# Evaluate the volume of a solid of revolution using shells.

A cylindrical shell $S$ formed by some revolution about the $y$-axis is given by the equation: $S=2\pi x f(x)dx$, where the circumference $C$ of the shell is $C=2\pi x$, the height of the shell ($H$) is $H=f(x)$, and the thickness of the shell $T$ is $T=dx$ such that the shell is $S=C \cdot H \cdot T$.

This implies the volume of the solid formed by the revolution of the region bounded by $f(x) = 4x-x^2$ and $g(x) = 8x-2x^2$ about $x=-2$ should be equal to: $\displaystyle2\pi\int_{0}^{4}{(x+2)(4x-x^2)}{dx} = \frac{256}{3}\pi$, as the circumference of a shell should be $C = 2\pi(x+2)$ with height $H = 4x-x^2$ and thickness $T=dx$.

Is this accurate?

Another way to do this problem, which acts as a check on your method, is to translate the entire graph to the right so that the axis of rotation is the $y$-axis, as is required by your original formula $S=2\pi xf(x)\,dx$. We do that by replacing $x$ in that formula by $x-2$ and moving the bounds from $0..4$ to $2..6$, so the integral in the original formula becomes
$$\int_2^62\pi x[4(x-2)-(x-2)^2]\,dx$$
This gives the same answer as your method, $\frac{256}3\pi$, so the check works and we can accept the answer with confidence.