Answer
$\displaystyle{V=162\pi}$
Work Step by Step
$\displaystyle{A\left(y\right)=\pi \left(2\sqrt{y}\right)^2}\\ \displaystyle{A\left(y\right)=\pi \left(4y\right)}\\$
$\displaystyle{V=\int_{0}^9A\left(y\right)\ dy}\\
\displaystyle{V=\int_{0}^9\pi \left(4y\right)\ dy}\\
\displaystyle{V=4\pi \int_{0}^9y\ dy}\\
\displaystyle{V=4\pi\left[\frac{1}{2}y^2\right]_{0}^9}\\
\displaystyle{V=4\pi\left(\left(\frac{1}{2}9^{2}\right)-\left(\frac{1}{2}0^{2}\right)\right)}\\
\displaystyle{V=162\pi}\\ $