Answer
$V= \pi \int_0^1\left(x^4-4x^2-x+4\sqrt{x}\right)dx$
Work Step by Step
Given
$$ y=\sqrt{x}, \quad y=x^2 ; \quad \text { about } y=2$$
First, we find the intersection points
\begin{aligned}
x^2&=\sqrt{x}\\
x^4-x&=0\\
x(x^3-1)&=0
\end{aligned}
Then
$$ x=0,\ \ \ x=1 $$
The volume of the solid using washer method is given by
\begin{aligned}
V&= \pi \int_a^b [R^2(x)-r^2(x)]dx
\end{aligned}
Here $$R=2- x^2,\ \ \ r= 2- \sqrt{x} $$
Hence
\begin{aligned}
V&= \pi \int_a^b [R^2(x)-r^2(x)]dx\\
&=\pi \int_0^1 [(2- x^2)^2-(2- \sqrt{x})^2]dx\\
&= \pi \int_0^1\left(x^4-4x^2-x+4\sqrt{x}\right)dx
\end{aligned}
