Answer
$\frac{128\pi}{3}$
Work Step by Step
Setup the integration using disk method about the y-axis.
Isolate x as a function of y
$x=\sqrt {16-y^2}$
$\pi \int_0^4 (\sqrt {16-y^2})^2dy$
$\pi \int_0^4 (16-y^2)dy$
$\pi[16y-\frac{1}{3}y^3]_0^4$
$ \pi(64-\frac{64}{3})$
$\frac{128\pi}{3}$
