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