Answer
$V=\int_{-\pi / 3}^{\pi / 3} 2 \pi\left(\frac{\pi}{2}-x\right)\left(\cos ^2 x-\frac{1}{4}\right) d x$
Work Step by Step
Given
$$y=\cos ^2 x, \quad|x| \leqslant \pi / 2, y=\frac{1}{4} ; \quad \text { about } x=\pi / 2$$
First, we find the intersection points
\begin{aligned}
\cos ^2 x&=\frac{1}{4}\\
\cos x &= \pm \frac{1}{2},\ \ \ |x|\leq \pi/2
\end{aligned}
Then
$$ x=\pm \frac{\pi}{3}$$
To find the volume of the solid when the bounded region rotates about $x=\pi/2$, we use the method of shell
\begin{aligned}
V&= 2\pi \int_a^b r(x)h(x)dx
\end{aligned}
Here
$$r(x)= \frac{\pi}{2}-x,\ \ h(x) =\cos ^2 x-\frac{1}{4}$$
Then
\begin{aligned}
V&= 2\pi \int_a^b r(x)h(x)dx\\
&= \int_{-\pi / 3}^{\pi / 3} 2 \pi\left(\frac{\pi}{2}-x\right)\left(\cos ^2 x-\frac{1}{4}\right) d x
\end{aligned}
