Chapter 07

January 27, 2005 11:45 L24-CH07 Sheet number 1 Page number 290 black

CHAPTER 7
Applications of the Deﬁnite Integral in
Geometry, Science, and Engineering

EXERCISE SET 7.1
2 2
1. A = (x + 1 − x)dx = (x /3 + x − x /2)
2 3 2
= 9/2
−1 −1

4
4 √
2. A = ( x + x/4)dx = (2x3/2 /3 + x2 /8) = 22/3
0 0

2 2
3. A = (y − 1/y 2 )dy = (y 2 /2 + 1/y) =1
1 1

2 2
4. A = (2 − y 2 + y)dy = (2y − y 3 /3 + y 2 /2) = 10/3
0 0

4 16
√
5. (a) A = (4x − x2 )dx = 32/3 (b) A = ( y − y/4)dy = 32/3
0 0

y

(4, 16)
y = 4x

y = x2
5
x
1

6. Eliminate x to get y 2 = 4(y + 4)/2, y 2 − 2y − 8 = 0, y
(y − 4)(y + 2) = 0; y = −2, 4 with corresponding (4, 4)
y2 = 4x
values of x = 1, 4.
y = 2x – 4
1 √ √ 4 √
(a) A = [2 x − (−2 x)]dx + [2 x − (2x − 4)]dx x
0 1
1 √ 4 √ (1, –2)
= 4 xdx + (2 x − 2x + 4)dx = 8/3 + 19/3 = 9
0 1

4
(b) A = [(y/2 + 2) − y 2 /4]dy = 9
−2

290


Exercise Set 7.1 291

1 √ 2
7. A = ( x − x2 )dx = 49/192 8. A = [0 − (x3 − 4x)]dx
1/4 0
2
y
= (4x − x3 )dx = 4
0
y
y = √x (1, 1)
x
y = x2 2
x
1
4
y = 2x3 – 4x

π/2
9. A = (0 − cos 2x)dx 10. Equate sec2 x and 2 to get sec2 x = 2,
π/4
π/2 y
=− cos 2x dx = 1/2 2
π/4 (#, 2) (3, 2)
y
1 y = sec2 x
1 y = cos 2 x x
x
3 6
√
sec x = ± 2, x = ±π/4
–1 π/4
A= (2 − sec2 x)dx = π − 2
−π/4

3π/4 √ 2
11. A = sin y dy = 2 12. A = [(x + 2) − x2 ]dx = 9/2
π/4 −1
y y

x = sin y (2, 4)
9

y = x2
(–1, 1)
3 x
x x=y–2

ln 2 e e
dy
13. A = e2x − ex dx 14. A = = ln y =1
0 1 y 1
ln 2
1 2x y
= e − ex = 1/2
2 e
0
y

y = e2x 1
4
x
1/e 1
2
y = ex
x
ln 2


292 Chapter 7

1
√
2 1 3
15. A= − |x| dx 16. √ = 2, x = ± , so
−1 1 + x2 1−x2 2
√
1 3/2
2 1
=2 −x dx A= √ 2− √ dx
0 1 + x2 − 3/2 1 − x2
1 √
−1
3/2
√
= 4 tan x−x 2
=π−1 = 2 − sin−1 x = 2 3 − 2π
0 √ 3
− 3/2
y
y
2 2 y=2

1.5
1
1 y= 1
1 – x2
0.5
x
x
–1 1
3 3
– 2 2

3 − x, x ≤ 1
17. y = 2 + |x − 1| = , y
1 + x, x≥1 (–5, 8)
y = –1x + 7
5
1
1
A= − x + 7 − (3 − x) dx (5, 6)
−5 5 y = 3–x
y = 1+x
5
1
+ − x + 7 − (1 + x) dx x
1 5
1 5
4 6
= x + 4 dx + 6 − x dx
−5 5 1 5
= 72/5 + 48/5 = 24

