2 − (x3)2] dx

= π

∫ 1

0

(x4 − x6) dx = 2π/35

1

1 (1, 1)

y = x2

y = x3

x

y

7. V = π

∫ 3

−1

(1 + y) dy = 8π

3

2x

y

x = √1 + y

8. V = π

∫ 3

0

[22 − (y + 1)] dy

= π

∫ 3

0

(3− y) dy = 9π/2

3 (2, 3)

x

y y = x2 – 1x = √y + 1

9. V =∫ 2

0

x4 dx = 32/5

2x

y

y = x2

10. V =∫ π/3

π/4

sec2 x dx =√

3− 1

3 4

-2

-1

1

2

x

y

y = sec x


296 Chapter 6

11. V = π

∫ 4

−4

[(25− x2)− 9] dx

= 2π∫ 4

0

(16− x2) dx = 256π/3

5

x

yy = √25 – x2

y = 3

12. V = π

∫ 3

−3

(9− x2)2 dx

= π

∫ 3

−3

(81− 18x2 + x4) dx = 1296π/5

–3 3

9

x

y

y = 9 – x2

13. V = π

∫ 4

0

[(4x)2 − (x2)2] dx

= π

∫ 4

0

(16x2 − x4) dx = 2048π/15

4

16 (4, 16)

x

y

y = x2y = 4x

14. V = π

∫ π/4

0

(cos2 x− sin2 x) dx

= π

∫ π/4

0

cos 2x dx = π/2

3

–1

1

x

yy = cos x

y = sin x

15. V = π

∫ ln 3

0

e2x dx =π

2e2x]ln 3

0= 4π 16. V = π

∫ 1

0

e−4x dx =π

4(1− e−4)

1

–1

1

x

y

17. V =∫ 2

−2

π1

4 + x2dx =

π

2tan−1(x/2)

]2−2

= π2/4

18. V =∫ 1

0

πe6x

1 + e6xdx =

π

6ln(1 + e6x)

]10

=π

6(ln(1 + e6)− ln 2)



19. V =∫ 1

0

(y1/3

)2

dy =35

11 xy

20. V =∫ 1

−1

(1− y2)2 dy = 2∫ 1

0

(1− y2)2 dy

= 2(

1− 23

+15

)=

1615

11

-1

xy

21. V = π

∫ 3π/4

π/4

csc2 y dy = 2π

–2 –1 1 2

3

6

9

x

y

x = csc y

22. V = π

∫ 1

0

(y − y4) dy = 3π/10

–1 1

–1

1 (1, 1)

x

y

x = √y

x = y2

23. V = π

∫ 2

−1

[(y + 2)2 − y4] dy = 72π/5

(4, 2)

x

y

x = y2

x = y + 2

(1, –1)

24. V = π

∫ 1

−1

[(2 + y2)2 − (1− y2)2

]dy

= π

∫ 1

−1

(3 + 6y2) dy = 10π

1 2

–1

1

x

y x = 2 + y2

x = 1 – y2

25. V =∫ 1

0

πe2y dy =π

2(e2 − 1

)26. V =

∫ 2

0

π

1 + y2dy = π tan−1 2

27. False. For example, consider the pyramid in Example 1, with the roles of the x- and y-axesinterchanged.

28. False. If the centers of the disks or washers don’t all lie on a line parallel to the x-axis, then Sisn’t a solid of revolution.


298 Chapter 6

29. False. For example, let S be the solid generated by rotating the region under y = ex over theinterval [0, 1]. Then A(x) = π(ex)2.

30. True. By Definition 5.8.1, the average value of A(x) is1

b− a

∫ b

a

A(x) dx =V

b− a.

31. V = π

∫ a

−a

b2

a2(a2 − x2) dx = 4πab2/3

–a a

b

x

y

bay = √a2 – x2

32. V = π

∫ 2

b

1x2

dx = π(1/b− 1/2);

π(1/b− 1/2) = 3, b = 2π/(π + 6)

33. V = π

∫ 0

−1

(x+ 1) dx

+ π

∫ 1

0

[(x+ 1)− 2x] dx

= π/2 + π/2 = π

–1 1

1

x

y

y = √2x

(1, √2)

y = √x + 1

34. V = π

∫ 4

0

x dx+ π

∫ 6

4

(6− x)2 dx

= 8π + 8π/3 = 32π/3

4 6

x

y

y = √x y = 6 – x

35. Partition the interval [a, b] with a = x0 < x1 < x2 < . . . < xn−1 < xn = b. Let x∗k be an arbitrarypoint of [xk−1, xk]. The disk in question is obtained by revolving about the line y = k the rectan-gle for which xk−1 < x < xk, and y lies between y = k and y = f(x); the volume of this disk is

∆Vk = π(f(x∗k)− k)2∆xk, and the total volume is given by V = π

∫ b

a

(f(x)− k)2 dx.

36. Assume for c < y < d that k ≤ v(y) ≤ w(y) (A similar proof holds for k ≥ v(y) ≥ w(y)). Partitionthe interval [c, d] withc = y0 < y1 < y2 < . . . < yn−1 < yn = d. Let y∗k be an arbitrary point of [yk−1, yk]. The washerin question is the region obtained by revolving the strip v(y∗k) < x < w(y∗k), yk−1 < y < yk aboutthe line x = k. The volume of this washer is ∆V = π[(v(y∗k) − k)2 − (w(y∗k) − k)2]∆yk, and thevolume of the solid obtained by rotating R is

V = π

∫ d

c

[(v(y)− k)2 − (w(y)− k)2] dy

37. (a) Intuitively, it seems that a line segment which is revolved about a line which is perpendicularto the line segment will generate a larger area, the farther it is from the line. This is because



the average point on the line segment will be revolved through a circle with a greater radius,and thus sweeps out a larger circle.Consider the line segment which connects a point (x, y) on the curve y =

√3− x to the point

(x, 0) beneath it. If this line segment is revolved around the x-axis we generate an area πy2.If on the other hand the segment is revolved around the line y = 2 then the area of theresulting (infinitely thin) washer is π[22− (2−y)2]. So the question can be reduced to askingwhether y2 ≥ [22− (2−y)2], y2 ≥ 4y−y2, or y ≥ 2. In the present case the curve y =

√3− x

always satisfies y ≤ 2, so V2 has the larger volume.

(b) The volume of the solid generated by revolving the area around the x-axis is

V1 = π

∫ 3

−1

(3− x) dx = 8π, and the volume generated by revolving the area around the line

y = 2 is V2 = π

∫ 3

−1

[22 − (2−√

3− x)2] dx =403π

38. (a) In general, points in the region R are farther from the y-axis than they are from the linex = 2.5, so by the reasoning in Exercise 33(a) the former should generate a larger volumethan the latter, i.e. the volume mentioned in Exercise 4 will be greater than that gotten byrevolving about the line x = 2.5.

(b) The original volume V1 of Exercise 4 is given by

V1 = π

∫ 2

1/2

(4− 1/y2) dy = 9π/2, and the other volume

V2 = π

∫ 2

1/2

[(1y− 2.5

)2

− (2− 2.5)2]dy =

(212− 10 ln 2

)π ≈ 3.568528194π,

and thus V1 is the larger volume.

39. V = π

∫ 3

0

(9− y2)2 dy

= π

∫ 3

0

(81− 18y2 + y4) dy

= 648π/5

9

3

x

yx = y2

40. V = π

∫ 9

0

[32 − (3−√x)2] dx

= π

∫ 9

0

(6√x− x) dx

= 135π/2

9x

y

y = √xy = 3

41. V = π

∫ 1

0

[(√x+ 1)2 − (x+ 1)2] dx

= π

∫ 1

0

(2√x− x− x2) dx = π/2

1

1x

y

x = y2

x = y

y = –1


300 Chapter 6

42. V = π

∫ 1

0

[(y + 1)2 − (y2 + 1)2] dy

= π

∫ 1

0

(2y − y2 − y4) dy = 7π/15

1

1

x

y

x = y2

x = y

x = –1

43. The region is given by the inequalities 0 ≤ y ≤ 1,√y ≤

x ≤ 3√y. For each y in the interval [0, 1] the cross-section

of the solid perpendicular to the axis x = 1 is a washerwith outer radius 1 − √y and inner radius 1 − 3

√y. The

area of this washer isA(y) = π[(1−√y)2 − (1− 3

√y)2]

= π(−2y1/2 + y + 2y1/3 − y2/3),so the volume is

V =∫ 1

0

A(y) dy = π

∫ 1

0

(−2y1/2 + y + 2y1/3 − y2/3) dy

= π

[−4

3y3/2 +

12y2 +

32y4/3 − 3

5y5/3

]10

=π

15

1

x

y

y

y=x 2

y=x 3

rotation axisx=1

44. The region is given by the inequalities 0 ≤ x ≤ 1, x3 ≤ y ≤ x2.For each x in the interval [0, 1] the cross-section of the solidperpendicular to the axis y = −1 is a washer with outer radius1 + x2 and inner radius 1 + x3. The area of this washer isA(x) = π

[(1 + x2)2 − (1 + x3)2

]= π(2x2 + x4 − 2x3 − x6), so

the volume is

V =∫ 1

0

A(x) dx = π

∫ 1

0

(2x2 + x4 − 2x3 − x6) dx

= π

[23x3 +

15x5 − 1

2x4 − 1

7x7

]10

=47π210

.1

1

x

y

x

y=x 2

y=x 3

rotation axis y=!1

45. A(x) = π(x2/4)2 = πx4/16,

V =∫ 20

0

(πx4/16) dx = 40, 000π ft3

46. V = π

∫ 1

0

(x− x4) dx = 3π/10



47. V =∫ 1

0

(x− x2)2 dx

=∫ 1

0

(x2 − 2x3 + x4) dx = 1/30

Square

(1, 1)

1

y = x

y = x2

x

y

48. A(x) =12π

(12√x

)2=

18πx,

V =∫ 4

0

18πx dx = π

4x

y

y = √x

49. On the upper half of the circle, y =√

1− x2, so:

(a) A(x) is the area of a semicircle of radius y, so

A(x) = πy2/2 = π(1− x2)/2; V =π

2

∫ 1

−1

(1− x2) dx = π

∫ 1

0

(1− x2) dx = 2π/3

1

–1

y = √1 – x2 x

yy

(b) A(x) is the area of a square of side 2y, so

A(x) = 4y2 = 4(1− x2); V = 4∫ 1

−1

(1− x2) dx = 8∫ 1

0

(1− x2) dx = 16/3

1

–1

y = √1 – x2 x

y2y

(c) A(x) is the area of an equilateral triangle with sides 2y, so

A(x) =√

34

(2y)2 =√

3y2 =√

3(1− x2);

V =∫ 1

−1

√3(1− x2) dx = 2

√3∫ 1

0

(1− x2) dx = 4√

3/3

x

y

1

–1

y = √1 – x2

2y

2y2y

50. The base of the tent is a hexagon of side r. An equation of the circle of radius r that lies in avertical x-y plane and passes through two opposite vertices of the base hexagon is x2 + y2 = r2.


302 Chapter 6

A horizontal, hexagonal cross section at height y above the base has area

A(y) =3√

32x2 =

3√

32

(r2 − y2), hence the volume is V =∫ r

0

3√

32

(r2 − y2) dy =√

3r3.

51. The two curves cross at x = b ≈ 1.403288534, so

V = π

∫ b

0

((2x/π)2 − sin16 x) dx+ π

∫ π/2

b

(sin16 x− (2x/π)2) dx ≈ 0.710172176.

52. Note that π2 sinx cos3 x = 4x2 for x = π/4. From the graph it is apparent that this is the firstpositive solution, thus the curves don’t cross on (0, π/4) and

V = π

∫ π/4

0

[(π2 sinx cos3 x)2 − (4x2)2] dx =148π5 +

172560

π6

53. V = π

∫ e

1

(1− (ln y)2) dy = π

54. V =∫ tan 1

0

π[x2 − x2 tan−1 x] dx =π

6[tan2 1− ln(1 + tan2 1)]

55. (a) V = π

∫ r

r−h(r2 − y2) dy = π(rh2 − h3/3) =

13πh2(3r − h)

(b) By the Pythagorean Theorem,

r2 = (r − h)2 + ρ2, 2hr = h2 + ρ2; from part (a),

V =πh

3(3hr − h2) =

πh

3

(32

(h2 + ρ2)− h2))

=16πh(3ρ2 + h2)

r

h

r x

y

x2 + y2 = r2

56. Find the volume generated by revolvingthe shaded region about the y-axis.

V = π

∫ −10+h

−10

(100− y2) dy =π

3h2(30− h)

Find dh/dt when h = 5 given that dV/dt = 1/2.

V =π

3(30h2 − h3),

dV

dt=π

3(60h− 3h2)

dh

dt,

12

=π

3(300− 75)

dh

dt,dh

dt= 1/(150π) ft/min

10h – 10

–10

h

x

y

x = √100 – y2

57. (a) The bulb is approximately a sphere of radius 1.25 cm attached to a cylinder of radius 0.625

cm and length 2.5 cm, so its volume is roughly43π(1.25)3 +π(0.625)2 ·2.5 ≈ 11.25 cm. (Other

answers are possible, depending on how we approximate the light bulb using familiar shapes.)

(b) ∆x =510

= 0.5; {y0, y1, · · · , y10} = {0, 2.00, 2.45, 2.45, 2.00, 1.46, 1.26, 1.25, 1.25, 1.25, 1.25};

left = π

9∑i=0

(yi2

)2

∆x ≈ 11.157;



right = π

10∑i=1

(yi2

)2

∆x ≈ 11.771; V ≈ average = 11.464 cm3

58. If x = r/2 then from y2 = r2 − x2 we get y =±√

3r/2. So the hole consists of a cylinder of radiusr/2 and length

√3r and two spherical caps of radius

r/2 and height (1 −√

3/2)r. The cylinder has volume

π(r

2

)2√3r =

π√

34

r3. From Exercise 55(a), each caphas volume

r2

x

y

x = !r2 – y2

!3r2

!3r2–

13π

[(1−√

32

)r

]2(3r −

(1−√

32

)r

)=

π

24(16− 9

√3)r3.

So the volume of the hole isπ√

34

r3 + 2π

24(16−9

√3)r3 =

π

6(8−3

√3)r3 and the volume remaining

is43πr3 − π

6(8− 3

√3)r3 =

π√

32

r3.

To obtain this by integrating, note that, for −√

3 ≤ y ≤√

3, the cross-section with y-coordinate yhas area A(y) = π[(r2 − y2)− r2/4] = π(3r2/4− y2), thus

V = π

∫ √3r/2

−√

3r/2

(3r2/4− y2) dy = 2π∫ √3r/2

0

(3r2/4− y2) dy =π√

32

r3

59. (a)

h

–4

x

y

0 ≤ h < 2

h – 4

(b)

–4

–2h

2 ≤ h ≤ 4

h – 4x

y

If the cherry is partially submerged then 0 ≤ h < 2 as shown in Figure (a); if it is totally sub-merged then 2 ≤ h ≤ 4 as shown in Figure (b). The radius of the glass is 4 cm and that of thecherry is 1 cm so points on the sections shown in the figures satisfy the equations x2 + y2 = 16and x2 + (y + 3)2 = 1. We will find the volumes of the solids that are generated when the shadedregions are revolved about the y-axis.

