Answer
$\frac{3}{16} \pi^2$
Work Step by Step
Given
$$y= \cos^{2} x\ \ \ \ [0,\pi/2] $$
To find the volume of the solid when the bounded area rotates about x-axis
\begin{aligned}
V & =\int_0^{\pi/2} \pi[f(x)]^2 d x\\
&=\pi \int_0^{\pi/2}\left(\cos ^2 x\right)^2 d x\\
&=\pi \int_0^{\pi/2}\left[\frac{1}{2}(1+\cos 2 x)\right]^2 d x \\
& =\frac{\pi}{4} \int_0^{\pi / 2}\left(1+\cos ^2 2 x+2 \cos 2 x\right) d x\\
&=\frac{\pi}{4} \int_0^{\pi / 2}\left[1+\frac{1}{2}(1+\cos 4 x)+2 \cos 2 x\right] d x \\
& =\frac{\pi}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{1}{4} \sin 4 x\right)+2\left(\frac{1}{2} \sin 2 x\right)\right]_0^{\pi / 2}\\
&=\frac{\pi}{4}\left[\left(\frac{3 \pi}{4}+\frac{1}{8} \cdot 0+0\right)-0\right]\\
&=\frac{3}{16} \pi^2
\end{aligned}
