Answer
$\displaystyle{V=\frac{13\pi}{6}}$
Work Step by Step
$\displaystyle{2=2y^2}\\ \displaystyle{y^2=1}\\ \displaystyle{y=-1\qquad y=1}\\ \displaystyle{y\geq0}\\ \displaystyle{\therefore y=1}$
$\displaystyle{V=\int_{0}^{1}(2\pi (2-y))\left(2-2y^2\right)\ dy}\\ \displaystyle{V=2\pi\int_{0}^{1}y^3-2y^2-y+2\ dy}\\ \displaystyle{V=2\pi\left[\frac{1}{4}y^4-\frac{2}{3}y^3-\frac{1}{2}y^2+2y\right]_{0}^{1}}\\ \displaystyle{V=2\pi\left(\left(\frac{1}{4}(1)^4-\frac{2}{3}(1)^3-\frac{1}{2}(1)^2+2(1)\right)-\left(0\right)\right)}\\ \displaystyle{V=\frac{13\pi}{6}}$
