CHAPTER 22
Volume
22.1
22.2
22.3
Fig. 22-1
Fig. 22-2
Derive the formula V= \irr-h for the volume of a right circular cone of height h and radius of base r.
Refer to Fig. 22-2. Consider the right triangle with vertices (0,0), (h, 0), and (h,r). If this is rotated about
the x-axis, a right circular cone of height h and radius of base r results. Note that the hypotenuse of the triangle
lies on the line y = (r/h)x. Then, by the disk formula,
173
Derive the formula V= jirr
3 for the volume of a sphere of radius r.
sphere of radius r results. By the disk formula, V= TT Jl y
2 dx = TT |I (r
2 - x
2
) dx = ir(r
2 x - j*
3 ) l
r
_ =
TKr
3 -ir
3 )-(-r
J +ir
3 )]=^r
3 .
In Problems 22.3-22.19, find the volume generated by revolving the given region about the given axis.
The region above the curve y = jc
3
, under the line y = \, and between *=0 and * = !; about the
AC-axis.
See Fig. 22-3. The upper curve is y = 1, and the lower curve is y = x
3 . We use the circular ring formula: V = w Jo' [I
2 - (x
3 )
2 ] dx = rr(x - fce
7
) ]
1
0 = TT(! - }) = f TT.
Fig. 22-3
22.4
The region of Problem 22.3, about the y-axis.
We integrate along the y-axis from 0 to 1. The upper curve is x = y
1 '
3 , the lower curve is the y-axis, and
we use the disk formula:
22.5
The region below the line
(See Fig. 22-4.)
y-2x, above the x-axis, and between jr = 0 and x = I; about the jc-axis.
We use the disk formula:
Consider the upper semicircle y = Vr2 - x2 (Fig. 22-1). If we rotate it about the x-axis, the