Answer
$\displaystyle \frac{3\pi^{2}}{8}$
Work Step by Step
Disks about the $x$-axis: $V=\displaystyle \int_{0}^{\pi}\pi(\sin^{2}x)^{2}dx$
$V=\displaystyle \pi\int_{0}^{\pi}\sin^{4}xdx$
Table of integrals:
$\color{blue}{73. \displaystyle \quad\int\sin^{n}udu=-\frac{1}{n}\sin^{n-1}u\cos u+\frac{n-1}{n}\int\sin^{\mathrm{n}-2}udu }$
$V=[-\displaystyle \frac{\pi}{4}\sin^{3}x\cos x]_{0}^{\pi}+\frac{3\pi}{4}\int_{0}^{\pi}\sin^{2}xdx$
The first term evaluates 0 for both $\pi$ and 0.
$=\displaystyle \frac{3\pi}{4}\int_{0}^{\pi}\sin^{2}xdx$
Table of integrals:
$\color{blue}{63. \quad \displaystyle \int\sin^{2}udu=\frac{1}{2}u-\frac{1}{4}\sin 2u+C }$
$=\displaystyle \frac{3\pi}{4}[\frac{1}{2}x-\frac{\pi}{4}\sin 2x]_{0}^{\pi}$
$=\displaystyle \frac{3\pi}{4}(\frac{1}{2}\pi-0)$
$=\displaystyle \frac{3\pi^{2}}{8}$
