Answer
$\frac{64}{15} \pi$
Work Step by Step
Given
$$ y=x^2,\ \ \ \ y= 2x$$
First, we find the intersection points
\begin{aligned}
x^2&=2x\\
x^2-2x&=0\\
x(x-2)&=0
\end{aligned}
Then
$$x=0,\ \ x=2 $$
Since
$$2x\geq x^2,\ \ \ \ 0\leq x\leq 2 $$
Then the outer radius $R= 2x$ and the inner $r= x^2$
Hence, by using washer method, we have
\begin{aligned}
V &=\pi\int_a^b[R^2-r^2]dx\\
& =\pi \int_0^2\left[(2 x)^2-\left(x^2\right)^2\right] d x\\
&=\pi \int_0^2\left(4 x^2-x^4\right) d x \\
& =\pi\left[\frac{4}{3} x^3-\frac{1}{5} x^5\right]_0^2\\
&=\pi\left(\frac{32}{3}-\frac{32}{5}\right) \\
& =32 \pi \cdot \frac{2}{15}\\
&=\frac{64}{15} \pi
\end{aligned}
