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