Answer
$\displaystyle{V=\frac{3\pi^2}{8}}$
Work Step by Step
$\displaystyle{A(x)=\pi\left(\sin^2x\right)^2}\\$
$\displaystyle{V=\int_0^\pi A(x)\ dx}\\
\displaystyle{V=\int_0^\pi \pi\left(\sin^2x\right)^2\ dx}\\
\displaystyle{V=\pi\int_0^\pi \left(\sin^2x\right)^2 \ dx}\\
\displaystyle{V=\pi\int_0^\pi \left(\frac{1-\cos 2 x}{2}\right)^{2} dx}\\
\displaystyle{V=\frac{\pi}{4} \int_0^\pi \left(1-2 \cos 2 x+\cos ^{2} 2 x\right) dx}\\
\displaystyle{V=\frac{\pi}{4} \int_0^\pi1-2 \cos 2 x+\frac{1}{2}\left(1+\cos 4x\right)\ dx}\\
\displaystyle{V=\frac{\pi}{4} \int_0^\pi\frac{1}{2}\cos 4x-2 \cos 2 x+\frac{3}{2}\ dx}\\
\displaystyle{V=\frac{\pi}{4} \left[\frac{1}{8}\sin 4x-\sin 2 x+\frac{3}{2}x\right]_0^\pi}\\
\displaystyle{V=\frac{3\pi^2}{8}}$
