The Disk and Shell MethodCharles A. CableDepartment of Mathematics, Allegheny College, Meadville, PA 16335The American Mathematical Monthly, February 1984, Volume 91, Number 2, p. 139.

In most calculus books there is little effort given to showing that the cylindrical shellmethod and disk method give the same value when computing the volume of a solidof 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, and is revolved about the y-axis.In this paper we shall show that the cylindrical shell and disk methods give the samevalue if the region revolved about the y-axis is bounded by andthe x-axis, provided is a differentiable function on and 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 acalculus course.Consider the solid of revolution K produced by revolving the region bounded by

and the x-axis, about the y-axis. We use the shell method, whichinvolves summing the volumes of cylindrical shells, to define the volume of K to be

If is differentiable on and hence continuous

there, this limit exists and is equal to Suppose the region is bounded by the function and the y-axis. Inthe disk method, which involves summing the volumes of disks, we consider

If is continuous on this limit exists and is equal to THEOREM. Let and let be differentiable, nonnegative and on with and where * Also let be continuous on

and let iff If R is the region bounded by the x-axis,and 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.Equivalently Proof: The region R can also be described as the region bounded by

and where

We observe from the way that is defined that it is continuous on If weevaluate the volume obtained by revolving the region R about the y-axis by using thedisk method, we find this to be This is equal to

Ed0

pb2 dy 2 E c0

pa2 dy 2 Edc

pfgsydg2 dy 5 pfb2d 2 a2cg 2 Edc

pfgsydg2 dy.

ed0pb2 dy 2 ed0pmsyd dy.

f0, dg.msyd

msyd 5 5agsyd if 0 y , c,if c y d.y 5 d,x 5 msyd, x 5 b, y 5 0

p fb2d 2 a2cg 2 edc p fgsydg2 dy 5 eba2pxf sxd dx.

x 5 bx 5 ay 5 f sxd,y 5 f sxd.x 5 gsydfa, bg

f9sxdc < d.f sbd 5 df sad 5 cfa, bg,1 2 1y 5 f sxd0 a , b,

edc pfgsydg2 dy.fc, dg,gsyd

limiPi0oni51 pfg syidg2 Dyi.

x 5 gsyd, y 5 c, y 5 deba2pxf sxd dx.

fa, bgf sxdlimiPi0oni51 2p xi f sxidDxi.y 5 f sxd, x 5 a, x 5 b

f sxdfa, bgf sxdy 5 f sxd, x 5 a, x 5 b

x 5 bx 5 ay 5 mx 1 b,

By the substitution formula, the latter expression is equal to

A straightforward application of the integration by parts formula and algebraicsimplification shows that

and the argument is complete.*The result is true in case but a slight alteration is needed in the argument.d < c,

Eba

2pxf sxd dx 5 pfb2d 2 a2cg 2 Epx2f9sxddx,

pfb2d 2 a2cg 2 ebapx2f9sxd dx.

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