Answer
$\frac{602\pi}{5}$
Work Step by Step
$V$ = $2\pi\int_0^{5}2ydy+2\pi\int_5^{9}y\sqrt {9-y}dy$
$2\pi\int_0^{5}2ydy$ = $2\pi(y^{2})|_0^5$ = $50\pi$
$2\pi\int_5^{9}y\sqrt {9-y}dy$
let
$u$ = $9-y$
$du$ = $-dy$
$2\pi\int_5^{9}y\sqrt {9-y}dy$ = $2\pi\int_0^{4}(9-u)\sqrt {u}du$ = $2\pi\int_0^{4}(9u^{\frac{1}{2}}-u^{\frac{3}{2}})du$ = $2\pi(6u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}})|_0^4$ = $\frac{352\pi}{5}$
Total volume = $50\pi+\frac{352\pi}{5}$ = $\frac{602\pi}{5}$
