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^2=8x` and x=2 revolved about the x=4?

Answer 1

Please see below.

Explanation:

Here is a sketch of the region. To use shells, we'll take a representative slice parallel to the axis of rotation. (Parallel to the line `x=4`.)

The slice is taken at some value of `x`.

The volume of the representative shell is `2pirhxx"thickness"`

In this case,

`"thickness" = dx`

the radius `r` is shown as a dotted black line segment from the slice at `x` to the line at `4`. So, `r = 4-x`

The height of the slice is the upper `y` value minus the lower `y` value.
Solving `y^2=8x`, we see that `y_"upper" = sqrt(8x)` and `y_"lower" = -sqrt(8x)`.
So, `h = sqrt(8x) - (-sqrt(8x)) = 2sqrt(8x)`

The volume of the representative shell is `2pi(4-x)(2sqrt(8x))dx`

`x` varies from `0` to `2`, so the volume of the solid is

`V = int_0^2 2 pi (4-x)(4sqrt(2x))dx`

`= 8pisqrt2 int_0^2 (4-x)sqrtx \ dx`

`= 896/15 pi`