Answer
$\displaystyle{V=\frac{128\pi}{3}}$
Work Step by Step
$\displaystyle{4=(y-3)^2}\\
\displaystyle{y=\pm2+3}\\
\displaystyle{y=1\qquad y=5}$
$\displaystyle{V=\int_{1}^{5}2\pi\left(1-y\right)\left(y^2-6y+9-4\right)\ dy}\\
\displaystyle{V=2\pi\int_{1}^{5}7y^2+5-y^3-11y\ dy}\\
\displaystyle{V=2\pi\left[\frac{7}{3}y^3+5y-\frac{1}{4}y^4-\frac{11}{2}y^2\right]_{1}^{5}}\\
\displaystyle{V=2\pi\left(\left(\frac{7}{3}(5)^3+5(5)-\frac{1}{4}(5)^4-\frac{11}{2}(5)^2\right)-\left(\frac{7}{3}(1)^3+5(1)-\frac{1}{4}(1)^4-\frac{11}{2}(1)^2\right)\right)}\\
\displaystyle{V=\frac{128\pi}{3}}$
