How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y=8-x^2#, #y=x^2# revolved about the x=2?

1 Answer
Jim H
Nov 16, 2016

Please see below.

Explanation:

In the following picture, the region is shown in blue. The points of intersection of the two curves are #(+-2,4)#. The axis of rotation (the line #x=2#) is in red.

To set up for the method of shells, a slice has been taken parallel to the axis of rotation. The thin side (the thickness of the shell) is #dx#. The slice has black borders and the radius of rotation is shown as a dashed black line.

enter image source here

The Volume of a representative shell is

#2 pi r h "thickness"#

Since the thickness is
#dx#

(the variable we will work with is #x#) we note that #x# varies from #-2# to #2#.

The radius of the shell is
#r = 2-x#.

The height is the upper curve minus the lower curve

#h = (8-x^2) - (x^2) = 8-2x^2#

So the volume of the solid of revolution is

#V = int_-2^2 2piunderbrace((2-x))_r underbrace((8-2x^2))_h overbrace(dx)^("thickness")#

# = 2pi[128/3] = (256 pi)/3#

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