Answer
$\dfrac{ 5\pi}{6}$
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}^{1} (2 \pi) \cdot y (2-y-y^2) dy$
Now, $V= 2\pi \times [y^2-\dfrac{y^3}{3}-\dfrac{y^4}{4}]_{0}^{1}$
or, $=2 \pi (-\dfrac{1}{3}-\dfrac{1}{4})$
or, $=\dfrac{ 5\pi}{6}$
