Answer
$\frac{\pi ^3}{8}-\frac{\pi }{2}$
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 y-axis
\begin{aligned}
V & =2\pi\int_0^{\pi / 2} xf(x) d x\\
&=2\pi \int_0^{\pi / 2}x\cos ^2 x d x
\end{aligned}
Integrate by parts
\begin{aligned}
u&= x\ \ \ \ \ \ \ \ \ & dv&= \cos^2(x)= \frac{1}{2}(1+\cos(2x))\\
du&=d x\ \ \ \ \ \ \ \ \ & v&=x+\frac{1}{2}\sin \left(2x\right)
\end{aligned}
Then
\begin{aligned}
V & =2\pi\int_0^{\pi / 2} xf(x) d x\\
&=2\pi \int_0^{\pi / 2}x\cos ^2 x d x \\
&= 2\pi \left( \frac{1}{2}x\left(x+\frac{1}{2}\sin \left(2x\right)\right)\bigg|_0^{\pi / 2}-\int _0^{\pi / 2}\frac{1}{2}\left(x+\frac{1}{2}\sin \left(2x\right)\right)dx\right)\\
&= 2\pi \left(\frac{1}{2}x\left(x+\frac{1}{2}\sin \left(2x\right)\right)-\frac{1}{2}\left(\frac{x^2}{2}-\frac{1}{4}\cos \left(2x\right)\right)\right)\bigg|_0^{\pi / 2}\\
&= 2\pi \left(\frac{\pi ^2-2}{16}-\frac{1}{8}\right)\\
&=\frac{\pi ^3}{8}-\frac{\pi }{2}
\end{aligned}
