Answer
$\displaystyle{V=\frac{2 \pi}{3} }$
Work Step by Step
$\displaystyle{A(x)=\pi(1-0)^{2}-\pi\left(1-x\right)^{2}}\\
\displaystyle{A(x)=\pi\left(2x-x^2\right)}$
$\begin{aligned} V &=\int_{0}^{1} A(x) \ dx \\ V &=\int_{0}^{1} \pi\left(2x-x^2\right) \ d x \\ V &=\pi \int_{0}^{1} 2x-x^2 \ dx \\ V &=\pi\left[x^2-\frac{1}{3} x^{3}\right]_{0}^{1} \\ V &=\pi\left(\left((1)^2-\frac{1}{3} (1)^{3}\right)-\left(0^2-\frac{1}{3} 0^{3}\right)\right)\\ V &=\frac{2 \pi}{3} \end{aligned}$
