Page 241 - Calculus Demystified

CHAPTER 8
228 the volume enclosed is Applications of the Integral
27 y 5/3 27 729π

V = π · y 2/3 dy = π · 5/3 = 5 .
0 0
Math Note: The reasoning we have used in the last example shows this: If the
curve x = g(y), c ≤ y ≤ d, is rotated about the y-axis then the volume enclosed
by the resulting surface is
d 2
V = c π ·[g(y)] dy.

You Try It: Calculate the volume enclosed when the curve y = x 1/3 ,32 ≤ x ≤
243, is rotated about the y-axis.
EXAMPLE 8.8
TEAMFLY
Set up, but do not evaluate, the integral that represents the volume gener-
ated when the planar region between y = x + 1 and y = 2x + 4 isrotated
2
about the x-axis.
SOLUTION
When the planar region is rotated about the x-axis, it will generate a donut-
shaped solid. Notice that the curves intersect at x =−1 and x = 3; hence
the intersection lies over the interval [−1, 3]. For each x in that interval, the
segment connecting (x, x + 1) to (x, 2x + 4) will be rotated about the x-axis.
2
It will generate a washer. See Fig. 8.18. The area of that washer is
A(x) = π ·[2x + 4] − π ·[x + 1].
2
2

[Notice that we calculate the area of a washer by subtracting the areas of two
circles—not by subtracting the radii and then squaring.]
It follows that the volume of the solid generated is
3 2 2
V = −1 π ·[2x + 4] − π ·[x + 1] dx.

8.2.2 THE METHOD OF CYLINDRICAL SHELLS

Our philosophy will now change. When we divide our region up into vertical strips,
we will now rotate each strip about the y-axis instead of the x-axis. Thus, instead
of generating a disk with each strip, we will now generate a cylinder.
Look at Fig. 8.19. When a strip of height h and thickness x, with distance r
from the y-axis, is rotated about the y-axis, the resulting cylinder has surface area
2πr · h and volume about 2πr · h · x. This is the expression that we will treat
in order to sum up the volumes of the cylinders.

®
Team-Fly