Answer
$\displaystyle{V=\frac{16\pi}{3}}$
Work Step by Step
$\displaystyle{2=1+\left(y-2\right)^2}\\ \displaystyle{\left(y^2-4y+4\right)-1=0}\\ \displaystyle{y^2-4y+3=0}\\ \displaystyle{(y-1)(y-3)=0}\\ y=1\qquad y=3$
$\displaystyle{V=\int_{1}^{3}(2\pi y)\left(2-\left(1+\left(y-2\right)^2\right)\right)\ dy}\\ \displaystyle{V=2\pi\int_{1}^{3}4y^2-y^3-3y\ dy}\\ \displaystyle{V=2\pi\left[\frac{4}{3}y^3-\frac{1}{4}y^4-\frac{3}{2}y^2\right]_{1}^{3}}\\ \displaystyle{V=2\pi\left(\left(\frac{4}{3}(3)^3-\frac{1}{4}(3)^4-\frac{3}{2}(3)^2\right)-\left(\frac{4}{3}(1)^3-\frac{1}{4}(1)^4-\frac{3}{2}(1)^2\right)\right)}\\ \displaystyle{V=\frac{16\pi}{3}}$