For 0 ≤ h < 2,

V = π

∫ h−4

−4

[(16− y2)− (1− (y + 3)2)] dy = 6π∫ h−4

−4

(y + 4) dy = 3πh2;

for 2 ≤ h ≤ 4,

V = π

∫ −2

−4

[(16− y2)− (1− (y + 3)2)] dy + π

∫ h−4

−2

(16− y2) dy

= 6π∫ −2

−4

(y + 4) dy + π

∫ h−4

−2

(16− y2) dy = 12π +13π(12h2 − h3 − 40)

=13π(12h2 − h3 − 4)


304 Chapter 6

so

V =

3πh2 if 0 ≤ h < 2

13π(12h2 − h3 − 4) if 2 ≤ h ≤ 4

60. x = h±√r2 − y2,

V = π

∫ r

−r

[(h+

√r2 − y2)2 − (h−

√r2 − y2)2

]dy

= 4πh∫ r

−r

√r2 − y2 dy

= 4πh(

12πr2)

= 2π2r2h

x

y

(x – h2) + y2 = r2

61. tan θ = h/x so h = x tan θ,

A(y) =12hx =

12x2 tan θ =

12

(r2 − y2) tan θ

because x2 = r2 − y2,

V =12

tan θ∫ r

−r(r2 − y2) dy

= tan θ∫ r

0

(r2 − y2) dy =23r3 tan θ

x

h

u

62. A(x) = (x tan θ)(2√r2 − x2)

= 2(tan θ)x√r2 − x2,

V = 2 tan θ∫ r

0

x√r2 − x2 dx

=23r3 tan θ

yx

√r2 – x2

x tan u

63. Each cross section perpendicular to they-axis is a square so

A(y) = x2 = r2 − y2,

18V =

∫ r

0

(r2 − y2) dy

V = 8(2r3/3) = 16r3/3

r

x = √r2 – y2

x

y

64. The regular cylinder of radius r and height h has the same circular cross sections as do those ofthe oblique cylinder, so by Cavalieri’s Principle, they have the same volume: πr2h.

65. Position an x-axis perpendicular to the bases of the solids. Let a be the smallest x-coordinate ofany point in either solid, and let b be the largest. Let A(x) be the common area of the cross-sections

of the solids at x-coordinate x. By equation (3), each solid has volume V =∫ b

a

A(x) dx, so they

are equal.

66. Equation (4) is obtained from equation (3) simply by interchanging the x- and y-axes. Equations(5) and (6) are special cases of equation (3) using particular formulas for A(x). Similarly, equations(7) and (8) are special case of equation (4), so they also follow from (3) by interchanging the axes.



EXERCISE SET 6.3

1. V =∫ 2

1

2πx(x2) dx = 2π∫ 2

1

x3 dx = 15π/2

2. V =∫ √2

0

2πx(√

4− x2 − x) dx = 2π∫ √2

0

(x√

4− x2 − x2) dx =8π3

(2−√

2)

3. V =∫ 1

0

2πy(2y − 2y2) dy = 4π∫ 1

0

(y2 − y3) dy = π/3

4. V =∫ 2

0

2πy[y − (y2 − 2)] dy = 2π∫ 2

0

(y2 − y3 + 2y) dy = 16π/3

5. V =∫ 1

0

2π(x)(x3) dx

= 2π∫ 1

0

x4 dx = 2π/5

–1 1

–1

1

x

y

y = x3

6. V =∫ 9

4

2πx(√x) dx

= 2π∫ 9

4

x3/2 dx = 844π/5

–9 –4 4 9

1

2

3

x

y

y = √x

7. V =∫ 3

1

2πx(1/x) dx = 2π∫ 3

1

dx = 4π

–3 –1 1 3

y = x1

x

y

8. V =∫ √π/2

0

2πx cos(x2) dx = π/√

2

√p2

x

y

y = cos (x2)

9. V =∫ 2

1

2πx[(2x− 1)− (−2x+ 3)] dx

= 8π∫ 2

1

(x2 − x) dx = 20π/3

(2, 3)

(2, –1)

(1, 1)

x

y

10. V =∫ 2

0

2πx(2x− x2) dx

= 2π∫ 2

0

(2x2 − x3) dx =83π

2

x

yy = 2x – x2


306 Chapter 6

11. V = 2π∫ 1

0

x

x2 + 1dx

= π ln(x2 + 1)]10

= π ln 2

–1 1

1

x

y

y = 1x2 + 1

12. V =∫ √3

1

2πxex2dx = πex

2]√3

1= π(e3 − e)

-√3 -1 1 √3

10

20

x

y

y = ex 2

13. V =∫ 1

0

2πy3 dy = π/2

1

x

y

x = y2

14. V =∫ 3

2

2πy(2y) dy = 4π∫ 3

2

y2 dy = 76π/3

23

x

y

x = 2y

15. V =∫ 1

0

2πy(1−√y) dy

= 2π∫ 1

0

(y − y3/2) dy = π/5 1x

y

x = !y

16. V =∫ 4

1

2πy(5− y − 4/y) dy

= 2π∫ 4

1

(5y − y2 − 4) dy = 9π

(1, 4)

(4, 1)x

y

x = 5 – y

x = 4/y

17. True. The surface area of the cylinder is 2π · [average radius] · [height], so by equation (1) thevolume equals the thickness times the surface area.

18. False. In the method of cylindrical shells we do not use cross-sections of the solid.

19. True. In 6.3.2 we integrate over an interval on the x-axis, which is perpendicular to the y-axis,which is the axis of revolution.

20. True. If f(x) = c for all x, then the Riemann sum equalsn∑k=1

2πx∗kf(x∗k)∆xk =n∑k=1

2πxk + xk−1

2c (xk − xk−1) = πc

n∑k=1

(x2k − x2

k−1) = πc (x2n − x2

0).



The volume equals∫ b

a

2πxf(x) dx =∫ b

a

2πxc dx = πcx2

]ba

= πc (b2 − a2).

Hence each Riemann sum equals the volume.

21. V = 2π∫ 2

1

xex dx = 2π(x− 1)ex]21

= 2πe2 22. V = 2π∫ π/2

0

x cosx dx = π2 − 2π

23. The volume is given by 2π∫ k

0

x sinx dx = 2π(sin k − k cos k) = 8; solve for k to get

k ≈ 1.736796.

24. (a)∫ b

a

2πx[f(x)− g(x)] dx (b)∫ d

c

2πy[f(y)− g(y)] dy

25. (a) V =∫ 1

0

2πx(x3 − 3x2 + 2x) dx = 7π/30

(b) much easier; the method of slicing would require that x be expressed in terms of y.

–1 1

x

y

y = x3 – 3x2 + 2x

26. Let a = x0 < x1 < x2 < . . . < xn−1 < xn = b be a partition of [a, b]. Let x∗k be the midpoint of[xk−1, xk].Revolve the strip xk−1 < x < xk, 0 < y < f(x∗k) about the line x = k. The result is acylindrical shell, a large coin with a very large hole through the center. The volume of the shell is∆Vk = 2π(x−k)f(x∗k)∆xk, just as the volume of a ring of average radius r, height y and thicknessh is 2πryh. Summing these volumes of cylindrical shells and taking the limit as max∆xk goes to

zero, we obtain V = 2π∫ b

a

(x− k)f(x) dx

27. (a) For x in [0,1], the cross-section with x-coordinate x has length x, and its distance from the

axis of revolution is 1− x, so the volume is∫ 1

0

2π(1− x)x dx.

(b) For y in [0,1], the cross-section with y-coordinate y has length 1 − y, and its distance from

the axis of revolution is 1 + y, so the volume is∫ 1

0

2π(1 + y)(1− y) dy.

28. (a) For x in [0,1], the cross-section with x-coordinate x has length√

1− x2, and its distance from

the axis of revolution is 1− x, so the volume is∫ 1

0

2π(1− x)√

1− x2 dx.

(b) For y in [0,1], the cross-section with y-coordinate y has length√

1− y2, and its distance from

the axis of revolution is 1 + y, so the volume is∫ 1

0

2π(1 + y)√

1− y2 dy.


308 Chapter 6

29. V =∫ 2

1

2π(x+ 1)(1/x3) dx

= 2π∫ 2

1

(x−2 + x−3) dx = 7π/4

–1 x 21x

y

y = 1/x3

x + 1

30. V =∫ 1

0

2π(1− y)y1/3 dy

= 2π∫ 1

0

(y1/3 − y4/3) dy = 9π/14

1

x

y

1 – y x = y1/3

31. x =h

r(r − y) is an equation of the line

through (0, r) and (h, 0) so

V =∫ r

0

2πy[h

r(r − y)

]dy

=2πhr

∫ r

0

(ry − y2) dy = πr2h/3

x

y(0, r)

(h, 0)

32. V =∫ k/4

0

2π(k/2− x)2√kx dx

= 2π√k

∫ k/4

0

(kx1/2 − 2x3/2) dx = 7πk3/60 x

y

x = k/2

k/2 – x

x = k/4

y = √kx

y = –√kx

33. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2 + (L/2)2 = R2.Use cylindrical shells to calculate the volume of the solid obtained by rotating about the y-axisthe region r < x < R, −

√R2 − x2 < y <

√R2 − x2:

V =∫ R

r

(2πx)2√R2 − x2 dx = −4

3π(R2 − x2)3/2

]Rr

=43π(L/2)3,

so the volume is independent of R.

34. V =∫ a

−a2π(b− x)(2

√a2 − x2) dx

= 4πb∫ a

−a

√a2 − x2 dx− 4π

∫ a

−ax√a2 − x2 dx

= 4πb · (area of a semicircle of radius a)− 4π(0)

= 2π2a2b

a–ax

y

√a2 – x2

–√a2 – x2

b – x

x = b

35. Vx = π

∫ b

1/2

1x2

dx = π(2− 1/b), Vy = 2π∫ b

1/2

dx = π(2b− 1);

Vx = Vy if 2− 1/b = 2b− 1, 2b2 − 3b+ 1 = 0, solve to get b = 1/2 (reject) or b = 1.

36. (a) V = 2π∫ b

1

x

1 + x4dx = π tan−1(x2)

]b1

= π[tan−1(b2)− π

4

]



(b) limb→+∞

V = π(π

2− π

4

)=

14π2

37. If the formula for the length of a cross-section perpendicular to the axis of revolution is simplerthan the formula for the length of a cross-section parallel to the axis of revolution, then the methodof disks/washers is probably easier. Otherwise the method of cylindrical shells probably is.

38. In the method of disks/washers, we integrate the area of a flat surface, perpendicular to the axisof revolution. The variable of integration measures distance along the axis of revolution.

In the method of cylindrical shells, we integrate the area of a curved surface surrounding the axis ofrevolution. The variable of integration measures distance perpendicular to the axis of revolution.

EXERCISE SET 6.4

1. By the Theorem of Pythagoras, the length is√

(2− 1)2 + (4− 2)2 =√

1 + 4 =√

5.

(a)dy

dx= 2, L =

∫ 2

1

√1 + 4 dx =

√5

(b)dx

dy=

12, L =

∫ 4

2

√1 + 1/4 dy = 2

√5/4 =

√5

2. By the Theorem of Pythagoras, the length is√

(1− 0)2 + (5− 0)2 =√

1 + 25 =√

26.

(a)dy

dx= 5, L =

∫ 1

0

√1 + 25 dx =

√26

(b)dx

dy=

15, L =

∫ 5

0

√1 + 1/25 dy = 5

√26/25 =

√26

3. f ′(x) =92x1/2, 1 + [f ′(x)]2 = 1 +

814x,

L =∫ 1

0

√1 + 81x/4 dx =

8243

(1 +

814x

)3/2]1

0

= (85√

85− 8)/243

4. g′(y) = y(y2 + 2)1/2, 1 + [g′(y)]2 = 1 + y2(y2 + 2) = y4 + 2y2 + 1 = (y2 + 1)2,

L =∫ 1

0

√(y2 + 1)2 dy =

∫ 1

0

(y2 + 1) dy = 4/3

5.dy

dx=

23x−1/3, 1 +

(dy

dx

)2= 1 +

49x−2/3 =

9x2/3 + 49x2/3

,

L =∫ 8

1

√9x2/3 + 43x1/3

dx =118

∫ 40

13

u1/2du, u = 9x2/3 + 4

=127u3/2

]4013

=127

(40√

40− 13√

13) =127

(80√

10− 13√

13)


310 Chapter 6

or (alternate solution)

x = y3/2,dx

dy=

32y1/2, 1 +

(dx

dy

)2= 1 +

94y =

4 + 9y4

,

L =12

∫ 4

1

√4 + 9y dy =

118

∫ 40

13

u1/2du =127

(80√

10− 13√

13)

6. f ′(x) =14x3 − x−3, 1 + [f ′(x)]2 = 1 +

(116x6 − 1

2+ x−6

)=

116x6 +

12

+ x−6 =(

14x3 + x−3

)2,

L =∫ 3

2

√(14x3 + x−3

)2dx =

∫ 3

2

(14x3 + x−3

)dx = 595/144

7. x = g(y) =124y3 + 2y−1, g′(y) =

18y2 − 2y−2,

1 + [g′(y)]2 = 1 +(

164y4 − 1

2+ 4y−4

)=

164y4 +

12

+ 4y−4 =(

18y2 + 2y−2

)2,

L =∫ 4

2

(18y2 + 2y−2

)dy = 17/6

8. g′(y) =12y3 − 1

2y−3, 1 + [g′(y)]2 = 1 +

(14y6 − 1

2+

14y−6

)=(

12y3 +

12y−3

)2,

L =∫ 4

1

(12y3 +

12y−3

)dy = 2055/64

9. False. The derivativedy

dx= − x√

1− x2is not defined at x = ±1, so it is not continuous on [−1, 1].

10. True. In a Riemann sum the k’th term has the form g(x∗k)∆xk for some function g.

11. True. If f(x) = mx+ c then the approximation equalsn∑k=1

√1 +m2 ∆xk =

n∑k=1

√1 +m2 (xk − xk−1) =

√1 +m2 (xn − x0) = (b− a)

√1 +m2

and the arc length is the distance from (a,ma+ c) to (b,mb+ c), which equals√(b− a)2 + [(mb+ c)− (ma+ c)]2 =

√(b− a)2 + [m(b− a)]2 = (b− a)

√1 +m2.

So each approximation equals the arc length.

