Answer
$$\frac{3\pi}{10}$$
Work Step by Step
Given $$x=y^2,\ \ x=\sqrt{y}, $$
We first find the intersection points
\begin{align*}
y^2&=\sqrt{y}\\
y^2-\sqrt{y}&=0\\
\sqrt{y}(y^{3/2}-1)&=0
\end{align*}
Then $y=0,\ 1$
Then the volume of the solid obtained by rotating the region enclosed by the graph of
$x=y^2,\ \ x=\sqrt{y}$ about the $y-$axis is given by
\begin{align*}
V&=\pi\int_{0}^{1}[f(y)]^2-[g(y)]^2dy\\
&= \pi \int_{0}^{1} [ y-y^4]dy\\
&=\pi \left( \frac{1}{2}y^2- \frac{1}{5}y^5\right)\bigg|_{0}^{1}\\
&=\frac{3\pi}{10}
\end{align*}
We use the Mathematica program to plot the solid.
