Question

How do you find the volume of the solid generated by revolving the region bounded by the graphs `x=y^2, x=4`, about the line x=6?

Answer 1

Please see below.

Explanation:

The graph of the region is shown below. A representative slice of thickness `dy` has been taken at height `y`. The radii of rotation are shown as dashed and dotted red lines.

When revolved about the line `x=6` the resulting slice has volume

`(piR^2-pir^2) * "thickness"` where `R` is the greater radius and `r` the lesser.

In this problem

`R = 6-y^2` and
`r = 6-4=2` and
`"thickness" = dy`.

`y` varies from `-2` to `2`. (Or go from `0` to `2` and double the result.)

So the volume of the solid of revolution is

`V = pi int_-2^2 ((6-y^2)^2 - (2)^2) dy = pi int_-2^2 (32-12y^2+y^4) dy`

`= (384pi)/5`

or about `241.3`.