Answer
$\displaystyle{V=\frac{16\pi}{3}}$
Work Step by Step
$\displaystyle{2y^2=y^2+1}\\ \displaystyle{y^2=1}\\ \displaystyle{y=-1\qquad y=1}\\$
$\displaystyle{V=\int_{-1}^{1}(2\pi (y+2))\left(y^2+1-\left(2y^2\right)\right)\ dy}\\ \displaystyle{V=2\pi\int_{-1}^{1}2+y-2y^2-y^3\ dy}\\ \displaystyle{V=2\pi\left[2y+\frac{1}{2}y^2-\frac{2}{3}y^3-\frac{1}{4}y^4\right]_{-1}^{1}}\\ \displaystyle{V=2\pi\left(\left(2(1)+\frac{1}{2}(1)^2-\frac{2}{3}(1)^3-\frac{1}{4}(1)^4\right)-\left(2(-1)+\frac{1}{2}(-1)^2-\frac{2}{3}(-1)^3-\frac{1}{4}(-1)^4\right)\right)}\\ \displaystyle{V=\frac{16\pi}{3}}$
