## Calculus 10th Edition

$16\pi$
Setup the integration using shell method about the line x=4 $2\pi \int _0^2 (4-x)(4x-x^2-x^2)dx$ $4\pi \int _0^2(4-x)(2x-x^2)dx$, Take a factor of 2 outside the integration $4\pi \int_0^2 (8x-4x^2-2x^2+x^3)dx$ $4\pi \int_0^2 (x^3 -6x^2 +8x)dx$ $4\pi [\frac{1}{4}x^4 - 2x^3 +4x^2]_0^2$ $4\pi(4-16+16)$ $16\pi$