2/5 1
18. A = (4x − x)dx 19. A = (x3 − 4x2 + 3x)dx
0 0
1 3
+ (−x + 2 − x)dx + [−(x3 − 4x2 + 3x)]dx
2/5 1
2/5 1 = 5/12 + 32/12 = 37/12
= 3x dx + (2 − 2x)dx = 3/5
0 2/5 4
y
( 2 , 8)
5 5 –1 4
y = –x + 2

y = 4x (1, 1)

x –8
y=x



20. Equate y = x3 − 2x2 and y = 2x2 − 3x 9
to get x3 − 4x2 + 3x = 0,
x(x − 1)(x − 3) = 0; x = 0, 1, 3
with corresponding values of y = 0, −1.9.
1
A= [(x3 − 2x2 ) − (2x2 − 3x)]dx –1 3
0
3 –2
+ [(2x3 − 3x) − (x3 − 2x2 )]dx
1
1 3
= (x3 − 4x2 + 3x)dx + (−x3 + 4x2 − 3x)dx
0 1
5 8 37
= + =
12 3 12

21. From the symmetry of the region 22. The region is symmetric about the origin so
5π/4 √ 2
A=2 (sin x − cos x)dx = 4 2 A=2 |x3 − 4x|dx = 8
π/4 0

1 3.1

0 o –3 3

–1 –3.1

0 1 1
23. A = (y 3 − y)dy + −(y 3 − y)dy 24. A = y 3 − 4y 2 + 3y − (y 2 − y) dy
−1 0 0
4
= 1/2
+ y 2 − y − (y 3 − 4y 2 + 3y) dy
1 1

= 7/12 + 45/4 = 71/6
4.1
–1 1

–1

–2.2 12.1
0


294 Chapter 7

√ y
25. The curves meet when x = ln 2, so
√ √ 2.5
ln 2 ln 2
2 1 2 1
A= (2x − xex ) dx = x2 − ex = ln 2 − 2
0 2 0 2 1.5

1

0.5
x
0.5 1

√ √
26. The curves meet for x = e−2 2/3
, e2 2/3
thus y
√ 20
2 2/3
e
3 1
A= √
− dx 15
e−2 2/3 x x 1 − (ln x)2
e2
√
2/3 √ 10
√ 2 2
= 3 ln x − sin−1 (ln x) √
= 4 2 − 2 sin−1 5
e−2 2/3 3
x
1 2 3

k
27. The area is given by (1/ 1 − x2 − x)dx = sin−1 k − k 2 /2 = 1; solve for k to get
0
k = 0.997301.

28. The curves intersect at x = a = 0 and x = b = 0.838422 so the area is
b
(sin 2x − sin−1 x)dx ≈ 0.174192.
a

29. Solve 3−2x = x6 +2x5 −3x4 +x2 to ﬁnd the real roots x = −3, 1; from a plot it is seen that the line
1
is above the polynomial when −3 < x < 1, so A = (3−2x−(x6 +2x5 −3x4 +x2 )) dx = 9152/105
−3

1 √
30. Solve x5 − 2x3 − 3x = x3 to ﬁnd the roots x = 0, ± 6 + 2 21. Thus, by symmetry,
√ √ 2
(6+2 21)/2
27 7 √
A=2 (x3 − (x5 − 2x3 − 3x)) dx = + 21
0 4 4

k
√ 9
√ y
31. 2 ydy = 2 ydy
0 k
y=9
k 9
1/2 1/2
y dy = y dy
0 k y=k

2 3/2 2
k = (27 − k 3/2 )
3 3
x
k 3/2 = 27/2
√
k = (27/2)2/3 = 9/ 3 4



k 2
32. x2 dx = x2 dx y
0 k x = √y
1 3 1
k = (8 − k 3 )
3 3
k3 = 4
√ x
k= 34
2
x=k

