Answer
a) $\dfrac{4 \pi}{15}$
and
b ) $\dfrac{7\pi}{30}$
Work Step by Step
a) $V= 2 \pi \int_{0}^{1} (y) \cdot (y-y^3) dy=\dfrac{-2y^3 \pi(3y^2-5)}{15}=\dfrac{4 \pi}{15}$
b) We need to use the shell model as follows:
$V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx \\= 2 \pi \int_{0}^{1} (1-y)\cdot (y-y^3) dy=\dfrac{2y^2 \pi(12y^3-15y^2-20y+30)}{30}=\dfrac{7\pi}{30}$
