Question

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?

Answer 1

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.

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`