Answer
$$2\pi$$
Work Step by Step
Given $$x=\sqrt{\sin y},\ \ x=0,\ \ \ y\in [0,\pi ] $$
Then the volume of the solid obtained by rotating the region enclosed by the graph of
$ x=\sqrt{y}$ about the $y-$axis is given by
\begin{align*}
V&=\pi\int_{0}^{\pi}[f(y)]^2dy\\
&= \pi \int_{0}^{\pi}\sin ydy\\
&=-\pi \cos y\bigg|_{0}^{\pi}\\
&=2\pi
\end{align*}
We use the Mathematica program to plot the solid.
