Answer
The volume of the resulting solid is given by the following:
$$
\int_{0}^{\pi} \pi\left(\sin ^{2} x\right)^{2} d x=\frac{3}{8} \pi^{2}
$$
Work Step by Step
Given the curve
$$
y=\sin ^{2} x
$$
The volume, when the region under the curve $ y$ from $0$ to $\pi$ is rotated
about the x-axis, is given by the following:
$$
V =\int_{0}^{\pi} \pi\left(y\right)^{2} d x
$$
Use disks about the $x$ -axis:
$$
\begin{aligned}
V &=\int_{0}^{\pi} \pi\left(\sin ^{2} x\right)^{2} d x\\
&=\pi \int_{0}^{\pi} \sin ^{4} x d x \quad\left[\begin{array}{c}
\text {we look at the Table of Integrals ,we see that } \\
\text { the closest entry is number $73$ with n=4}:
\end{array}\right] \\
&\stackrel{73}{=} \pi\left\{\left[-\frac{1}{4} \sin ^{3} x \cos x\right]_{0}^{\pi}+\frac{3}{4} \int_{0}^{\pi} \sin ^{2} x d x\right\} \\
& \quad\quad\quad\quad\quad\quad\quad\quad \quad\left[\begin{array}{c}
\text {we look at the Table of Integrals ,we see that } \\
\text { the closest entry is number $63$ }:
\end{array}\right] \\
& \stackrel{63}{=} \pi\left\{0+\frac{3}{4}\left[\frac{1}{2} x-\frac{1}{4} \sin 2 x\right]_{0}^{\pi}\right\}=\pi\left[\frac{3}{4}\left(\frac{1}{2} \pi-0\right)\right] \\
&=\frac{3}{8} \pi^{2}
\end{aligned}
$$
