The Disk and Shell Method Charles A. Cable Department of Mathematics, Allegheny College, Meadville, PA 16335 The American Mathematical Monthly, February 1984, Volume 91, Number 2, p. 139.
I
n most calculus books there is little effort given to showing that the cylindrical shell method and disk method give the same value when computing the volume of a solid of revolution. Indeed it is not obvious that these two distinct methods should give the same result. In some texts this is demonstrated when the trapezoid bounded by the x-axis, y 5 mx 1 b, x 5 a and x 5 b is revolved about the y-axis. In this paper we shall show that the cylindrical shell and disk methods give the same value if the region revolved about the y-axis is bounded by y 5 f sxd, x 5 a, x 5 b and the x-axis, provided f sxd is a differentiable function on fa, bg and f sxd is one-to-one. The proof is simple and uses two theorems which the students have recently learned (substitution formula and integration by parts.) This proof can easily be included in a calculus course.
Consider the solid of revolution K produced by revolving the region bounded by y 5 f sxd, x 5 a, x 5 b and the x-axis, about the y-axis. We use the shell method, which involves summing the volumes of cylindrical shells, to define the volume of K to be n limiPi→0 i51 2p xi f sxi dDxi. If f sxd is differentiable on fa, bg and hence continuous
o
there, this limit exists and is equal to ea 2p xf sxd dx. b
Suppose the region is bounded by the function x 5 gs yd, y 5 c, y 5 d and the y-axis. In the disk method, which involves summing the volumes of disks, we consider limiPi→0
o
n p i51
f g s yi dg2 Dyi.
If gs yd is continuous on fc, dg, this limit exists and is equal to ec p f gs ydg2 dy. d
THEOREM. Let 0 ø a , b, and let y 5 f sxd be differentiable, nonnegative and 1 2 1 on fa, bg, with f sad 5 c and f sbd 5 d where c < d.* Also let f9sxd be continuous on fa, bg and let x 5 gs yd iff y 5 f sxd. If R is the region bounded by y 5 f sxd, the x-axis, x 5 a and x 5 b and R is revolved about the y-axis, then the value obtained by using the disk method is equal to the value obtained by using the cylindrical shell method. d b Equivalently p fb2d 2 a2cg 2 ec p f gs ydg2 dy 5 ea 2p xf sxd dx. Proof: The region R can also be described as the region bounded by x 5 ms yd, x 5 b, y 5 0 and y 5 d, where ms yd 5
5ags yd
if 0 ø y , c, if c ø y ø d.
We observe from the way that ms yd is defined that it is continuous on f0, dg. If we evaluate the volume obtained by revolving the region R about the y-axis by using the d d disk method, we find this to be e0 pb2 dy 2 e0 p ms yd dy. This is equal to
E pb d
0
2
dy 2
E pa c
0
2
dy 2
E p fgs ydg dy 5 p fb d 2 a cg 2 E p fgs ydg dy. d
c
2
2
2
d
c
2
By the substitution formula, the latter expression is equal to
p fb2d 2 a2cg 2 ea p x2f9sxd dx. b
A straightforward application of the integration by parts formula and algebraic simplification shows that
E 2pxf sxd dx 5 p fb d 2 a cg 2 E p x f9sxddx, b
2
2
2
a
and the argument is complete. *The result is true in case d <
c, but a slight alteration is needed in the argument.
2
