# How do you find the volume of the solid generated by revolving the region bounded by #y=2x^2#, #y=0#, #x=2#, revolving on #y=8#?

##### 1 Answer

It looks like this:

graph{(y-2x^2)(y-8)(x-2)(sqrt(x)/sqrtx) <= 0 [-0.5, 5, -0.35, 8.0]}

I'm assuming it's also in the domain of

Let's make it easier to work with though; due to the symmetry of the revolved solid, if you just take the portion of the revolved solid that would be above the line

**Different solid---same volume.**

And no need to do any subtractions to specify a range for the graph---we can just use

...since you are essentially stacking "circles" with radii that vary according to a function to generate a solid. These "circles" are perpendicular to the revolution axis. The "circle" is defined by