12. False. We only need f to be continuous on [a, b] and differentiable on (a, b).

13. dy/dx =secx tanx

secx= tanx,

√1 + (y′)2 =

√1 + tan2 x = secx when 0 < x < π/4, so

L =∫ π/4

0

secx dx = ln(1 +√

2)

14. dy/dx =cosxsinx

= cotx,√

1 + (y′)2 =√

1 + cot2 x = cscx when π/4 < x < π/2, so

L =∫ π/2

π/4

cscx dx = − ln(√

2− 1) = − ln

(√2− 1√2 + 1

(√

2 + 1)

)= ln(1 +

√2)



15. (a)

(–1, 1)

(8, 4)

x

y (b) dy/dx does not exist at x = 0.

(c) x = g(y) = y3/2, g′(y) =32y1/2,

L =∫ 1

0

√1 + 9y/4 dy (portion for − 1 ≤ x ≤ 0)

+∫ 4

0

√1 + 9y/4 dy (portion for 0 ≤ x ≤ 8)

=827

(138

√13− 1

)+

827

(10√

10− 1) = (13√

13 + 80√

10− 16)/27

16. First we apply equation (3) with a = 1, b = 2, f(x) = x2, and f ′(x) = 2x. The arc length is

L =∫ 2

1

√1 + (2x)2 dx =

∫ 2

1

√1 + 4x2 dx.

Next we apply equation (5) with c = 1, d = 4, g(y) =√y, and g′(y) = 1

2y−1/2.The arc length is

L =∫ 4

1

√1 +

(12y−1/2

)2

dy =∫ 4

1

√1 +

14y

dy

To see that these are equal, let y = x2, dy = 2x dx in the second integral:∫ 4

1

√1 +

14y

dy =∫ 2

1

√1 +

14x2

2x dx =∫ 2

1

√1 + 4x2 dx.

17. (a) The function y = f(x) = x2 is inverse to the function x =g(y) =

√y : f(g(y)) = y for 1/4 ≤ y ≤ 4, and g(f(x)) = x for

1/2 ≤ x ≤ 2. Geometrically this means that the graphs ofy = f(x) and x = g(y) are symmetric to each other with respectto the line y = x and hence have the same arc length.

1 2 3 4

1

2

3

4

x

y

(b) L1 =∫ 2

1/2

√1 + (2x)2 dx and L2 =

∫ 4

1/4

√1 +

(1

2√x

)2

dx

Make the change of variables x =√y in the first integral to obtain

L1 =∫ 4

1/4

√1 + (2

√y)2

12√ydy =

∫ 4

1/4

√(1

2√y

)2

+ 1 dy = L2

(c) L1 =∫ 4

1/4

√1 +

(1

2√y

)2

dy, L2 =∫ 2

1/2

√1 + (2y)2 dy

(d) For L1, ∆x =320, xk =

12

+ k320

=3k + 10

20, and thus

L1 ≈10∑k=1

√(∆x)2 + [f(xk)− f(xk−1)]2

=10∑k=1

√(320

)2

+(

(3k + 10)2 − (3k + 7)2

400

)2

≈ 4.072396336


312 Chapter 6

For L2, ∆x =1540

=38, xk =

14

+3k8

=3k + 2

8, and thus

L2 ≈10∑k=1

√√√√(38

)2

+

[√3k + 2

8−√

3k − 18

]2

≈ 4.071626502

(e) Each polygonal path is shorter than the curve segment, so both approximations in (d) aresmaller than the actual length. Hence the larger one, the approximation for L1, is better.

(f) For L1, ∆x =320

, the midpoint is x∗k =12

+(k − 1

2

)320

=6k + 17

40, and thus

L1 ≈10∑k=1

320

√1 +

(2

6k + 1740

)2

≈ 4.072396336.

For L2,∆x =1540

, and the midpoint is x∗k =14

+(k − 1

2

)1540

=6k + 1

16, and thus

L2 ≈10∑k=1

1540

√1 +

(4

6k + 116

)−1

≈ 4.066160149

(g) L1 =∫ 2

1/2

√1 + (2x)2 dx ≈ 4.0729, L2 =

∫ 4

1/4

√1 +

(1

2√x

)2

dx ≈ 4.0729

18. (a) The function y = f(x) = x8/3 is inverse to the function x =g(y) = y3/8 : f(g(y)) = y for 10−8 ≤ y ≤ 1 and g(f(x)) = xfor 10−3 ≤ x ≤ 1. Geometrically this means that the graphsof y = f(x) and x = g(y) are symmetric to each other withrespect to the line y = x.

0.2 0.6 1

0.2

0.6

1

x

y

(b) L1 =∫ 1

10−3

√1 +

(83x5/3

)2

dx,

L2 =∫ 1

10−8

√1 +

(38x−5/8

)2

dx

In the expression for L1 make the change of variable y = x8/3. Then

L1 =∫ 1

10−8

√1 +

(83y5/8

)2 38y−5/8 dy =

∫ 1

10−8

√(38y−5/8

)2

+ 1 dy = L2

(c) L1 =∫ 1

10−8

√1 +

(38y−5/8

)2

dy, L2 =∫ 1

10−3

√1 +

(83y5/3

)2

dy

(d) For L1, ∆x =999

10000, xk =

11000

+ k999

10000, and thus

L1 ≈10∑k=1

√(∆x)2 + [f(xk)− f(xk−1)]2

=10∑k=1

√√√√( 99910000

)2

+

[(1

1000+

999k10000

)8/3

−(

11000

+999(k − 1)

10000

)8/3]2

≈ 1.524983407

For L2, ∆y =99999999

1000000000, yk = 10−8 + k

999999991000000000

, and thus



L2 ≈10∑k=1

√(∆y)2 + [g(yk)− g(yk−1)]2 ≈ 1.518667833


(f) For L1, ∆x =999

10000, the midpoint is x∗k = 10−3 +

(k − 1

2

)999

10000, and thus

L1 ≈10∑k=1

99910000

√1 +

(83

(x∗k)5/3)2

≈ 1.524166463.

For L2,∆y =99999999

1000000000, the midpoint is y∗k = 10−8 +

(k − 1

2

)99999999

1000000000and thus

L2 ≈10∑k=1

√1 + g′(y∗k)2 ∆y ≈ 1.347221106

(g) L1 =∫ 1

10−3

√1 +

(83x5/3

)2

≈ 1.525898203, L2 =∫ 1

10−8

√1 +

(38y−5/8

)2

dy ≈ 1.526898203

19. (a) The function y = f(x) = tanx is inverse to the function x =g(y) = tan−1 x : f(g(y)) = y for 0 ≤ y ≤

√3, and g(f(x)) = x

for 0 ≤ x ≤ π/3. Geometrically this means that the graphs ofy = f(x) and x = g(y) are symmetric to each other with respectto the line y = x.

0.5 1 1.5 2

0.5

1

1.5

2

x

y

(b) L1 =∫ π/3

0

√1 + sec4 x dx, L2 =

∫ √3

0

√1 +

1(1 + x2)2

dx

In the expression for L1 make the change of variabley = tanx to obtain

L1 =∫ √3

0

√1 + (

√1 + y2)4

11 + y2

dy =∫ √3

0

√1

(1 + y2)2+ 1 dy = L2

(c) L1 =∫ √3

0

√1 +

1(1 + y2)2

dy, L2 =∫ π/3

0

√1 + sec4 y dy

(d) For L1, ∆xk =π

30, xk = k

π

30, and thus

L1 ≈10∑k=1

√(∆xk)2 + [f(xk)− f(xk−1)]2

=10∑k=1

√( π30

)2

+ [tan(kπ/30)− tan((k − 1)π/30)]2 ≈ 2.056603923

For L2, ∆xk =√

310, xk = k

√3

10, and thus

L2 ≈10∑k=1

√√√√(√310

)2

+

[tan−1

(k

√3

10

)− tan−1

((k − 1)

√3

10

)]2

≈ 2.056724591



314 Chapter 6

(f) For L1, ∆xk =π

30, the midpoint is x∗k =

(k − 1

2

)π

30, and thus

L1 ≈10∑k=1

π

30

√1 + sec4

[(k − 1

2

)π

30

]≈ 2.050944217.

For L2,∆xk =√

310

, and the midpoint is x∗k =(k − 1

2

) √3

10, and thus

L2 ≈10∑k=1

√3

10

√1 +

1((x∗k)2 + 1)2

≈ 2.057065139

(g) L1 =∫ π/3

0

√1 + sec4 x dx ≈ 2.0570

L2 =∫ √3

0

√1 +

1(12 + y2)2

dx ≈ 2.0570

20. 0 ≤ m ≤ f ′(x) ≤M , so m2 ≤ [f ′(x)]2 ≤M2, and 1 +m2 ≤ 1 + [f ′(x)]2 ≤ 1 +M2; thus√

1 +m2 ≤√

1 + [f ′(x)]2 ≤√

1 +M2,∫ b

a

√1 +m2 dx ≤

∫ b

a

√1 + [f ′(x)]2 dx ≤

∫ b

a

√1 +M2 dx, and

(b− a)√

1 +m2 ≤ L ≤ (b− a)√

1 +M2

21. f ′(x) = secx tanx, 0 ≤ secx tanx ≤ 2√

3 for 0 ≤ x ≤ π/3 soπ

3≤ L ≤ π

3

√13.

22. The distance is∫ 4.6

0

√1 + (2.09− 0.82x)2 dx ≈ 6.65 m

23. L =∫ π

0

√1 + (k cosx)2 dx 1 2 1.84 1.83 1.832

3.8202 5.2704 5.0135 4.9977 5.0008

k

L

Experimentation yields the values in the table, which by the Intermediate-Value Theorem showthat the true solution k to L = 5 lies between k = 1.83 and k = 1.832, so k = 1.83 to two decimalplaces.

24. (a)

100 200

–1.6

–1.2

–0.8

–0.4

xy (b) The maximum deflection occurs at

x = 96 inches (the midpointof the beam) and is about 1.42 in.

(c) The length of the centerline is∫ 192

0

√1 + (dy/dx)2 dx ≈ 192.03 in.

25. y = 0 at x = b = 12.54/0.41 ≈ 30.585; distance =∫ b

0

√1 + (12.54− 0.82x)2 dx ≈ 196.31 yd



26. Let Pk be the point on the curve with coordinates (x(tk), y(tk)). The length of the curve isapproximately the length of the polygonal path P0P1 · · ·Pn, which equalsn∑k=1

√(x(tk)− x(tk−1))2 + (y(tk)− y(tk−1))2.

27. (dx/dt)2 + (dy/dt)2 = (t2)2 + (t)2 = t2(t2 + 1), L =∫ 1

0

t(t2 + 1)1/2dt = (2√

2− 1)/3

28. (dx/dt)2 + (dy/dt)2 = [2(1 + t)]2 + [3(1 + t)2]2 = (1 + t)2[4 + 9(1 + t)2],

L =∫ 1

0

(1 + t)[4 + 9(1 + t)2]1/2dt = (80√

10− 13√

13)/27

29. (dx/dt)2 + (dy/dt)2 = (−2 sin 2t)2 + (2 cos 2t)2 = 4, L =∫ π/2

0

2 dt = π

30. (dx/dt)2 + (dy/dt)2 = (− sin t+ sin t+ t cos t)2 + (cos t− cos t+ t sin t)2 = t2,

L =∫ π

0

t dt = π2/2

31. (dx/dt)2 + (dy/dt)2 = [et(cos t− sin t)]2 + [et(cos t+ sin t)]2 = 2e2t,

L =∫ π/2

0

√2etdt =

√2(eπ/2 − 1)

32. (dx/dt)2 + (dy/dt)2 = (2et cos t)2 + (−2et sin t)2 = 4e2t, L =∫ 4

1

2et dt = 2(e4 − e)

33. (a) (dx/dt)2 + (dy/dt)2 = 4 sin2 t+ cos2 t = 4 sin2 t+ (1− sin2 t) = 1 + 3 sin2 t,

L =∫ 2π

0

√1 + 3 sin2 t dt = 4

∫ π/2

0

√1 + 3 sin2 t dt

(b) 9.69

(c) distance traveled =∫ 4.8

1.5

√1 + 3 sin2 t dt ≈ 5.16 cm

34. (dx/dt)2 + (dy/dt)2 = (−a sin t)2 + (b cos t)2 = a2 sin2 t+ b2 cos2 t

= a2(1− cos2 t) + b2 cos2 t = a2 − (a2 − b2) cos2 t

= a2

[1− a2 − b2

a2cos2 t

]= a2[1− k2 cos2 t],

L =∫ 2π

0

a√

1− k2 cos2 t dt = 4a∫ π/2

0

√1− k2 cos2 t dt

35. The length of the curve is approximated by the length of a polygon whose vertices lie on the graphof y = f(x). Each term in the sum is the length of one edge of the approximating polygon. By thedistance formula, the length of the k’th edge is

√(∆xk)2 + (∆yk)2, where ∆xk is the change in x

along the edge and ∆yk is the change in y along the edge. We use the Mean-Value Theorem toexpress ∆yk as f ′(x∗k)∆xk. Factoring the ∆xk out of the square root yields the k’th term in thesum.


316 Chapter 6

36. To apply Formula (4), we need to have a formula for dy/dx as a function of x. This may not bepossible if either

(A) the curve is defined by giving x as a function of y, or(B) the curve is defined by giving x and y as functions of a parameter t, or(C) the curve is defined implicitly by giving an equation satisfied by x and y.

In case (A), we may use Formula (5) instead of (4); for (B) we may use the result of Exercise 26. Incase (C), we may have to settle for an approximation, by finding approximate coordinates of manypoints P0, P1, · · ·, Pn on the curve, and computing the length of the polygonal path P0P1 · · ·Pn.

Even in cases where Formula (4) can be applied, we may be unable to evaluate the integral inclosed form, so we’ll have to use methods for approximate integration, as discussed in Section 7.7.