2
33. (a) A = (2x − x2 )dx = 4/3
0
(b) y = mx intersects y = 2x − x2 where mx = 2x − x2 , x2 + (m − 2)x = 0, x(x + m − 2) = 0 so
x = 0 or x = 2 − m. The area below the curve and above the line is
2−m 2−m 2−m
1 1 1
(2x − x2 − mx)dx = (2 − m)x2 − x3
[(2 − m)x − x2 ]dx = = (2 − m)3
0 0 2 3 0 6
√
so (2 − m)3 /6 = (1/2)(4/3) = 2/3, (2 − m)3 = 4, m = 2 − 4.
3

3
34. The line through (0, 0) and (5π/6, 1/2) is y = x; y
5π y = sin x

5π/6
√
1
( 56c , 1 )
2
3 3 5
A= sin x − x dx = − π+1 x
0 5π 2 24
c

35. The curves intersect at x = 0 and, by Newton’s Method, at x ≈ 2.595739080 = b, so
b b
A≈ (sin x − 0.2x)dx = − cos x + 0.1x2 ≈ 1.180898334
0 0

36. By Newton’s Method, the points of intersection are at x ≈ ±0.824132312, so with
b b
b = 0.824132312 we have A ≈ 2 (cos x − x2 )dx = 2(sin x − x3 /3) ≈ 1.094753609
0 0

37. By Newton’s Method the points of intersection are x = x1 ≈ 0.4814008713 and
x2
ln x
x = x2 ≈ 2.363938870, and A ≈ − (x − 2) dx ≈ 1.189708441.
x1 x

38. By Newton’s Method the points of intersection are x = ±x1 where x1 ≈ 0.6492556537, thus
x1
2
A≈2 − 3 + 2 cos x dx ≈ 0.826247888
0 1 + x2

39. The x-coordinates of the points of intersection are a ≈ −0.423028 and b ≈ 1.725171; the area is
b
(2 sin x − x2 + 1)dx ≈ 2.542696.
a

40. Let (a, k), where π/2 < a < π, be the coordinates of the point of intersection of y = k with
y = sin x. Thus k = sin a and if the shaded areas are equal,
a a
(k − sin x)dx = (sin a − sin x) dx = a sin a + cos a − 1 = 0
0 0
Solve for a to get a ≈ 2.331122, so k = sin a ≈ 0.724611.


296 Chapter 7

60
41. (v2 (t) − v1 (t)) dt = s2 (60) − s2 (0) − (s1 (60) − s1 (0)), but they are even at time t = 60, so
0
s2 (60)
= s1 (60). Consequently the integral gives the diﬀerence s1 (0)−s2 (0) of their starting points
in meters.
T
42. Since a1 (0) = a2 (0) = 0, A = (a2 (t)−a1 (t)) dt = v2 (T )−v1 (T ) is the diﬀerence in the velocities
0
of the two cars at time T .

43. (a) It gives the area of the region that is between f and g when f (x) > g(x) minus the area of
the 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.

1 1
n x2 n 1
44. (b) lim 1/n
(x − x) dx = lim x(n+1)/n − = lim − = 1/2
n→+∞ 0 n→+∞ n+1 2 0
n→+∞ n+1 2

45. Solve x1/2 + y 1/2 = a1/2 for y to get y
a

y = (a1/2 − x1/2 )2 = a − 2a1/2 x1/2 + x
a
A= (a − 2a1/2 x1/2 + x)dx = a2 /6 x
0 a
√
46. Solve for y to get y = (b/a) a2 − x2 for the upper half of the ellipse; make use of symmetry to
a
b 4b a 4b 1 2
get A = 4 a2 − x2 dx = a2 − x2 dx = · πa = πab.
0 a a 0 a 4

47. Let A be the area between the curve and the x-axis and AR the area of the rectangle, then
b b
k kbm+1
A= kxm dx = xm+1 = , AR = b(kbm ) = kbm+1 , so A/AR = 1/(m + 1).
0 m+1 0 m+1

EXERCISE SET 7.2
3 1
1. V = π (3 − x)dx = 8π 2. V = π [(2 − x2 )2 − x2 ]dx
−1 0
1
=π (4 − 5x2 + x4 )dx
0

