Answer
$\dfrac{16 \pi (3\sqrt 2+5)}{15})$
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 (\sqrt y-(-y)) dy$
Now, $V=2 \pi (\dfrac{2y^2}{5}\sqrt y+\dfrac{y^2}{3})_{0}^{2}$
or, $=16 \pi \times (\dfrac{3\sqrt 2+5}{15})$
or, $=\dfrac{16 \pi (3\sqrt 2+5)}{15})$