EXERCISE SET 6.5

1. S =∫ 1

0

2π(7x)√

1 + 49 dx = 70π√

2∫ 1

0

x dx = 35π√

2

2. f ′(x) =1

2√x

, 1 + [f ′(x)]2 = 1 +1

4x

S =∫ 4

1

2π√x

√1 +

14x

dx = 2π∫ 4

1

√x+ 1/4 dx = π(17

√17− 5

√5)/6

3. f ′(x) = −x/√

4− x2, 1 + [f ′(x)]2 = 1 +x2

4− x2=

44− x2

,

S =∫ 1

−1

2π√

4− x2(2/√

4− x2) dx = 4π∫ 1

−1

dx = 8π

4. y = f(x) = x3 for 1 ≤ x ≤ 2, f ′(x) = 3x2,

S =∫ 2

1

2πx3√

1 + 9x4 dx =π

27(1 + 9x4)3/2

]21

= 5π(29√

145− 2√

10)/27

5. S =∫ 2

0

2π(9y + 1)√

82 dy = 2π√

82∫ 2

0

(9y + 1) dy = 40π√

82

6. g′(y) = 3y2, S =∫ 1

0

2πy3√

1 + 9y4 dy = π(10√

10− 1)/27

7. g′(y) = −y/√

9− y2, 1 + [g′(y)]2 =9

9− y2, S =

∫ 2

−2

2π√

9− y2 · 3√9− y2

dy = 6π∫ 2

−2

dy = 24π

8. g′(y) = −(1− y)−1/2, 1 + [g′(y)]2 =2− y1− y

,

S =∫ 0

−1

2π(2√

1− y)√

2− y√1− y

dy = 4π∫ 0

−1

√2− y dy = 8π(3

√3− 2

√2)/3

9. f ′(x) =12x−1/2 − 1

2x1/2, 1 + [f ′(x)]2 = 1 +

14x−1 − 1

2+

14x =

(12x−1/2 +

12x1/2

)2,

S =∫ 3

1

2π(x1/2 − 1

3x3/2

)(12x−1/2 +

12x1/2

)dx =

π

3

∫ 3

1

(3 + 2x− x2) dx = 16π/9



10. f ′(x) = x2 − 14x−2, 1 + [f ′(x)]2 = 1 +

(x4 − 1

2+

116x−4

)=(x2 +

14x−2

)2,

S =∫ 2

1

2π(

13x3 +

14x−1

)(x2 +

14x−2

)dx = 2π

∫ 2

1

(13x5 +

13x+

116x−3

)dx = 515π/64

11. x = g(y) =14y4 +

18y−2, g′(y) = y3 − 1

4y−3,

1 + [g′(y)]2 = 1 +(y6 − 1

2+

116y−6

)=(y3 +

14y−3

)2,

S =∫ 2

1

2π(

14y4 +

18y−2

)(y3 +

14y−3

)dy =

π

16

∫ 2

1

(8y7 + 6y + y−5) dy = 16,911π/1024

12. x = g(y) =√

16− y; g′(y) = − 12√

16− y, 1 + [g′(y)]2 =

65− 4y4(16− y)

,

S =∫ 15

0

2π√

16− y

√65− 4y

4(16− y)dy = π

∫ 15

0

√65− 4y dy = (65

√65− 5

√5)π

6

13. f ′(x) = cosx, 1 + [f ′(x)]2 = 1 + cos2 x,

S =∫ π

0

2π sinx√

1 + cos2 x dx = 2π(√

2 + ln(√

2 + 1)) ≈ 14.42

14. x = g(y) = tan y, g′(y) = sec2 y, 1 + [g′(y)]2 = 1 + sec4 y;

S =∫ π/4

0

2π tan y√

1 + sec4 y dy ≈ 3.84

15. f ′(x) = ex, 1 + [f ′(x)]2 = 1 + e2x, S =∫ 1

0

2πex√

1 + e2x dx ≈ 22.94

16. x = g(y) = ln y, g′(y) = 1/y, 1 + [g′(y)]2 = 1 + 1/y2; S =∫ e

1

2π√

1 + 1/y2 ln y dy ≈ 7.054965608

17. True, by equation (1) with r1 = 0, r2 = r, and l =√r2 + h2.

18. True. The lateral surface area of the cylinder is 2πr1 + r2

2l = π(r1 + r2)l; by equation (1) this

equals the area of the frustum.

19. True. If f(x) = c for all x then f ′(x) = 0 so the approximation isn∑k=1

2πc ∆xk = 2πc(b− a). Since

the surface is the lateral surface of a cylinder of length b−a and radius c, its area is also 2πc(b−a).

20. True. A true Riemann sum only involves one point x∗k in each interval, not two.

21. n = 20, a = 0, b = π,∆x = (b− a)/20 = π/20, xk = kπ/20,

S ≈ π20∑k=1

[sin(k − 1)π/20 + sin kπ/20]√

(π/20)2 + [sin(k − 1)π/20− sin kπ/20]2 ≈ 14.39


318 Chapter 6

22. We use equation (2) with x changed to y and with f(y) = ln y.n = 20, a = 1, b = e, ∆y = (b− a)/20 = (e− 1)/20, yk = 1 + k(e− 1)/20,

S =20∑k=1

π[ln yk−1 + ln yk]√

(∆y)2 + [ln yk − ln yk−1]2 ≈ 7.05

23. S =∫ b

a

2π[f(x) + k]√

1 + [f ′(x)]2 dx

24. Yes, since the area of a frustum was used to figure out how to define surface area in general.

25. f(x) =√r2 − x2, f ′(x) = −x/

√r2 − x2, 1 + [f ′(x)]2 = r2/(r2 − x2),

S =∫ r

−r2π√r2 − x2(r/

√r2 − x2) dx = 2πr

∫ r

−rdx = 4πr2

26. g(y) =√r2 − y2, g′(y) = −y/

√r2 − y2, 1 + [g′(y)]2 = r2/(r2 − y2),

S =∫ r

r−h2π√r2 − y2

√r2/(r2 − y2) dy = 2πr

∫ r

r−hdy = 2πrh

27. Suppose the two planes are y = y1 and y = y2, where −r ≤ y1 ≤ y2 ≤ r. Then the area of thezone equals the area of a spherical cap of height r− y1 minus the area of a spherical cap of heightr − y2. By Exercise 26, this is 2πr(r − y1) − 2πr(r − y2) = 2πr(y2 − y1), which only depends onthe radius r and the distance y2 − y1 between the planes.

28. 2πk√

1 + [f ′(x)]2 ≤ 2πf(x)√

1 + [f ′(x)]2 ≤ 2πK√

1 + [f ′(x)]2, so∫ b

a

2πk√

1 + [f ′(x)]2 dx ≤∫ b

a

2πf(x)√

1 + [f ′(x)]2 dx ≤∫ b

a

2πK√

1 + [f ′(x)]2 dx,

2πk∫ b

a

√1 + [f ′(x)]2 dx ≤ S ≤ 2πK

∫ b

a

√1 + [f ′(x)]2 dx, 2πkL ≤ S ≤ 2πKL

29. Note that 1 ≤ secx ≤ 2 for 0 ≤ x ≤ π/3. Let L be the arc length of the curve y = tanx

for 0 < x < π/3. Then L =∫ π/3

0

√1 + sec2 x dx, and by Exercise 24, and the inequaities above,

2πL ≤ S ≤ 4πL. But from the inequalities for secx above, we can show that√

2π/3 ≤ L ≤√

5π/3.

Hence, combining the two sets of inequalities, 2π(√

2π/3) ≤ 2πL ≤ S ≤ 4πL ≤ 4π√

5π/3. Toobtain the inequalities in the text, observe that

2π2

3< 2π

√2π3≤ 2πL ≤ S ≤ 4πL ≤ 4π

√5π3

<4π2

3

√13.

30. (a) 1 ≤√

1 + [f ′(x)]2 so 2πf(x) ≤ 2πf(x)√

1 + [f ′(x)]2,∫ b

a

2πf(x) dx ≤∫ b

a

2πf(x)√

1 + [f ′(x)]2 dx, 2π∫ b

a

f(x) dx ≤ S, 2πA ≤ S

(b) 2πA = S if f ′(x) = 0 for all x in [a, b] so f(x) is constant on [a, b].

31. Let a = t0 < t1 < . . . < tn−1 < tn = b be a partition of [a, b]. Then the lateral area of the frustum

of slant height ` =√

∆x2k + ∆y2

k and radii y(t1) and y(t2) is π(y(tk) + y(tk−1))`. Thus the area of

the frustum Sk is given by Sk = π(y(tk−1) + y(tk))√

[x(tk)− x(tk−1)]2 + [y(tk)− y(tk−1)]2 with

the limit as max ∆tk → 0 of S =∫ b

a

2πy(t)√x′(t)2 + y′(t)2 dt



32. Let a = t0 < t1 < . . . < tn−1 < tn = b be a partition of [a, b]. Then the lateral area of the frustum

of slant height ` =√

∆x2k + ∆y2

k and radii x(t1) and x(t2) is π(x(tk) +x(tk−1))`. Thus the area of

the frustum Sk is given by Sk = π(x(tk−1) + x(tk))√

[x(tk)− x(tk−1)]2 + [y(tk)− y(tk−1)]2 with

the limit as max ∆tk → 0 of S =∫ b

a

2πx(t)√x′(t)2 + y′(t)2 dt

33. x′ = 2t, y′ = 2, (x′)2 + (y′)2 = 4t2 + 4

S = 2π∫ 4

0

(2t)√

4t2 + 4dt = 8π∫ 4

0

t√t2 + 1dt =

8π3

(17√

17− 1)

34. x′ = −2 cos t sin t, y′ = 5 cos t, (x′)2 + (y′)2 = 4 cos2 t sin2 t+ 25 cos2 t,

S = 2π∫ π/2

0

5 sin t√

4 cos2 t sin2 t+ 25 cos2 t dt =π

6(145√

29− 625)

35. x′ = 1, y′ = 4t, (x′)2 + (y′)2 = 1 + 16t2, S = 2π∫ 1

0

t√

1 + 16t2 dt =π

24(17√

17− 1)

36. x′ = −2 sin t cos t, y′ = 2 sin t cos t, (x′)2 + (y′)2 = 8 sin2 t cos2 t

S = 2π∫ π/2

0

cos2 t√

8 sin2 t cos2 t dt = 4√

2π∫ π/2

0

cos3 t sin t dt =√

2π

37. x′ = −r sin t, y′ = r cos t, (x′)2 + (y′)2 = r2,

S = 2π∫ π

0

r sin t√r2 dt = 2πr2

∫ π

0

sin t dt = 4πr2

38. In each case we approximate a curve by a polygonal path and use a known formula (for lengthor surface area) to derive a more general formula. Both derivations involve the length of a linesegment, which is approximated using the Mean-Value Theorem, introducing

√1 + [f ′(x)]2 into

the resulting formulas.

39. Suppose we approximate the k’th frustum by the lateral surface of a cylinder of width ∆xk andradius f(x∗k), where x∗k is between xk−1 and xk. The area of this surface is 2πf(x∗k) ∆xk. Pro-ceeding as before, we would conclude that S =

∫ ba

2πf(x) dx, which is too small. Basically, when|f ′(x)| > 0, the area of the frustum is larger than the area of the cylinder, and ignoring this resultsin an incorrect formula.

EXERCISE SET 6.6

1. W =∫ 3

0

F (x) dx =∫ 3

0

(x+ 1) dx =[

12x2 + x

]30

= 7.5 ft·lb

2. W =∫ 5

0

F (x) dx =∫ 2

0

40 dx−∫ 5

2

403

(x− 5) dx = 80 + 60 = 140 J

3. Since W =∫ b

a

F (x) dx = the area under the curve, it follows that d < 2.5 since the area increases

faster under the left part of the curve. In fact, Wd =∫ d

0

F (x) dx = 40d, and

W =∫ 5

0

F (x) dx = 140, so d = 7/4.


320 Chapter 6

4. The total work is∫ b

a

F (x) dx. The average value of F over [a, b] is1

b− a

∫ b

a

F (x) dx, which

equals the work divided by the length of the interval.

5. distance traveled =∫ 5

0

v(t) dt =∫ 5

0

4t5dt =

25t2]50

= 10 ft. The force is a constant 10 lb, so the

work done is 10 · 10 = 100 ft·lb.

6. Hooke’s law says that F (x) = kx where x is the distance the spring is stretched beyond its naturallength. Since F (x) = 6 when x = 4 1

2 − 4 = 12 , we have k = 12. Stretching the spring to a length

of 6 meters corresponds to x = 6− 4 = 2, so

W =∫ 2

0

F (x) dx =∫ 2

0

12x dx = 6x2]20

= 24 N·m = 24 J

7. F (x) = kx, F (0.2) = 0.2k = 100, k = 500 N/m, W =∫ 0.8

0

500x dx = 160 J

8. (a) F (x) = kx, F (0.05) = 0.05k = 45, k = 900 N/m

(b) W =∫ 0.03

0

900x dx = 0.405 J (c) W =∫ 0.10

0.05

900x dx = 3.375 J

9. W =∫ 1

0

kx dx = k/2 = 10, k = 20 lb/ft

10. False. The distance that the car moves is 0, so no work is done.

11. False. The work depends on the force and the distance, not on the elapsed time.

12. True. If W1 =∫ D

0

kx dx =kx2

2

]D0

=kD2

2, then W2 =

∫ 2D

0

kx dx =kx2

2

]2D0

= 2kD2 = 4W1.

13. True. By equation (6), work and energy have the same units in any system of units.

14. W =∫ 6

0

(9− x)62.4(25π) dx

= 1560π∫ 6

0

(9− x) dx = 56,160π ft·lb 9 - x

x

0

6

95

15. W =∫ 9/2

0

(9− x)62.4(25π) dx

= 1560π∫ 9/2

0

(9− x) dx = 47,385π ft·lb 9 - x

x

0

4.5

95



16. r/10 = x/15, r = 2x/3,

W =∫ 10

0

(15− x)62.4(4πx2/9) dx

=83.2

3π

∫ 10

0

(15x2 − x3) dx

= 208, 000π/3 ft·lb

15 – x

x

0

10

1510

r

17. w/4 = x/3, w = 4x/3,

W =∫ 2

0

(3− x)(9810)(4x/3)(6) dx

= 78480∫ 2

0

(3x− x2) dx

= 261, 600 J

3 – x

x

0

2

34

w(x)

18. w = 2√

4− x2

W =∫ 2

−2

(3− x)(50)(2√

4− x2)(10) dx

= 3000∫ 2

−2

√4− x2 dx− 1000

∫ 2

−2

x√

4− x2 dx

= 3000[π(2)2/2]− 0 = 6000π ft·lb

3

2

0

–2

3 – x

x

w(x)

2

19. (a) W =∫ 9

0

(10− x)62.4(300) dx

= 18,720∫ 9

0

(10− x) dx

= 926,640 ft·lb

(b) To empty the pool in one hour would require926,640/3600 = 257.4 ft·lb of work per secondso hp of motor = 257.4/550 = 0.468.

0

109 10 – xx

20 15

20. W =∫ 9

0

x(62.4)(300) dx = 18,720∫ 9

0

x dx = (81/2)18,720 = 758,160 ft·lb

21. W =∫ 100

0

15(100− x) dx

= 75, 000 ft·lb100

0

100 – x

x

Pulley

Chain

22. The total time of winding the rope is (20 ft)/(2 ft/s) = 10 s. During the time interval from time tto time t+ ∆t the work done is ∆W = F (t) ·∆x.

The distance ∆x = 2∆t, and the force F (t) is given by the weight w(t) of the bucket, rope andwater at time t. The bucket and its remaining water together weigh (3 + 20)− t/2 lb, and the rope


322 Chapter 6

is 20− 2t ft long and weighs 4(20− 2t) oz or 5− t/2 lb. Thus at time t the bucket, water and ropetogether weigh w(t) = 23− t/2 + 5− t/2 = 28− t lb.

