Answer
$8\pi$
Work Step by Step
Setup the integration using shell method about the line x=4
$ 2\pi \int _0^2 (4-x)(2x-x^2) dx$
$ 2\pi \int _0^2 (8x-4x^2 - 2x^2 +x^3)dx$
$2\pi \int_0^2 (x^3 - 6x^2 +8x)dx$, Integrate
$2\pi [\frac{1}{4}x^4 - 2x^3 + 4x^2]_0^2$, Evaluate the definite integral
$2\pi(4-16+16)$
$8\pi$
