Answer
$\displaystyle{V=\frac{27\pi}{2}}$
Work Step by Step
$\displaystyle{4-y=y^2-4y+4}\\
\displaystyle{0=y^2-4y+y}\\
\displaystyle{(y)(y-3)=0}\\
y=0\qquad y=3$
$\displaystyle{V=\int_{0}^{3}(2\pi y)\left(\left(4-y\right)-\left(y^2-4y+4\right)\right)\ dy}\\
\displaystyle{V=2\pi\int_{0}^{3}3y^2-y^3\ dy}\\
\displaystyle{V=2\pi\left[y^3-\frac{1}{4}y^4\right]_{0}^{3}}\\
\displaystyle{V=2\pi\left(\left((3)^3-\frac{1}{4}(3)^4\right)-(0)\right)}\\
\displaystyle{V=\frac{27\pi}{2}}$