The amount of work done in the time interval from time t to time t+ ∆t is thus∆W = (28− t)2∆t, and the total work done is

W = limn→+∞

∑(28− t)2∆t =

∫ 10

0

(28− t)2 dt = 2(28t− t2/2)∣∣∣100

= 460 ft·lb.

23. When the rocket is x ft above the ground

total weight = weight of rocket + weight of fuel

= 3 + [40− 2(x/1000)]

= 43− x/500 tons,

W =∫ 3000

0

(43− x/500) dx = 120, 000 ft·tons

3000

0

xRocket

24. Let F (x) be the force needed to hold charge A at position x. Then

F (x) =c

(a− x)2, F (−a) =

c

4a2= k, so c = 4a2k.

W =∫ 0

−a4a2k(a− x)−2 dx = 2ak J

0–a ax

BA

25. (a) 150 = k/(4000)2, k = 2.4× 109, w(x) = k/x2 = 2,400,000,000/x2 lb

(b) 6000 = k/(4000)2, k = 9.6× 1010, w(x) =(9.6× 1010

)/(x+ 4000)2 lb

(c) W =∫ 5000

4000

9.6(1010)x−2 dx = 4,800,000 mi·lb = 2.5344× 1010 ft·lb

26. (a) 20 = k/(1080)2, k = 2.3328× 107, weight = w(x+ 1080) = 2.3328 · 107/(x+ 1080)2 lb

(b) W =∫ 10.8

0

[2.3328 · 107/(x+ 1080)2] dx = 213.86 mi·lb = 1,129,188 ft·lb

27. W =12mv2

f −12mv2

i =12

4.00× 105(v2f − 202). But W = F · d = (6.40× 105) · (3.00× 103), so

19.2× 108 = 2.00× 105v2f − 8.00× 107, 19200 = 2v2

f − 800, vf = 100 m/s.

28. W = F · d = (2.00× 105)(1.50× 105) = 3× 1010 J; from the Work-Energy Relationship (6),

v2f = 2W/m+ v2

i = 2(3× 1010)/(2× 103) + (1× 104)2 = 1.3× 108, so vf ≈ 11, 402 m/s.

29. (a) The kinetic energy would have decreased by12mv2 =

12

4 · 106(15000)2 = 4.5× 1014 J

(b) (4.5× 1014)/(4.2× 1015) ≈ 0.107 (c)100013

(0.107) ≈ 8.24 bombs

30. “Pushing/pulling” problems usually involve a single rigid object being moved; see Examples 1-4and 6. “Pumping” problems usually involve a liquid or flexible solid, different parts of which movedifferent distances; see Example 5 and Exercises 14-21. Exercise 22 is an example of a combinationof the two categories: The bucket is rigid, the water is liquid, and the rope is a flexible solid.



31. The work-energy relationship involves 4 quantities, the work W , the mass m, and the initial andfinal velocities vi and vf . In any problem in which 3 of these are given, the work-energy relationshipcan be used to compute the fourth.In cases where the force is constant, we may combine equation (1) with the work-energy relationship

to get Fd =12mv2

f −12mv2

i . In this form there are 5 quantities, the force F , the distance d, the

mass m, and the initial and final velocities vi and vf . So if any 4 of these are given, the work-energyrelationship can be used to compute the fifth.

EXERCISE SET 6.7

1. (a) m1 and m3 are equidistant from x = 5, but m3 has a greater mass, so the sum is positive.(b) Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum

is 5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35a = 0, a = 50/7. Thefulcrum should be placed 50/7 units to the right of m1.

2. (a) The sum must be negative, since m1,m2 and m3 are all to the left of the fulcrum, and themagnitude of the moment of m1 about x = 4 is by itself greater than the moment of m aboutx = 4 (i.e. 40 > 28), so even if we replace the masses of m2 and m3 with 0, the sum isnegative.

(b) At equilibrium, 10(0− 4) + 3(2− 4) + 4(3− 4) +m(6− 4) = 0,m = 25

3. By symmetry, the centroid is (1/2, 1/2). We confirm this using Formulas (8) and (9) with a = 0,

b = 1, f(x) = 1: The area is 1, so x =∫ 1

0

x dx =12

and y =∫ 1

0

12dx =

12

, as expected.

4. By symmetry, the centroid is (0, 0). We confirm this using Formulas (10) and (11) with

a = −1, b = 1, f(x) = 1 − |x|, g(x) = |x| − 1: The area is 2, so x =∫ 1

−1

x(2 − 2|x|) dx =

12

(∫ 0

−1

x(2 + 2x) dx+∫ 1

0

x(2− 2x) dx)

=12

([x2 +

23x3

]0−1

+[x2 − 2

3x3

]10

)=

12

(−1

3+

13

)=

0 and y =12

∫ 1

−1

12[(1− |x|)2 − (|x| − 1)2

]dx =

12

∫ 1

−1

0 dx = 0, as expected.

5. By symmetry, the centroid is (1, 1/2). We confirm this using Formulas (8) and (9) with a = 0,

b = 2, f(x) = 1: The area is 2, so x =12

∫ 2

0

x dx = 1 and y =12

∫ 2

0

12dx =

12

, as expected.

6. By symmetry, the centroid is (0, 0). We confirm this using Formulas (10) and (11) with

a = −1, b = 1, f(x) =√

1− x2, g(x) = −√

1− x2: The area is π, so x =1π

∫ 1

−1

x · 2√

1− x2 dx =

− 23π

(1− x2)3/2]1−1

= 0 and y =1π

∫ 1

−1

12· 0 dx = 0, as expected.

7. By symmetry, the centroid lies on the line y = 1 − x. To find x we use Formula (8) with a = 0,

b = 1, f(x) = x: The area is12

, so x = 2∫ 1

0

x2 dx =23

. Hence y = 1− 23

=13

and the centroid is(23,

13

).


324 Chapter 6

8. We use Formulas (8) and (9) with a = 0, b = 1, f(x) = x2: The area is∫ 1

0

x2 dx =13

, so

x = 3∫ 1

0

x3 dx =34

and y = 3∫ 1

0

12x4 dx =

310

. The centroid is(

34,

310

).

9. We use Formulas (10) and (11) with a = 0, b = 1, f(x) = 2− x2, g(x) = x: The area is∫ 1

0

(2− x2 − x) dx =[2x− 1

3x3 − 1

2x2

]10

=76

, so

x =67

∫ 1

0

x(2− x2 − x) dx =67

[x2 − 1

4x4 − 1

3x3

]10

=514

and

y =67

∫ 1

0

12

[(2− x2)2 − x2] dx =37

∫ 1

0

(4− 5x2 + x4) dx =37

[4x− 5

3x3 +

15x5

]10

=3835

.

The centroid is(

514,

3835

).

10. By symmetry the centroid lies on the line y = x. To find x, we use Formula (8) with a = 0, b = 1,

f(x) =√

1− x2: The area isπ

4, so x =

4π

∫ 1

0

x√

1− x2 dx =4π

[−1

3(1− x2)3/2

]10

=4

3π. The

centroid is(

43π,

43π

).

11. We use Formulas (8) and (9) with a = 0, b = 2, f(x) = 1− x

2: The area is 1, so

x =∫ 2

0

x(

1− x

2

)dx =

[12x2 − 1

6x3

]20

=23

and

y =∫ 2

0

12

(1− x

2

)2

dx =18

∫ 2

0

(4− 4x+ x2) dx =18

[4x− 2x2 +

13x3

]20

=13

.

The centroid is(

23,

13

).

12. By symmetry, x = 1. To find y we use the analogue of Formula (10) with the roles of x and yreversed: The triangle is described by 0 ≤ y ≤ 1, y ≤ x ≤ 2− y. The area is 1, so

y =∫ 1

0

y [(2− y)− y] dy =∫ 1

0

(2y − 2y2) dy =[y2 − 2

3y3

]10

=13

. The centroid is(

1,13

).

13. The graphs of y = x2 and y = 6− x meet when x2 = 6− x, so x = −3 or x = 2. We use Formulas(10) and (11) with a = −3, b = 2, f(x) = 6− x, g(x) = x2:

The area is∫ 2

−3

(6− x− x2) dx =[6x− 1

2x2 − 1

3x3

]2−3

=1256

, so

x =6

125

∫ 2

−3

x(6− x− x2) dx =6

125

[3x2 − 1

3x3 − 1

4x4

]2−3

= −12

and

y =6

125

∫ 2

−3

12

[(6− x)2 − (x2)2] dx =3

125

∫ 2

−3

(36− 12x+ x2 − x4) dx

=3

125

[36x− 6x2 +

13x3 − 1

5x5

]2−3

= 4. The centroid is(−1

2, 4)

.

14. We use Formulas (10) and (11) with a = 0, b = 2, f(x) = x + 6, g(x) = x2: The area is∫ 2

0

(x+ 6− x2) dx =[

12x2 + 6x− 1

3x3

]20

=343

, so

x =334

∫ 2

0

x(x+ 6− x2) dx =334

[13x3 + 3x2 − 1

4x4

]20

=1617

and



y =334

∫ 2

0

12

[(x+ 6)2 − (x2)2] dx =368

∫ 2

0

(x2 + 12x+ 36− x4) dx

=368

[13x3 + 6x2 + 36x− 1

5x5

]20

=34685

. The centroid is(

1617,

34685

).

15. The curves meet at (−1, 1) and (2, 4). We use Formulas (10) and (11) with a = −1, b = 2,

f(x) = x+ 2, g(x) = x2: The area is∫ 2

−1

(x+ 2− x2) dx =[

12x2 + 2x− 1

3x3

]2−1

=92

, so

x =29

∫ 2

−1

x(x+ 2− x2) dx =29

[13x3 + x2 − 1

4x4

]2−1

=12

and

y =29

∫ 2

−1

12[(x+ 2)2 − (x2)2

]dx =

19

∫ 2

−1

(x2 + 4x+ 4− x4) dx

=19

[13x3 + 2x2 + 4x− 1

5x5

]2−1

=85

. The centroid is(

12,

85

).

16. By symmetry, x = 0. To find y we use Formula (11) with a = −1, b = 1, f(x) = 1, g(x) = x2: The

area is∫ 1

−1

(1− x2) dx =[x− 1

3x3

]1−1

=43

, so

y =34

∫ 1

−1

12[12 − (x2)2

]dx =

38

∫ 1

−1

(1− x4) dx =38

[x− 1

5x5

]1−1

=35

. The centroid is(

0,35

).

17. By symmetry, y = x. To find x we use Formula (10) with a = 0, b = 1, f(x) =√x, g(x) = x2:

The area is∫ 1

0

(√x− x2) dx =

[23x3/2 − 1

3x3

]10

=13

, so

x = 3∫ 1

0

x(√x− x2) dx = 3

[25x5/2 − 1

4x4

]10

=920

. The centroid is(

920,

920

).

18. We use Formulas (12) and (13) with c = 1, d = 2, w(y) =1y

: The area is∫ 2

1

1ydy = ln 2, so

x =1

ln 2

∫ 2

1

12

(1y

)2

dy =1

2 ln 2

[−1y

]21

=1

4 ln 2and y =

1ln 2

∫ 2

1

y · 1ydy =

1ln 2

. The centroid is(1

4 ln 2,

1ln 2

).

19. We use the analogue of Formulas (10) and (11) with the roles of x and y reversed: The region

is described by 1 ≤ y ≤ 2, y−2 ≤ x ≤ y. The area is∫ 2

1

(y − y−2) dy =[

12y2 + y−1

]21

= 1, so

x =∫ 2

1

12

[y2 − (y−2)2] dy =12

∫ 2

1

(y2 − y−4) dy =12

[13y3 +

13y−3

]21

=4948

and

y =∫ 2

1

y(y − y−2) dy =[

13y3 − ln y

]21

=73− ln 2. The centroid is

(4948,

73− ln 2

).

20. By symmetry, y = x. To find x we use Formula (10) with a = 1, b = 4, f(x) = 5 − x,

g(x) = 4x−1: The area is∫ 4

1

(5 − x − 4x−1) dx =[5x− 1

2x2 − 4 lnx

]41

=15− 16 ln 2

2, so

x =2

15− 16 ln 2

∫ 4

1

x(5− x− 4x−1) dx =2

15− 16 ln 2

[52x2 − 1

3x3 − 4x

]41

=9

15− 16 ln 2.

The centroid is(

915− 16 ln 2

,9

15− 16 ln 2

).


326 Chapter 6

21. An isosceles triangle is symmetric across the median to its base. So, if the density is constant, itwill balance on a knife-edge under the median. Hence the centroid lies on the median.

22. An ellipse is symmetric across both its major axis and its minor axis. So, if the density is constant,it will balance on a knife-edge under either axis. Hence the centroid lies on both axes, so it is atthe intersection of the axes.

23. The region is described by 0 ≤ x ≤ 1, 0 ≤ y ≤√x. The area is A =

∫ 1

0

√x dx =

23

, so the mass

is M = δA = 2 · 23

=43

. By Formulas (8) and (9), x =32

∫ 1

0

x√x dx =

32

[25x5/2

]10

=35

and

y =32

∫ 1

0

12

(√x)2 dx =

34

∫ 1

0

x dx =38

. The center of gravity is(

35,

38

).

24. The region is described by −1 ≤ y ≤ 1, y4 ≤ x ≤ 1. The area is

A =∫ 1

−1

(1− y4) dy =[y − 1

5y5

]1−1

=85

, so the mass is M = δA = 15 · 85

= 24.

By symmetry, y = 0. By the analogue of Formula (11) with the roles of x and y reversed,

x =58

∫ 1

−1

12

[12 − (y4)2] dy =516

∫ 1

−1

(1− y8) dy =516

[y − 1

9y9

]1−1

=59

.

The center of gravity is(

59, 0)

.

25. The region is described by 0 ≤ y ≤ 1, −y ≤ x ≤ y. The area is A = 1, so the mass is M = δA =3 ·1 = 3. By symmetry, x = 0. By the analogue of Formula (10) with the roles of x and y reversed,

y =∫ 1

0

y[y − (−y)] dy =∫ 1

0

2y2 dy =23y3

]10

=23

. The center of gravity is(

0,23

).

26. The region is described by −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2. The area is A =∫ 1

−1

(1 − x2) dx =[x− 1

3x3

]1−1

=43

, so the mass is M = δA = 3 · 43

= 4. By symmetry, x = 0. By Formula (9),

y =34

∫ 1

−1

12

(1 − x2)2 dx =38

∫ 1

−1

(1 − 2x2 + x4) dx =38

[x− 2

3x3 +

15x5

]1−1

=25

. The center of

gravity is(

0,25

).

27. The region is described by 0 ≤ x ≤ π, 0 ≤ y ≤ sinx. The area is A =∫ π

0

sinx dx = 2, so the mass

is M = δA = 4 · 2 = 8. By symmetry, x =π

2. By Formula (9), y =

12

∫ π

0

12

(sinx)2 dx =π

8. The

center of gravity is(π

2,π

8

).