= 38π/15

2 2
1
3. V = π (3 − y)2 dy = 13π/6 4. V = π (4 − 1/y 2 )dy = 9π/2
0 4 1/2

2
5. V = x4 dx = 32/5 y
0
y = x2

x
2



π/3 √
6. V = sec2 x dx = 3−1 y
π/4 y = sec x
2
1
x
3 4
-1
-2

π/2 √ 1
7. V = π cos x dx = (1 − 2/2)π 8. V = π [(x2 )2 − (x3 )2 ]dx
π/4 0
y 1
1 y = √cos x =π (x4 − x6 )dx = 2π/35
0
x
y
3 6

–1 1 (1, 1)
y = x2
y = x3
x
1

4 3
9. V = π [(25 − x2 ) − 9]dx 10. V = π (9 − x2 )2 dx
−4 −3
4 3
= 2π (16 − x2 )dx = 256π/3 =π (81 − 18x2 + x4 )dx = 1296π/5
0 −3
y y
5 y = √25 – x2 9
y=3 y = 9 – x2

x x
–3 3

4 π/4
11. V = π [(4x)2 − (x2 )2 ]dx 12. V = π (cos2 x − sin2 x)dx
0 0
4 π/4
=π (16x2 − x4 )dx = 2048π/15 =π cos 2x dx = π/2
0 0
y y
16 (4, 16) y = cos x
1
y = 4x
y = x2 y = sin x x
x 3
4
–1


298 Chapter 7

ln 3 ln 3 1
π 2x π
13. V = π e2x dx = e = 4π 14. V = π e−4x dx = (1 − e−4 )
0 2 0 0 4

y
1

x
1

–1

2 2
1 π
15. V = π 2
dx = tan−1 (x/2) = π 2 /4
−2 4+x 2 −2

1 1
e6x π π
16. V = π 6x
dx = ln(1 + e6x ) = (ln(1 + e6 ) − ln 2)
0 1+e 6 0 6

1 2 1
3 2 1 8
17. V = y 1/3 dy = 18. V = (1 − y 2 )2 dy = 1 − + =
0 5 0 3 5 15

y y

0.75 0.75
x x

0.75 0.75
0.25 0.25
0.25 0.25

3 3
19. V = π (1 + y)dy = 8π 20. V = π [22 − (y + 1)]dy
−1 0
y 3
=π (3 − y)dy = 9π/2
3 0

x = √y + 1
y y = x2 – 1
x = √1 + y
x 3 (2, 3)
2

x



3π/4 1
21. V = π csc2 y dy = 2π 22. V = π (y − y 4 )dy = 3π/10
π/4 0

y y
9 x = y2
1
(1, 1)
6 x = csc y x = √y
x
–1 1

3
x –1
–2 –1 1 2

2 1
23. V = π [(y + 2)2 − y 4 ]dy = 72π/5 24. V = π (2 + y 2 )2 − (1 − y 2 )2 dy
−1 −1
y 1
=π (3 + 6y 2 )dy = 10π
−1
x = y2 (4, 2) y x = 2 + y2
x= y+2 x = 1 – y2
x 1

(1, –1) x
1 2

–1

1 2
π 2 π
25. V = πe2y dy = e −1 26. V = dy = π tan−1 2
0 2 0 1 + y2

a 2
b2 2 1
27. V = π (a − x2 )dx = 4πab2 /3 28. V = π dx = π(1/b − 1/2);
−a a2 b x2
y π(1/b − 1/2) = 3, b = 2π/(π + 6)
b y = √a2 – x 2
b
a

x
–a a

0
29. V = π (x + 1)dx y
−1 (1, √2)
1
1 y = √x + 1
y = √2x
+π [(x + 1) − 2x]dx x
0 –1 1
= π/2 + π/2 = π


