Answer
The integral describes the volume of solid obtained by rotating the region
bounded by $f(x)= \sqrt{\sin (x)},\ \ y=0, \ x=0,\ \ x=\pi $ about the $x$-axis.
Work Step by Step
Given
$$ \pi \int_0^\pi \sin x d x$$
Compare the integral with
$$V= \int_a^b f^2(x)dx $$
\begin{aligned}
\pi \int_0^\pi \sin x d x&=\pi \int_0^\pi(\sqrt{\sin x})^2 d x
\end{aligned}
The integral describes the volume of solid obtained by rotating the region
bounded by $f(x)= \sqrt{\sin (x)},\ \ y=0, \ x=0,\ \ x=\pi $ about the $x$-axis.