28. The region is described by 0 ≤ x ≤ 1, 0 ≤ y ≤ ex. The area is A =∫ 1

0

ex dx = e− 1, so the mass

is M = δA =1

e− 1· (e − 1) = 1. By Formulas (8) and (9), x =

1e− 1

∫ 1

0

xex dx =1

e− 1and

y =1

e− 1

∫ 1

0

12

(ex)2 dx =e+ 1

4. The center of gravity is

(1

e− 1,e+ 1

4

).



29. The region is described by 1 ≤ x ≤ 2, 0 ≤ y ≤ lnx. The area isA =∫ 2

1

lnx dx = 2 ln 2−1 = ln 4−1,

so the mass is M = δA = ln 4 − 1. By Formulas (8) and (9), x =1

ln 4− 1

∫ 2

1

x lnx dx =

1ln 4− 1

(ln 4− 3

4

)=

4 ln 4− 34(ln 4− 1)

and y =1

ln 4− 1

∫ 2

1

12

(lnx)2 dx =(ln 2)2 − ln 4 + 1

ln 4− 1. The center

of gravity is(

4 ln 4− 34(ln 4− 1)

,(ln 2)2 − ln 4 + 1

ln 4− 1

).

30. The region is described by 0 ≤ x ≤ π

4, sinx ≤ y ≤ cosx. The area is

A =∫ π/4

0

(cosx−sinx) dx =√

2−1, so the mass is M = δA = (1+√

2)(√

2−1) = 1. By Formulas

(10) and (11), x =1√

2− 1

∫ π/4

0

x(cosx− sinx) dx =(π√

2− 4)(√

2 + 1)4

and

y =1√

2− 1

∫ π/4

0

12

(cos2 x−sin2 x) dx =√

2 + 14

. The center of gravity is

((π√

2− 4)(√

2 + 1)4

,

√2 + 14

).

31. True, by symmetry. 32. True, by symmetry. 33. True, by symmetry.

34. False. Rotating the square does not change its area or its centroid, so the Theorem of Pappusimplies that the volume is also unchanged.

35. By symmetry, y = 0. We use Formula (10) with a replaced by 0, b replaced by a, f(x) =bx

a, and

g(x) = −bxa

: The area is ab, so x =1ab

∫ a

0

x

(bx

a−(−bxa

))dx =

2a2

∫ a

0

x2 dx =2a2· a

3

3=

2a3

.

The centroid is(

2a3, 0)

.

36. Let M be a median of the triangle, joining one vertex to the midpoint P of the opposite side.Establish a coordinate system so that the origin is at P , the side containing P lies along they-axis, and the rest of the triangle is to the right of the y-axis. Then the coordinates of the verticesare (0,−a), (0, a), and (b, c), where a > 0 and b > 0.

The upper edge has equation y = a +c− ab

x =cx

b+a(b− x)

band the lower edge has equation

y = −a+c+ a

bx =

cx

b− a(b− x)

b, where x runs from 0 to b. The triangle’s area is

12· 2a · b = ab.

By Formulas (8) and (9),

x =1ab

∫ b

0

x

[(cx

b+a(b− x)

b

)−(cx

b− a(b− x)

b

)]dx =

1ab

∫ b

0

2ab

(bx− x2) dx

=2b2

[b

2x2 − 1

3x3

]b0

=b

3and

y =1ab

∫ b

0

12

[(cx

b+a(b− x)

b

)2

−(cx

b− a(b− x)

b

)2]dx =

12ab

∫ b

0

4acx(b− x)b2

dx

=1

2ab· 4acb2

[b

2x2 − 1

3x3

]b0

=c

3.

The centroid is(b

3,c

3

). Note that this lies on the line y =

c

bx, which is the median M . Since we

picked M arbitrarily, the centroid lies on all 3 medians, so it is their intersection.


328 Chapter 6

37. We will assume that a, b, and c are positive; the other cases are similar.

The region is described by 0 ≤ y ≤ c, −a− b− ac

y ≤ x ≤ a+b− ac

y. By symmetry, x = 0. To find

y, we use the analogue of Formula (10) with the roles of x and y reversed. The area is c(a+ b), so

y =1

c(a+ b)

∫ c

0

y

[(a+

b− ac

y

)−(−a− b− a

cy

)]dy =

1c(a+ b)

∫ c

0

(2ay +

2(b− a)c

y2

)dy

=1

c(a+ b)

[ay2 +

2(b− a)3c

y3

]c0

=c(a+ 2b)3(a+ b)

. The centroid is(

0,c(a+ 2b)3(a+ b)

).

38. A parallelogram is symmetric about the intersection of its diagonals: It is identical to its 180◦

rotation about that point. By symmetry, the intersection of the diagonals is the centroid.

39. x = 0 from the symmetry of the region, πa2/2 is the area of the semicircle, 2πy is the distancetraveled by the centroid to generate the sphere so 4πa3/3 = (πa2/2)(2πy), y = 4a/(3π)

40. (a) V =[

12πa2

] [2π(a+

4a3π

)]=

13π(3π + 4)a3

(b) the distance between the centroid and the line is√

22

(a+

4a3π

)so

V =[

12πa2

] [2π√

22

(a+

4a3π

)]=

16

√2π(3π + 4)a3

41. x = k so V = (πab)(2πk) = 2π2abk

42. y = 4 from the symmetry of the region,

A =∫ 2

−2

∫ 8−x2

x2dy dx = 64/3 so V = (64/3)[2π(4)] = 512π/3

43. The region generates a cone of volume13πab2 when it is revolved about the x-axis, the area of the

region is12ab so

13πab2 =

(12ab

)(2πy), y = b/3. A cone of volume

13πa2b is generated when the

region is revolved about the y-axis so13πa2b =

(12ab

)(2πx), x = a/3. The centroid is (a/3, b/3).

44. The centroid of a region is defined to be the center of gravity of a lamina of constant densityoccupying the region. So assume that both R1 and R2 have density δ. By equation (6), themoment of R1 about the y-axis is δA1x1, and the moment of R2 about the y-axis is δA2x2. Themoment of the union R is the sum of these, δ(A1x1 + A2x2). The mass of R is the sum of themasses of R1 and R2, δ(A1+A2). Again using equation (6), the x-coordinate of the centroid of R isδ(A1x1 +A2x2)δ(A1 +A2)

=A1x1 +A2x2

A1 +A2. Similarly, the y-coordinate of the centroid of R is

A1y1 +A2y2

A1 +A2.

In words, the centroid of R lies on the line segment joining the centroids of R1 and R2, and its

distance from the centroid of R1 isA2

A1 +A2times the distance between the centroids of R1 and

R2. For example, if A1 = 2A2 then the centroid of R is one third of the way from the centroid ofR1 to the centroid of R2.

If R is decomposed into n regions R1, · · ·, Rn of areas A1, · · ·, An and centroids (x1, y1), · · ·,

(xn, yn), then the centroid of R is(A1x1 + · · ·+AnxnA1 + · · ·+An

,A1y1 + · · ·+AnynA1 + · · ·+An

).



45. The Theorem of Pappus says that V = 2πAd, where A is the area of a region in the plane, d is thedistance from the region’s centroid to an axis of rotation, and V is the volume of the resulting solidof revolution. In any problem in which 2 of these quantities are given, the Theorem of Pappus canbe used to compute the third.

EXERCISE SET 6.8

1. (a) F = ρhA = 62.4(5)(100) = 31,200 lbP = ρh = 62.4(5) = 312 lb/ft2

(b) F = ρhA = 9810(10)(25) = 2,452,500 NP = ρh = 9810(10) = 98.1 kPa

2. (a) F = PA = 6 · 105(160) = 9.6× 107 N (b) F = PA = 100(60) = 6000 lb

3. F =∫ 2

0

62.4x(4) dx

= 249.6∫ 2

0

x dx = 499.2 lb

2

0 4

x

4. F =∫ 3

1

9810x(4) dx

= 39,240∫ 3

1

x dx

= 156,960 N

3

1

04

x

5. F =∫ 5

0

9810x(2√

25− x2) dx

= 19,620∫ 5

0

x(25− x2)1/2 dx

= 8.175× 105 N

50

5

x y = √25 – x2

y

2√25 – x2

6. By similar triangles

w(x)4

=2√

3− x2√

3, w(x) =

2√3

(2√

3− x),

F =∫ 2√

3

0

62.4x[

2√3

(2√

3− x)]dx

=124.8√

3

∫ 2√

3

0

(2√

3x− x2) dx = 499.2 lb

2√3

0 4

4 4

xw(x)


330 Chapter 6

7. By similar triangles,

w(x)6

=10− x

8, w(x) =

34

(10− x),

F =∫ 10

2

9810x[

34

(10− x)]dx

= 7357.5∫ 10

2

(10x− x2) dx = 1,098,720 N10

2

0

xw(x)

8

6

8. w(x) = 16 + 2u(x), but

u(x)4

=12− x

8so u(x) =

12

(12− x),

w(x) = 16 + (12− x) = 28− x,

F =∫ 12

4

62.4x(28− x) dx

= 62.4∫ 12

4

(28x− x2) dx = 77,209.6 lb.

12

4 4

16

4

0

xw(x)

u(x)

9. Yes: if ρ2 = 2ρ1 then F2 =∫ b

a

ρ2h(x)w(x) dx =∫ b

a

2ρ1h(x)w(x) dx = 2∫ b

a

ρ1h(x)w(x) dx = 2F1.

10. F =∫ 2

0

50x(2√

4− x2) dx

= 100∫ 2

0

x(4− x2)1/2 dx

= 800/3 lb

20

x

y

2√4 – x2

y = √4 – x2

11. Find the forces on the upper and lower halves and add them:

w1(x)√2a

=x√

2a/2, w1(x) = 2x

F1 =∫ √2a/2

0

ρx(2x) dx = 2ρ∫ √2a/2

0

x2 dx =√

2ρa3/6,

w2(x)√2a

=√

2a− x√2a/2

, w2(x) = 2(√

2a− x)

0

x

x

√2a

√2a√2a/2

w1(x)

w2(x)aa

aa

F2 =∫ √2a

√2a/2

ρx[2(√

2a− x)] dx = 2ρ∫ √2a

√2a/2

(√

2ax− x2) dx =√

2ρa3/3,

F = F1 + F2 =√

2ρa3/6 +√

2ρa3/3 = ρa3/√

2 lb

12. False. The units of pressure are the units of force divided by the units of area. In SI, the units offorce are newtons and the units of pressure are newtons per square meter, or pascals.

13. True. By equation (6), the fluid force equals ρhA. For a cylinder, hA is the volume, so ρhA is theweight of the water.

14. False. Consider a tank of height h, whose horizontal cross-sections are a by b rectangles. Byequation (6), the fluid force on the bottom of the tank is ρhab. By equation (8), the fluid force on



either a by h side of the tank is∫ h

0

ρxa dx =ρh2a

2. So if h > 2b, then the fluid force on the side

is larger than the fluid force on the bottom.

15. False. Let the height of the tank be h, the area of the base be A, and the volume of the tank beV . Then the fluid force on the base is ρhA and the weight of the water is ρV . So if hA > V , thenthe force exceeds the weight. This is true, for example, for a conical tank with its vertex at the

top, for which V =hA

3.

16. Suppose that a flat surface is immersed, at an angle θ with the vertical, in a fluid of weight densityρ, and that the submerged portion of the surface extends from x = a to x = b along an x-axiswhose positive direction is down.

Following the derivation of equation (8), we divide the interval [a, b] into n subintervals a = x0 <x1 < . . . < xn−1 < xn = b. As in that derivation, we have Fk = ρh(x∗k)Ak, for some point x∗kbetween xk−1 and xk.

Because the surface is tilted, the k’th strip is approximatelya rectangle with width w(x∗k) and length ∆xk sec θ; its area isAk ≈ w(x∗k)∆xk sec θ. So Fk ≈ ρh(x∗k)w(x∗k)∆xk sec θ. Follow-ing the argument in the text we arrive at the desired equation

F =∫ b

a

ρh(x)w(x) sec θ dx.

x=xk

x=xk -1

!xk"

! x ksec

"

17. Place the x-axis pointing down with its origin at the top of the pool, so that h(x) = x and w(x) =10. The angle between the bottom of the pool and the vertical is θ = tan−1(16/(8−4)) = tan−1 4,

so sec θ =√

17. Hence F =∫ 8

4

62.4h(x)w(x) sec θ dx = 624√

17∫ 8

4

x dx = 14976√

17 ≈ 61748 lb.

18. If we lower the water level by k ft, k < 4, then the force is computed as in Exercise 17, but with

h(x) = x − k, so F =∫ 8

4

62.4h(x)w(x) sec θ dx = 624√

17∫ 8

4

(x − k) dx = 624√

17(24 − 4k) lb.

For this to be half of 14976√

17, we need k = 3, so we should lower the water level by 3 ft. (Notethat this is plausible, since this lowers the average depth from 6 ft to 3 ft, cutting the volume andweight of the water in half.)

19. h(x) = x sin 60◦ =√

3x/2,

θ = 30◦, sec θ = 2/√

3,

F =∫ 100

0

9810(√

3x/2)(200)(2/√

3) dx

= 200 · 9810∫ 100

0

x dx

= 9810 · 1003 = 9.81× 109 N

xh(x)

200

100

0

60°

20. F =∫ h+2

h

ρ0x(2) dx

= 2ρ0

∫ h+2

h

x dx

= 4ρ0(h+ 1)

h + 2

h

0

2

2x

h


332 Chapter 6

21. (a) The force on the window is F =∫ h+2

hρ0x(2) dx = 4ρ0(h+1) so (assuming that ρ0 is constant)

dF/dt = 4ρ0(dh/dt) which is a positive constant if dh/dt is a positive constant.

(b) If dh/dt = 20 then dF/dt = 80ρ0 lb/min from part (a).

22. (a) Let h1 and h2 be the maximum and minimum depths of the disk Dr. The pressure P (r) onone side of the disk satisfies inequality (5):ρh1 ≤ P (r) ≤ ρh2. But

limr→0+

h1 = limr→0+

h2 = h, and hence

ρh = limr→0+

ρh1 ≤ limr→0+

P (r) ≤ limr→0+

ρh2 = ρh, so limr→0+

P (r) = ρh.

(b) The disks Dr in part (a) have no particular direction (the axes of the disks have arbitrarydirection). Thus P , the limiting value of P (r), is independent of direction.

23. h =P

ρ=

14.7 lb/in2

4.66× 10−5 lb/in3≈ 315, 000 in ≈ 5 mi. The answer is not reasonable. In fact the atmo-

sphere is thinner at higher altitudes, and it’s difficult to define where the “top” of the atmosphereis.