300 Chapter 7

4 6
30. V = π x dx + π (6 − x)2 dx y
0 4

= 8π + 8π/3 = 32π/3 y = √x y=6–x

x
4 6

31. Partition the interval [a, b] with a = x0 < x1 < x2 < . . . < xn−1 < xn = b. Let x∗ be an arbitrary
k
point 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
b
∆Vk = π(f (x∗ ) − k)2 ∆xk , and the total volume is given by V = π
k (f (x) − k)2 dx.
a

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

volume of the solid obtained by rotating R is
d
V =π [(v(y) − k)2 − (w(y) − k)2 ] dy
c

33. (a) Intuitively, it seems that a line segment revolved about a line which is perpendicular to 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 πy 2 .
If on the other hand the segment is revolved around the line y = 2 then the area of the
resulting (inﬁnitely thin) washer is π[22 − (2 − y)2 ]. So the question can be reduced to√ asking
whether y 2 ≥ [22 − (2 − y)2 ], y 2 ≥ 4y − y 2 , 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
3
V1 = π (3 − x) dx = 8π, and the volume generated by revolving the area around the line
−1
3 √ 40
y = 2 is V2 = π [22 − (2 − 3 − x)2 ] dx = π
−1 3

34. (a) In general, points in the region R are farther from the y-axis than they are from the line
x = 2.5, so by the reasoning in Exercise 33(a) the former should generate a larger volume
than the latter, i.e. the volume mentioned in Exercise 4 will be greater than that gotten by
revolving about the line x = 2.5.
(b) The original volume V1 of Exercise 4 is given by
2
V1 = π (4 − 1/y 2 ) dy = 9π/2, and the other volume
1/2
2 2
1 21
V2 = π − 2.5 − (2 − 2.5)2 dy = − 10 ln 2 π ≈ 3.568528194π,
1/2 y 2
and thus V1 is the larger volume.



3 9 √
35. V = π (9 − y 2 )2 dy 36. V = π [32 − (3 − x)2 ]dx
0 0
3 9 √
=π (81 − 18y 2 + y 4 )dy =π (6 x − x)dx
0 0

= 648π/5 = 135π/2
y y
3 x = y2

y=3
x y = √x
9 x
9

1 √ y
37. V = π [( x + 1)2 − (x + 1)2 ]dx x=y
0 1 x = y2
1 √ x
=π (2 x − x − x2 )dx = π/2 1
0
y = –1

1 y
38. V = π [(y + 1)2 − (y 2 + 1)2 ]dy x=y
0
1
1
=π (2y − y 2 − y 4 )dy = 7π/15
0
x
1
x = y2
x = –1

1
39. A(x) = π(x2 /4)2 = πx4 /16, 40. V = π (x − x4 )dx = 3π/10
0
20
V = (πx4 /16)dx = 40, 000π ft3
0

1√
1 2
1 1
41. V = (x − x2 )2 dx 42. A(x) = π x = πx,
0 2 2 8
1 4
1
= (x2 − 2x3 + x4 )dx = 1/30 V = πx dx = π
0 0 8
Square y
y
y = x (1, 1) y = √x
y = x2
1 x 4
x


302 Chapter 7

√
43. On the upper half of the circle, y = 1 − x2 , so:
(a) A(x) is the area of a semicircle of radius y, so
1 1
π
A(x) = πy 2 /2 = π(1 − x2 )/2; V = (1 − x2 ) dx = π (1 − x2 ) dx = 2π/3
2 −1 0

y
–1 y

1 x
y = √1 – x2

(b) A(x) is the area of a square of side 2y, so
1 1
A(x) = 4y 2 = 4(1 − x2 ); V = 4 (1 − x2 ) dx = 8 (1 − x2 ) dx = 16/3
−1 0

–1 y
2y

1 x
y = √1 – x 2

(c) A(x) is the area of an equilateral triangle with sides 2y, so
√
3 √ √
A(x) = (2y)2 = 3y 2 = 3(1 − x2 );
4
1 √ √ 1 √
V = 3(1 − x2 ) dx = 2 3 (1 − x2 ) dx = 4 3/3
−1 0

