Answer
$\displaystyle{V=\frac{117\pi}{5}}$
Work Step by Step
$\displaystyle{y^2-2y+1=y+1}\\
\displaystyle{y^2-3y=0}\\
\displaystyle{y(y-3)=0}\\
\displaystyle{y=0\qquad y=3}$
$A(y)=\pi\left(y+1+1\right)^2-\pi\left((y-1)^2+1\right)^2\\
A(y)=\pi\left(4y^3+12y-7y^2-y^4\right)$
$\displaystyle{V=\int_{0}^{3}A(y)\ dy}\\
\displaystyle{V=\int_{0}^{3}\pi\left(4y^3+12y-7y^2-y^4\right)\ dy}\\
\displaystyle{V=\pi\int_{0}^{3}4y^3+12y-7y^2-y^4\ dy}\\
\displaystyle{V=\pi\left[y^4+6y^2-\frac{7}{3}y^3-\frac{1}{5}y^5\right]_{0}^{3}}\\
\displaystyle{V=\pi\left(\left((3)^4+6(3)^2-\frac{7}{3}(3)^3-\frac{1}{5}(3)^5\right)-\left(0\right)\right)}\\
\displaystyle{V=\frac{117\pi}{5}}$