24. According to equation (6), if the density is constant then the fluid force on a horizontal surfaceof area A at depth h equals the weight of the water above it. It is plausible to assume that thisis also true if the density is not constant. To compute this weight, partition the interval [0, h]with 0 = x0 < x1 < · · · < xn−1 < xn = h. Let x∗k be an arbitrary point of [xk−1, xk]. Thevolume of water which is above the flat surface and at depth between xk−1 and xk is A ∆xk soits weight is approximately ρ(x∗k)A ∆xk. Adding these estimates, we find that the total weight is

approximatelyn∑k=1

ρ(x∗k)A ∆xk. Taking the limit as n → +∞ and the lengths of the subintervals

all approach zero gives the total weight, and hence the total force on the surface:∫ h

0

ρ(x)Adx.

Dividing by A gives the pressure: P =∫ h

0

ρ(x) dx.

EXERCISE SET 6.9

1. (a) sinh 3 ≈ 10.0179(b) cosh(−2) ≈ 3.7622(c) tanh(ln 4) = 15/17 ≈ 0.8824(d) sinh−1(−2) ≈ −1.4436(e) cosh−1 3 ≈ 1.7627

(f) tanh−1 34≈ 0.9730

2. (a) csch(−1) ≈ −0.8509(b) sech(ln 2) = 0.8(c) coth 1 ≈ 1.3130

(d) sech−1 12≈ 1.3170

(e) coth−1 3 ≈ 0.3466(f) csch−1(−

√3) ≈ −0.5493

3. (a) sinh(ln 3) =12

(eln 3 − e− ln 3) =12

(3− 1

3

)=

43

(b) cosh(− ln 2) =12

(e− ln 2 + eln 2) =12

(12

+ 2)

=54

(c) tanh(2 ln 5) =e2 ln 5 − e−2 ln 5

e2 ln 5 + e−2 ln 5=

25− 1/2525 + 1/25

=312313



(d) sinh(−3 ln 2) =12

(e−3 ln 2 − e3 ln 2) =12

(18− 8)

= −6316

4. (a)12

(eln x + e− ln x) =12

(x+

1x

)=x2 + 1

2x, x > 0

(b)12

(eln x − e− ln x) =12

(x− 1

x

)=x2 − 1

2x, x > 0

(c)e2 ln x − e−2 ln x

e2 ln x + e−2 ln x=x2 − 1/x2

x2 + 1/x2=x4 − 1x4 + 1

, x > 0

(d)12

(e− ln x + eln x) =12

(1x

+ x

)=

1 + x2

2x, x > 0

5. sinhx0 coshx0 tanhx0 coth x0 sech x0 csch x0

(a) 2√

5 2/√

5√

5/2 1/√

5 1/2

(b) 3/4 5/4 3/5 5/3 4/5 4/3

(c) 4/3 5/3 4/5 5/4 3/5 3/4

(a) cosh2 x0 = 1 + sinh2 x0 = 1 + (2)2 = 5, coshx0 =√

5

(b) sinh2 x0 = cosh2 x0 − 1 =2516− 1 =

916

, sinhx0 =34

(because x0 > 0)

(c) sech2x0 = 1− tanh2 x0 = 1−(

45

)2

= 1− 1625

=925

, sech x0 =35

,

coshx0 =1

sech x0=

53

, fromsinhx0

coshx0= tanhx0 we get sinhx0 =

(53

)(45

)=

43

6.d

dxcschx =

d

dx

1sinhx

= − coshxsinh2 x

= − cothx csch x for x 6= 0

d

dxsech x =

d

dx

1coshx

= − sinhxcosh2 x

= − tanhx sech x for all x

d

dxcothx =

d

dx

coshxsinhx

=sinh2 x− cosh2 x

sinh2 x= − csch2x for x 6= 0

7.d

dxcosh−1 x =

d

dxln(x+

√x2 − 1) =

1x+√x2 − 1

(1 +

2x2√x2 − 1

)=

1x+√x2 − 1

√x2 − 1 + x√x2 − 1

=1√

x2 − 1d

dxtanh−1 x =

d

dx

[12

ln(

1 + x

1− x

)]=

12· 1

1+x1−x

· (1− x) · 1− (1 + x)(−1)(1− x)2

=2

2(1 + x)(1− x)=

11− x2

8. y = sinh−1 x if and only if x = sinh y; 1 =dy

dx

dx

dy=dy

dxcosh y; so

d

dx[sinh−1 x] =

dy

dx=

1cosh y

=1√

1 + sinh2 y=

1√1 + x2

for all x.

Let x ≥ 1. Then y = cosh−1 x if and only if x = cosh y; 1 =dy

dx

dx

dy=dy

dxsinh y, so


334 Chapter 6

d

dx[cosh−1 x] =

dy

dx=

1sinh y

=1√

cosh2 y − 1=

1x2 − 1

for x ≥ 1.

Let −1 < x < 1. Then y = tanh−1 x if and only if x = tanh y; thus

1 =dy

dx

dx

dy=dy

dxsech2y =

dy

dx(1− tanh2 y) = 1− x2, so

d

dx[tanh−1 x] =

dy

dx=

11− x2

.

9. 4 cosh(4x− 8) 10. 4x3 sinh(x4) 11. − 1x

csch2(lnx)

12. 2sech22xtanh 2x

13.1x2

csch(1/x) coth(1/x) 14. −2e2x sech(e2x) tanh(e2x)

15.2 + 5 cosh(5x) sinh(5x)√

4x+ cosh2(5x)16. 6 sinh2(2x) cosh(2x)

17. x5/2 tanh(√x) sech2(

√x) + 3x2 tanh2(

√x)

18. −3 cosh(cos 3x) sin 3x 19.1√

1 + x2/9

(13

)= 1/

√9 + x2

20.1√

1 + 1/x2(−1/x2) = − 1

|x|√x2 + 1

21. 1/[(cosh−1 x)

√x2 − 1

]

22. 1/[√

(sinh−1 x)2 − 1√

1 + x2

]23. −(tanh−1 x)−2/(1− x2)

24. 2(coth−1 x)/(1− x2) 25.sinhx√

cosh2 x− 1=

sinhx| sinhx|

={

1, x > 0−1, x < 0

26. (sech2x)/√

1 + tanh2 x 27. − ex

2x√

1− x+ ex sech−1√x

28. 10(1 + x csch−1x)9(− x

|x|√

1 + x2+ csch−1x

)

29.17

sinh7 x+ C 30.12

sinh(2x− 3) + C 31.23

(tanhx)3/2 + C

32. −13

coth(3x) + C 33. ln(coshx) + C 34. −13

coth3 x+ C

35. −13

sech3x

]ln 3

ln 2

= 37/375 36. ln(coshx)]ln 3

0= ln 5− ln 3

37. u = 3x,13

∫1√

1 + u2du =

13

sinh−1 3x+ C

38. x =√

2u,∫ √

2√2u2 − 2

du =∫

1√u2 − 1

du = cosh−1(x/√

2) + C

39. u = ex,∫

1u√

1− u2du = − sech−1(ex) + C



40. u = cos θ, −∫

1√1 + u2

du = − sinh−1(cos θ) + C

41. u = 2x,∫

du

u√

1 + u2= −csch−1|u|+ C = −csch−1|2x|+ C

42. x = 5u/3,∫

5/3√25u2 − 25

du =13

∫1√

u2 − 1du =

13

cosh−1(3x/5) + C

43. tanh−1 x]1/20

= tanh−1(1/2)− tanh−1(0) =12

ln1 + 1/21− 1/2

=12

ln 3

44. sinh−1 t]√3

0= sinh−1

√3− sinh−1 0 = ln(

√3 + 2)

45. True. coshx− sinhx =ex + e−x

2− ex − e−x

2= e−x is positive for all x.

46. True. tanhx and sech x are bounded; the other 4 hyperbolic functions are not.

47. True. Only sinhx has this property.

48. False. For example, cos2 x + sin2 x = 1, but the corresponding identity for hyperbolic functionshas a minus sign: cosh2 x− sinh2 x = 1.

49. A =∫ ln 3

0

sinh 2x dx =12

cosh 2x]ln 3

0

=12

[cosh(2 ln 3)− 1],

but cosh(2 ln 3) = cosh(ln 9) =12

(eln 9 + e− ln 9) =12

(9 + 1/9) = 41/9 so A =12

[41/9− 1] = 16/9.

50. V = π

∫ ln 2

0

sech2x dx = π tanhx]ln 2

0

= π tanh(ln 2) = 3π/5

51. V = π

∫ 5

0

(cosh2 2x− sinh2 2x) dx = π

∫ 5

0

dx = 5π

52.∫ 1

0

cosh ax dx = 2,1a

sinh ax]10

= 2,1a

sinh a = 2, sinh a = 2a;

let f(a) = sinh a− 2a, then an+1 = an −sinh an − 2ancosh an − 2

, a1 = 2.2, . . . , a4 ≈ a5 ≈ 2.177318985.

53. y′ = sinhx, 1 + (y′)2 = 1 + sinh2 x = cosh2 x

L =∫ ln 2

0

coshx dx = sinhx]ln 2

0

= sinh(ln 2) =12

(eln 2 − e− ln 2) =12

(2− 1

2

)=

34

54. y′ = sinh(x/a), 1 + (y′)2 = 1 + sinh2(x/a) = cosh2(x/a)

L =∫ x1

0

cosh(x/a) dx = a sinh(x/a)]x1

0

= a sinh(x1/a)

55. (a) limx→+∞

sinhx = limx→+∞

12

(ex − e−x) = +∞− 0 = +∞


336 Chapter 6

(b) limx→−∞

sinhx = limx→−∞

12

(ex − e−x) = 0−∞ = −∞

(c) limx→+∞

tanhx = limx→+∞

ex − e−x

ex + e−x= 1

(d) limx→−∞

tanhx = limx→−∞

ex − e−x

ex + e−x= −1

(e) limx→+∞

sinh−1 x = limx→+∞

ln(x+√x2 + 1) = +∞

(f) limx→1−

tanh−1 x = limx→1−

12

[ln(1 + x)− ln(1− x)] = +∞

56. Since limx→+∞

sinhxex/2

= 1 and limx→+∞

coshxex/2

= 1, limx→+∞

tanhx =lim

x→+∞

sinhxex/2

limx→+∞

coshxex/2

= 1.

Since limx→−∞

sinhxe−x/2

= −1 and limx→−∞

coshxe−x/2

= 1, limx→−∞

tanhx =lim

x→−∞

sinhxe−x/2

limx→+∞

coshxe−x/2

= −1.

57. sinh(−x) =12

(e−x − ex) = −12

(ex − e−x) = − sinhx

cosh(−x) =12

(e−x + ex) =12

(ex + e−x) = coshx

58. (a) coshx+ sinhx =12

(ex + e−x) +12

(ex − e−x) = ex

(b) coshx− sinhx =12

(ex + e−x)− 12

(ex − e−x) = e−x

(c) sinhx cosh y + coshx sinh y =14

(ex − e−x)(ey + e−y) +14

(ex + e−x)(ey − e−y)

=14

[(ex+y − e−x+y + ex−y − e−x−y) + (ex+y + e−x+y − ex−y − e−x−y)]

=12

[e(x+y) − e−(x+y)] = sinh(x+ y)

(d) Let y = x in part (c).

(e) The proof is similar to part (c), or: treat x as variable and y as constant, and differentiatethe result in part (c) with respect to x.

(f) Let y = x in part (e).

(g) Use cosh2 x = 1 + sinh2 x together with part (f).

(h) Use sinh2 x = cosh2 x− 1 together with part (f).

59. (a) Divide cosh2 x− sinh2 x = 1 by cosh2 x.

(b) tanh(x+ y) =sinhx cosh y + coshx sinh ycoshx cosh y + sinhx sinh y

=

sinhxcoshx

+sinh ycosh y

1 +sinhx sinh ycoshx cosh y

=tanhx+ tanh y

1 + tanhx tanh y

(c) Let y = x in part (b).



60. (a) Let y = cosh−1 x; then x = cosh y =12

(ey + e−y), ey − 2x + e−y = 0, e2y − 2xey + 1 = 0,

ey =2x±

√4x2 − 42

= x ±√x2 − 1. To determine which sign to take, note that y ≥ 0

so e−y ≤ ey, x = (ey + e−y)/2 ≤ (ey + ey)/2 = ey, hence ey ≥ x thus ey = x +√x2 − 1,

y = cosh−1 x = ln(x+√x2 − 1).

(b) Let y = tanh−1 x; then x = tanh y =ey − e−y

ey + e−y=e2y − 1e2y + 1

, xe2y + x = e2y − 1,

1 + x = e2y(1− x), e2y = (1 + x)/(1− x), 2y = ln1 + x

1− x, y =

12

ln1 + x

1− x.

61. (a)d

dx(cosh−1 x) =

1 + x/√x2 − 1

x+√x2 − 1

= 1/√x2 − 1

(b)d

dx(tanh−1 x) =

d

dx

[12

(ln(1 + x)− ln(1− x))]

=12

(1

1 + x+

11− x

)= 1/(1− x2)

62. Let y = sech−1x then x = sech y = 1/ cosh y, cosh y = 1/x, y = cosh−1(1/x); the proofs for theremaining two are similar.

63. If |u| < 1 then, by Theorem 6.9.6,∫

du

1− u2= tanh−1 u+ C.

For |u| > 1,∫

du

1− u2= coth−1 u+ C = tanh−1(1/u) + C.

64. (a)d

dx(sech−1|x|) =

d

dx(sech−1

√x2) = − 1√

x2√

1− x2

x√x2

= − 1x√

1− x2

(b) Similar to solution of part (a)

65. (a) limx→+∞

(cosh−1 x− lnx) = limx→+∞

[ln(x+√x2 − 1)− lnx]

= limx→+∞

lnx+√x2 − 1x

= limx→+∞

ln(1 +√

1− 1/x2) = ln 2

(b) limx→+∞

coshxex

= limx→+∞

ex + e−x

2ex= limx→+∞

12

(1 + e−2x) = 1/2

66. For |x| < 1, y = tanh−1 x is defined and dy/dx = 1/(1− x2) > 0; y′′ = 2x/(1− x2)2 changes signat x = 0, so there is a point of inflection there.

67. Let x = −u/a,∫

1√u2 − a2

du = −∫

a

a√x2 − 1

dx = − cosh−1 x+ C = − cosh−1(−u/a) + C.

− cosh−1(−u/a) = − ln(−u/a+√u2/a2 − 1) = ln

[a

−u+√u2 − a2

u+√u2 − a2

u+√u2 − a2

]

= ln∣∣∣u+

√u2 − a2

∣∣∣− ln a = ln |u+√u2 − a2|+ C1

so∫

1√u2 − a2

du = ln∣∣∣u+

√u2 − a2

∣∣∣+ C2.

68. Using sinhx+ coshx = ex (Exercise 58a), (sinhx+ coshx)n = (ex)n = enx = sinhnx+ coshnx.

69.∫ a

−aetx dx =

1tetx]a−a

=1t(eat − e−at) =

2 sinh att

for t 6= 0.