2y 2y
–1 y

2y
1 x
y = √1 – x2

44. The base of the dome is a hexagon of side r. An equation of the circle of radius r that lies in a
vertical x-y plane and passes through two opposite vertices of the base hexagon is x2 + y 2 = r2 .
A horizontal, hexagonal cross section at height y above the base has area
√ √ √
3 3 2 3 3 2 r
3 3 2 √
A(y) = x = (r − y ), hence the volume is V =
2
(r − y 2 ) dy = 3r3 .
2 2 0 2

45. The two curves cross at x = b ≈ 1.403288534, so
b π/2
V =π ((2x/π)2 − sin16 x) dx + π (sin16 x − (2x/π)2 ) dx ≈ 0.710172176.
0 b

46. Note that π 2 sin x cos3 x = 4x2 for x = π/4. From the graph it is apparent that this is the ﬁrst
positive solution, thus the curves don’t cross on (0, π/4) and
π/4
1 5 17 6
V =π [(π 2 sin x cos3 x)2 − (4x2 )2 ] dx = π + π
0 48 2560



e
47. V = π (1 − (ln y)2 ) dy = π
1

tan 1
π
48. V = π[x2 − x2 tan−1 x] dx = [tan2 1 − ln(1 + tan2 1)]
0 6
r
1 2
49. (a) V = π (r2 − y 2 ) dy = π(rh2 − h3 /3) = πh (3r − h)
r−h 3
y
(b) By the Pythagorean Theorem,
h
r2 = (r − h)2 + ρ2 , 2hr = h2 + ρ2 ; from Part (a), x2 + y2 = r2
r x
πh πh 3 2 r
V = (3hr − h2 ) = (h + ρ2 ) − h2 )
3 3 2
1
= πh(h2 + 3ρ2 )
6

50. Find the volume generated by revolving y
the shaded region about the y-axis.
h – 10 10 x
−10+h
π
V =π (100 − y 2 )dy = h2 (30 − h)
−10 3 h
x = √100 – y 2
–10
Find dh/dt when h = 5 given that dV /dt = 1/2.
π dV π dh
V = (30h2 − h3 ), = (60h − 3h2 ) ,
3 dt 3 dt
1 π dh dh
= (300 − 75) , = 1/(150π) ft/min
2 3 dt dt

5
51. (b) ∆x = = 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};
10
9 2
yi
left = π ∆x ≈ 11.157;
i=0
2
10 2
yi
right = π ∆x ≈ 11.771; V ≈ average = 11.464 cm3
i=1
2

√
52. If x = r/2 then from y 2 = r2 − x2 we get y = ± 3r/2 y
√ √ √3r
as limits of integration; for − 3 ≤ y ≤ 3, 2
x = √r 2 – y 2
A(y) = π[(r − y ) − r /4] = π(3r /4 − y ), thus
2 2 2 2 2 x
√
3r/2 r
2
V =π √ (3r2 /4 − y 2 )dy
− 3r/2 – √3r
√ 2
3r/2 √
= 2π (3r2 /4 − y 2 )dy = 3πr3 /2
0


304 Chapter 7

y y
53. (a) (b)
h –4
x x

h–4 –2
h
h
–4 –4
0≤h<2 2≤h≤4

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 the
cherry is 1 cm so points on the sections shown in the ﬁgures satisfy the equations x2 + y 2 = 16
and x2 + (y + 3)2 = 1. We will ﬁnd the volumes of the solids that are generated when the shaded
regions are revolved about the y-axis.
For 0 ≤ h < 2,
h−4 h−4
V =π [(16 − y 2 ) − (1 − (y + 3)2 )]dy = 6π (y + 4)dy = 3πh2 ;
−4 −4

for 2 ≤ h ≤ 4,
−2 h−4
V =π [(16 − y 2 ) − (1 − (y + 3)2 )]dy + π (16 − y 2 )dy
−4 −2
−2 h−4
1
= 6π (y + 4)dy + π (16 − y 2 )dy = 12π + π(12h2 − h3 − 40)
−4 −2 3
1
= π(12h2 − h3 − 4)
3
so 
 3πh
2
if 0 ≤ h < 2
V =
 1 π(12h2 − h3 − 4) if 2 ≤ h ≤ 4
3

