Answer
$\dfrac{ 16\pi}{3}$
Work Step by Step
We need to use the shell model as follows:
$V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx$
$ \implies V= \int_{0}^{2} (2 \pi) \cdot y (2+y-y^2) dy$
Now, $V= 2\pi \times [y^2+\dfrac{y^3}{3}-\dfrac{y^4}{4}]_{0}^{2}$
or, $=2 \pi (4+\dfrac{8}{3}-4)$
or, $=\dfrac{ 16\pi}{3}$