338 Chapter 6

70. (a) y′ = sinh(x/a), 1 + (y′)2 = 1 + sinh2(x/a) = cosh2(x/a)

L = 2∫ b

0

cosh(x/a) dx = 2a sinh(x/a)]b0

= 2a sinh(b/a)

(b) The highest point is at x = b, the lowest at x = 0,so S = a cosh(b/a)− a cosh(0) = a cosh(b/a)− a.

71. From part (b) of Exercise 70, S = a cosh(b/a) − a so 30 = a cosh(200/a) − a. Let u = 200/a,then a = 200/u so 30 = (200/u)[coshu − 1], coshu − 1 = 0.15u. If f(u) = coshu − 0.15u − 1,

then un+1 = un −coshun − 0.15un − 1

sinhun − 0.15;u1 = 0.3, . . . , u4 ≈ u5 ≈ 0.297792782 ≈ 200/a so

a ≈ 671.6079505. From part (a), L = 2a sinh(b/a) ≈ 2(671.6079505) sinh(0.297792782) ≈ 405.9 ft.

72. From part (a) of Exercise 70, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Letu = 50/a, then a = 50/u so (50/u) sinhu = 60, sinhu = 1.2u. If f(u) = sinhu − 1.2u, then

un+1 = un −sinhun − 1.2uncoshun − 1.2

;u1 = 1, . . . , u5 ≈ u6 ≈ 1.064868548 ≈ 50/a so a ≈ 46.95415231.

From part (b), S = a cosh(b/a)− a ≈ 46.95415231[cosh(1.064868548)− 1] ≈ 29.2 ft.

73. Set a = 68.7672, b = 0.0100333, c = 693.8597, d = 299.2239.

(a) 650

0–300 300

(b) L = 2∫ d

0

√1 + a2b2 sinh2 bx dx

≈ 1480.2798 ft

(c) x ≈ ±283.6249 ft (d) 82◦

74. (a)

1

1

2

t

r (b) r = 1 when t ≈ 0.673080 s.

(c) dr/dt ≈ 4.48 m/s.

75. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D,then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y)has length a; thus a2 = x2 + (D − y)2, D = y +

√a2 − x2 = a sech−1(x/a).

(b) Find D when a = 15, x = 10: D = 15 sech−1(10/15) = 15 ln

(1 +

√5/9

2/3

)≈ 14.44 m.

(c) dy/dx = − a2

x√a2 − x2

+x√

a2 − x2=

1√a2 − x2

[−a

2

x+ x

]= − 1

x

√a2 − x2,


Review Exercises, Chapter 6 339

1 + [y′]2 = 1 +a2 − x2

x2=a2

x2; with a = 15, L =

∫ 15

5

√225x2

dx =∫ 15

5

15xdx = 15 lnx

]155

=

15 ln 3 ≈ 16.48 m.

76. First we would need to show that the line segment from the origin to P meets the right branch ofthe hyperbola only at P , so that the shaded region in Figure 6.9.3b is well-defined. (This is easy.)

Next we’d need to show that the area of the shaded region approaches +∞ as the point P movesupward and to the right along the curve, so that cosh t and sinh t will be defined for all t > 0(and hence, by symmetry, for all t.) (This is not quite as easy.)

77. Since (cosh t, sinh t) lies on the hyperbola x2 − y2 = 1, we have cosh2 t− sinh2 t = 1.

Since it lies on the right half of the hyperbola, cosh t > 0.

From the symmetry of the hyperbola, cosh(−t) = cosh t and sinh(−t) = − sinh t.

Next, we can obtain the derivatives of the hyperbolic functions. Define tanh t =sinh tcosh t

and

sech t =1

cosh t. Suppose that h is a small positive number, (x0, y0) = (cosh t, sinh t), and

(x1, y1) = (cosh(t + h), sinh(t + h)). Then h/2 = (t + h)/2 − t/2 is approximately the areaof the triangle with vertices (0, 0), (x0, y0), and (x1, y1), which equals 1

2 (x0y1 − x1y0). Hencecosh t sinh(t+ h)− cosh(t+ h) sinh t ≈ h. Dividing by cosh t cosh(t+ h) implies

tanh(t+ h)− tanh t ≈ h

cosh t cosh(t+ h)≈ h sech2t.

Taking the limit as h→ 0 givesd

dttanh t = sech2 t.

Dividing cosh2 t− sinh2 t = 1 by cosh2 t gives 1− tanh2 t =1

cosh2 t, so cosh t = (1− tanh2 t)−1/2.

Henced

dtcosh t = −1

2(1− tanh2 t)−3/2(−2) tanh t · d

dttanh t = cosh3 t tanh t sech2 t = sinh t and

d

dtsinh t =

d

dt(cosh t tanh t) = cosh t · d

dttanh t+ tanh t · d

dtcosh t = cosh t sech2 t+ tanh t sinh t

=1 + sinh2 t

cosh t= cosh t.

REVIEW EXERCISES, CHAPTER 6

6. (a) A =∫ 2

0

(2 + x− x2) dx (b) A =∫ 2

0

√y dy +

∫ 4

2

[√y − (y − 2)] dy

(c) V = π

∫ 2

0

[(2 + x)2 − x4] dx

(d) V = 2π∫ 2

0

y√y dy + 2π

∫ 4

2

y[√y − (y − 2)] dy

(e) V = 2π∫ 2

0

x(2 + x− x2) dx (f) V = π

∫ 2

0

y dy +∫ 4

2

π(y − (y − 2)2) dy

(g) V = π

∫ 2

0

[(2 + x+ 3)2 − (x2 + 3)2] dx (h) V = 2π∫ 2

0

[2 + x− x2](5− x) dx

7. (a) A =∫ b

a

(f(x)− g(x)) dx+∫ c

b

(g(x)− f(x)) dx+∫ d

c

(f(x)− g(x)) dx

(b) A =∫ 0

−1

(x3 − x) dx+∫ 1

0

(x− x3) dx+∫ 2

1

(x3 − x) dx =14

+14

+94

=114


340 Chapter 6

8. distance =∫|v| dt, so

(a) distance =∫ 60

0

(3t− t2/20) dt = 1800 ft.

(b) If T ≤ 60 then distance =∫ T

0

(3t− t2/20) dt =32T 2 − 1

60T 3 ft.

9. Find where the curves cross: set x3 = x2 + 4; by observation x = 2 is a solution. Then

V = π

∫ 2

0

[(x2 + 4)2 − (x3)2] dx =4352105

π.

10. V = 2∫ L/2

0

π16R2

L4(x2 − L2/4)2 =

4π15LR2 11. V =

∫ 4

1

(√x− 1√

x

)2

dx = 2 ln 2 +32

12. (a) π

∫ 1

0

(sin−1 x)2 dx. (b) 2π∫ π/2

0

y(1− sin y) dy.

13. By implicit differentiationdy

dx= −

(yx

)1/3, so 1 +

(dy

dx

)2= 1 +

(yx

)2/3=x2/3 + y2/3

x2/3=

4x2/3

,

L =∫ −1

−8

2(−x)1/3

dx = 9.

14. (a) L =∫ ln 10

0

√1 + (ex)2 dx (b) L =

∫ 10

1

√1 +

1y2dy

15. A =∫ 16

9

2π√

25− x

√1 +

(−1

2√

25− x

)2

dx = π

∫ 16

9

√101− 4x dx =

π

6

(653/2 − 373/2

)

16. (a) S =∫ 8/27

0

2π · 3x1/3√

1 + x−4/3 dx (b) S =∫ 2

0

2πy3

27

√1 + y4/81 dy

(c) S =∫ 2

0

2π(y + 2)√

1 + y4/81 dy

17. (a) F = kx,12

= k14

, k = 2, W =∫ 1/4

0

kx dx = 1/16 J

(b) 25 =∫ L

0

kx dx = kL2/2, L = 5 m

18. F = 30x+ 2000, W =∫ 150

0

(30x+ 2000) dx = 15 · 1502 + 2000 · 150 = 637,500 lb·ft

19. The region is described by −4 ≤ y ≤ 4,y2

4≤ x ≤ 2 +

y2

8. By symmetry, y = 0. To find x, we use

the analogue of Formula (11) in Section 6.7. The area is

A =∫ 4

−4

(2 +

y2

8− y2

4

)dy =

∫ 4

−4

(2− y2

8

)dy =

[2y − y3

24

]4−4

=323

. So

x =332

∫ 4

−4

12

[(2 +

y2

8

)2

−(y2

4

)2]dy =

364

∫ 4

−4

(4 +

y2

2− 3y4

64

)dy

=364

[4y +

y3

6− 3y5

320

]4−4

=85

. The centroid is(

85, 0)

.


Making Connections, Chapter 6 341

20. The region is described by −a ≤ x ≤ a, 0 ≤ y ≤ b

a

√a2 − x2. By symmetry, x = 0. To find y, we

use Formula (9) in Section 6.7. The area is A =∫ a

−a

(b

a

√a2 − x2

)dx. This is

b

atimes the area

of a half-disc of radius a, so A =πab

2. Hence

y =2πab

∫ a

−a

12

(b

a

√a2 − x2

)2

dx =b

πa3

∫ a

−a(a2 − x2) dx =

b

πa3

[a2x− x3

3

]a−a

=4b3π

.

The centroid is(

0,4b3π

).

21. (a) F =∫ 1

0

ρx3 dx N

(b) By similar triangles,w(x)

4=x

2, w(x) = 2x, so

F =∫ 4

1

ρ(1 + x)2x dx lb/ft2.0

42

h(x) = 1 + x

xw(x)

(c) A formula for the parabola is y =8

125x2 − 10,

so F =∫ 0

−10

9810|y|2√

1258

(y + 10) dy N.

22. y′ = a cosh ax, y′′ = a2 sinh ax = a2y

23. (a) cosh 3x= cosh(2x+ x) = cosh 2x coshx+ sinh 2x sinhx

= (2 cosh2 x− 1) coshx+ (2 sinhx coshx) sinhx

= 2 cosh3 x− coshx+ 2 sinh2 x coshx

= 2 cosh3 x− coshx+ 2(cosh2 x− 1) coshx = 4 cosh3 x− 3 coshx

(b) from Theorem 6.9.2 with x replaced byx

2: coshx = 2 cosh2 x

2− 1,

2 cosh2 x

2= coshx+ 1, cosh2 x

2=

12

(coshx+ 1),

coshx

2=

√12

(coshx+ 1) (because coshx

2> 0)

(c) from Theorem 6.9.2 with x replaced byx

2: coshx = 2 sinh2 x

2+ 1,

2 sinh2 x

2= coshx− 1, sinh2 x

2=

12

(coshx− 1), sinhx

2= ±

√12

(coshx− 1)

MAKING CONNECTIONS, CHAPTER 6

1. (a) By equation (2) of Section 6.3, the volume is V =∫ 1

0

2πxf(x2) dx. Making the substitution

u = x2, du = 2x dx gives V =∫ 1

0

2πf(u) · 12du = π

∫ 1

0

f(u) du = πA1.

(b) By the Theorem of Pappus, the volume in (a) equals 2πA2x, where x = a is the x-coordinate

of the centroid of R. Hence a =πA1

2πA2=

A1

2A2.


342 Chapter 6

2. (a) At depth x feet below the surface, the radius is 10 − x

3ft, so the area is π

(10− x

3

)2

=π

9(30− x)2 ft2. Hence the weight of a thin layer at depth x ft with height ∆x ft is approxi-

mately 62.4π

9(30− x)2∆x lb. The work needed to lift this layer to the top is approximately

62.4π

9x(30− x)2∆x ft·lb. Hence the total work needed to empty the tank is∫ 15

0

62.4π

9x(30−x)2 dx = 62.4

π

9

∫ 15

0

(900x−60x2 +x3) dx = 62.4π

9

[450x2 − 20x3 +

14x4

]150

= 321750π ≈ 1010807 ft·lb.(b) When the piston has risen x feet from the bottom of the tank, its radius is 5 +

x

3ft, so its

area is π(

5 +x

3

)2

=π

9(15 + x)2 ft2. The depth of the water above the piston is 15 − x ft,

so the fluid force pushing down on the piston is 62.4(15 − x)π

9(15 + x)2 lb; this also equals

the force needed to raise the piston. So the total work done raising the piston is∫ 15

0

62.4(15− x)π

9(15 + x)2 dx = 62.4

π

9

∫ 15

0

(3375 + 225x− 15x2 − x3) dx

= 62.4π

9

[3375x+

2252x2 − 5x3 − 1

4x4

]150

= 321750π ft·lb.

3. The area of the annulus with inner radius r and outer radius r+ ∆r is π(r+ ∆r)2−πr2 ≈ 2πr∆r,

so its mass is approximately 2πrf(r)∆r. Hence the total mass of the lamina is∫ a

0

2πrf(r) dr.

4. Let the x-axis point downward, with x = 0 at the surface of the fluid. Let the y-axis be perpen-dicular to the x-axis and in the plane of the submerged surface. Suppose the surface has area Aand is described by a ≤ x ≤ b, g(x) ≤ y ≤ f(x). In Formula (8) of Section 6.8 we have h(x) = xand w(x) = f(x)− g(x), so the fluid force on the submerged surface is

F =∫ b

a

ρx(f(x)− g(x)) dx = ρA · 1A

∫ b

a

x(f(x)− g(x)) dx = ρAx = A · ρx,

by Formula (10) of Section 6.7. Since ρx is the pressure at the centroid, the fluid force equals thearea times the pressure at the centroid.

If the same surface were horizontal at the depth of the centroid, then Formula (6) of Section 6.8implies that the fluid force would be ρxA, the same as above.

5. (a) Consider any solid obtained by sliding a horizontal region, of any shape, some distancevertically. Thus the top and bottom faces, and every horizontal cross-section in between, areall congruent. This includes all of the cases described in part (a) of the problem.

Suppose such a solid, whose base has area A, is floating in a fluid so that its base is a distanceh below the surface. The pressure at the base is ρh, so the fluid exerts an upward force onthe base of magnitude ρhA. The fluid also exerts forces on the sides of the solid, but thoseare horizontal, so they don’t contribute to the buoyancy. Hence the buoyant force equalsρhA. Since the part of the solid which is below the surface has volume hA, the buoyant forceequals the weight of fluid which would fill that volume; i.e. the weight of the fluid displacedby the solid.

(b) Now consider a solid which is the union of finitely many solids of the type described above.The buoyant force on such a solid is the sum of the buoyant forces on its constituents, whichequals the sum of the weights of the fluid displaced by them, which equals the weight of thefluid displaced by the whole solid. So the Archimedes Principle applies to the union.

Any solid can be approximated by such unions, so it is plausible that the Archimedes Principleapplies to all solids.

november 10, 2008 16:14 ism et chapter 6 sheet number 1...

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