Answer
$\frac{3\pi}{10}$
Work Step by Step
Given
$$y=x^2 ,\ \ \ y=\sqrt{x} $$
first, we find the intersection points
\begin{aligned}
x^2&=\sqrt{x}\\
x^3&=x\\
x^3-x&=0\\
x(x^2-1) &=0 \\
\end{aligned}
then $x=0,\ x=1,\ x=-1$ , we refuse $x=-1 $ .
Now, we need to find the volume of the generated solid when the region rotates about y-axis , using slicing
\begin{aligned}
V&= \int _{a}^{b}A(y)dy
\end{aligned}
where
\begin{aligned}
A(y)&= \pi [f^2(y)-g^2(y)]\\
&= \pi [(\sqrt{y})^2- (y^2)^2]
\end{aligned}
Hence
\begin{aligned}
V&=\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{aligned}
-----------------------------
Using cylindrical shells.
\begin{aligned} V&= 2\pi\int r(x)h(x)dx\end{aligned}
Here
\begin{aligned} r(x)& = x\\
h(x) &= \sqrt{x}-x^2\\
\end{aligned}
Then
\begin{aligned} V&= 2\pi\int r(x)h(x)dx\\
&=2\pi \int_0^2x(\sqrt{x}-x^2)dx\\
&= 2\pi \int_0^2(x^{3/2}-x^3)dx\\
&= 2\pi \left(\frac{2}{5} x^{5/2}-\frac{1 }{4}x^4\right)\bigg|_0^1\\
&= \frac{3\pi}{10}\end{aligned}
