Calculus 2nd Edition Rogawski

322 CHAPTER 6 APPLICATIONS OF THE INTEGRAL
55. Find the volume of the cone obtained by rotating the region under
the segment joining (0,h)and (r, 0) about the y-axis.
y
56. The torus (doughnut-shaped solid) in Figure 15 is obtained by rotating
the circle (x − a) 2 + y 2 = b 2 around the y-axis (assume that
a>b). Show that it has volume 2π 2 ab 2 . Hint: Evaluate the integral
by interpreting it as the area of a circle.
−a
1
−1
a
x
y
FIGURE 16 The hyperbola with equation y 2 − x 2 = 1.
a
a + b
x
59. A “bead” is formed by removing a cylinder of radius r from the
center of a sphere of radius R (Figure 17). Find the volume of the bead
with r = 1 and R = 2.
y
y
FIGURE 15 Torus obtained by rotating a circle about the y-axis.
h
57. Sketch the hypocycloid x 2/3 + y 2/3 = 1 and find the volume
of the solid obtained by revolving it about the x-axis.
r
R
x
x
58. The solid generated by rotating the region between the branches
of the hyperbola y 2 − x 2 = 1 about the x-axis is called a hyperboloid
(Figure 16). Find the volume of the hyperboloid for −a ≤ x ≤ a.
FIGURE 17 A bead is a sphere with a cylinder removed.
Further Insights and Challenges
60. Find the volume V of the bead (Figure 17) in terms of r
and R. Then show that V = π 6 h3 , where h is the height of the bead.
This formula has a surprising consequence: Since V can be expressed
in terms of h alone, it follows that two beads of height 1 cm, one formed
from a sphere the size of an orange and the other from a sphere the size
of the earth, would have the same volume! Can you explain intuitively
how this is possible?
1
c
y
R
y = f(x)
61. The solid generated by rotating the region inside the ellipse with
equation ( x
a
) 2 +
( y
b
) 2 = 1 around the x-axis is called an ellipsoid.
Show that the ellipsoid has volume 4 3 πab2 . What is the volume if the
ellipse is rotated around the y-axis?
a
FIGURE 18 The tractrix.
2
x
62. The curve y = f (x) in Figure 18, called a tractrix, has the following
property: the tangent line at each point (x, y) on the curve has
slope
dy
dx = √ −y
1 − y 2
Let R be the shaded region under the graph of 0 ≤ x ≤ a in Figure 18.
Compute the volume V of the solid obtained by revolving R around
the x-axis in terms of the constant c = f (a). Hint: Use the substitution
u = f (x) to show that
∫ 1 √
V = π u 1 − u 2 du
c
63. Verify the formula
∫ x2
x 1
(x − x 1 )(x − x 2 )dx = 1 6 (x 1 − x 2 ) 3 3
Then prove that the solid obtained by rotating the shaded region in Figure
19 about the x-axis has volume V = π 6 BH2 , with B and H as in the
figure. Hint: Let x 1 and x 2 be the roots of f (x) = ax + b − (mx + c) 2 ,
where x 1