Answer
$8 \pi$
Work Step by Step
Given
$$y=x^2, \quad 0 \leqslant x \leqslant 2, \quad y=4, \quad x=0$$
The volume of the solid when the region rotate about y -axis given by
\begin{aligned}
V&= \int_a^b 2\pi r(x) h(x)dx
\end{aligned}
where $r(x)$ is the radius of shell and $h(x)$ is the height.
Here
\begin{aligned}
r(x)&= x,\\
h(x)&= 4-x^2
\end{aligned}
Hence
\begin{aligned}
V &=\int_0^2 2 \pi x\left(4-x^2\right) d x\\
&=2 \pi \int_0^2\left(4 x-x^3\right) d x \\
&=2 \pi\left[2 x^2-\frac{1}{4} x^4\right]_0^2\\
&=2 \pi(8-4) \\
&=8 \pi
\end{aligned}