Question

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by y=x^8 and y = 256 in the first quadrant about the y-axis.

Answer 1

Using the shell method, the volume integral would be

$\displaystyle 2\pi \int_0^2 x(256-x^8)\,\mathrm dx$

That is, each shell has a radius of x (the distance from a given x in the interval [0, 2] to the axis of revolution, x = 0) and a height equal to the difference between the boundary curves y = x ⁸ and y = 256. Each shell contributes an infinitesimal volume of 2π (radius) (height) (thickness), so the total volume of the overall solid would be obtained by integrating over [0, 2].

The volume itself would be

$\displaystyle 2\pi \int_0^2 x(256-x^8)\,\mathrm dx = 2\pi \left(128x^2-\frac1{10}x^{10}\right)\bigg|_{x=0}^{x=2} = \boxed{\frac{4096\pi}5}$

Using the disk method, the integral for volume would be

$\displaystyle \pi \int_0^{256} \left(\sqrt[8]{y}\right)^2\,\mathrm dy = \pi \int_0^{256} \sqrt[4]{y}\,\mathrm dy$

where each disk would have a radius of x = ⁸√y (which comes from solving y = x ⁸ for x) and an infinitesimal height, such that each disk contributes an infinitesimal volume of π (radius)² (height). You would end up with the same volume, 4096π/5.