54. x=h± r2 − y2 ,
y
r
(x – h 2) + y 2 = r 2
V =π (h + r2 − y 2 )2 − (h − r2 − y 2 )2 dy
−r
r x
= 4πh r2 − y 2 dy
−r

1 2
= 4πh πr = 2π 2 r2 h
2

55. tan θ = h/x so h = x tan θ,
1 1 1
A(y) = hx = x2 tan θ = (r2 − y 2 ) tan θ
2 2 2 h
because x = r − y ,
2 2 2
u
r
1 x
V = tan θ (r − y )dy
2 2
2 −r
r
2 3
= tan θ (r2 − y 2 )dy = r tan θ
0 3



56. A(x) = (x tan θ)(2 r2 − x2 ) 57. Each cross section perpendicular to the
y-axis is a square so
= 2(tan θ)x r2 − x2 ,
r
A(y) = x2 = r2 − y 2 ,
V = 2 tan θ x r2 − x2 dx 1 r
0 V = (r2 − y 2 )dy
8 0
2 3
= r tan θ
3 V = 8(2r3 /3) = 16r3 /3
x tan u y
y
x
x = √r 2 – y 2
√r 2 – x 2

r
x

58. The regular cylinder of radius r and height h has the same circular cross sections as do those of
the oblique clinder, so by Cavalieri’s Principle, they have the same volume: πr2 h.

EXERCISE SET 7.3
2 2
1. V = 2πx(x2 )dx = 2π x3 dx = 15π/2
1 1
√ √
2 2
8π √
2. V = 2πx( 4 − x2 − x)dx = 2π (x 4 − x2 − x2 )dx = (2 − 2)
0 0 3
1 1
3. V = 2πy(2y − 2y 2 )dy = 4π (y 2 − y 3 )dy = π/3
0 0

2 2
4. V = 2πy[y − (y 2 − 2)]dy = 2π (y 2 − y 3 + 2y)dy = 16π/3
0 0

1 9 √
5. V = 2π(x)(x3 )dx 6. V = 2πx( x)dx
0 4
1 9
= 2π x4 dx = 2π/5 = 2π x3/2 dx = 844π/5
0 4
y y
3 y = √x
y= x3
2
1
1
x
x
–1 1
–9 –4 4 9
–1

3 3
7. V = 2πx(1/x)dx = 2π dx = 4π y
1 1 1
y= x

x
–3 –1 1 3


306 Chapter 7

√
π/2 √
8. V = 2πx cos(x2 )dx = π/ 2 y
0 y = cos (x2)

x
√p
2

2 2
9. V = 2πx[(2x − 1) − (−2x + 3)]dx 10. V = 2πx(2x − x2 )dx
1 0
2 2
8
= 8π (x2 − x)dx = 20π/3 = 2π (2x2 − x3 )dx = π
1 0 3
y y
(2, 3) y = 2x – x 2

(1, 1)
x
x 2

(2, –1)

√ √
1 3 3
x x2 x2
11. V = 2π 2+1
dx 12. V = 2πxe dx = πe = π(e3 − e)
0 x 1 1
1 y
= π ln(x2 + 1) = π ln 2
0 20 2
y y = ex
1 y= 1
x2 + 1
10

x
x
–1 1
-√3 -1 1 √3

1 3 3
13. V = 2πy 3 dy = π/2 14. V = 2πy(2y)dy = 4π y 2 dy = 76π/3
0 2 2
y y
3
1 x = y2 2 x = 2y
x
x

1
√
15. V = 2πy(1 − y)dy y
0 y = √x
1 x
= 2π (y − y 3/2 )dy = π/5 1
0

Chapter 07

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Chapter 07