Answer
$\frac{144\pi}{7}$
Work Step by Step
Setup the integration using disk method about the line y=4
$ \pi \int_0^2 (4-\frac{1}{2}x^3)^2 dx$
$\pi \int_0^2(16-4x^3 + \frac{1}{4}x^6)dx$
$\pi [16x -x^4 + \frac{1}{28}x^7]_0^2$
$\pi(32-16+\frac{128}{28})-(0)$
$\frac{144\pi}{7}$
