Answer
$\displaystyle{V=8\pi}\\$
Work Step by Step
$\displaystyle{V=\int_{0}^{2}(2\pi y)\left(y^2\right)\ dy}\\
\displaystyle{V=2\pi\int_{0}^{2}y^3\ dy}\\
\displaystyle{V=2\pi\left[\frac{1}{4}y^4\right]_{0}^{2}}\\
\displaystyle{V=2\pi\left(\left(\frac{1}{4}(2)^4\right)-(0)\right)}\\
\displaystyle{V=8\pi}\\$