Using the shell method, the volume integral would be
That is, each shell has a radius of <em>x</em> (the distance from a given <em>x</em> in the interval [0, 2] to the axis of revolution, <em>x</em> = 0) and a height equal to the difference between the boundary curves <em>y</em> = <em>x</em> ⁸ and <em>y</em> = 256. Each shell contributes an infinitesimal volume of 2<em>π</em> (radius) (height) (thickness), so the total volume of the overall solid would be obtained by integrating over [0, 2].
The volume itself would be
Using the disk method, the integral for volume would be
where each disk would have a radius of <em>x</em> = ⁸√<em>y</em> (which comes from solving <em>y</em> = <em>x</em> ⁸ for <em>x</em>) and an infinitesimal height, such that each disk contributes an infinitesimal volume of <em>π</em> (radius)² (height). You would end up with the same volume, 4096<em>π</em>/5.
