Answer
(a) $$\dfrac{4 \pi}{15}$$
b)$$\dfrac{7\pi}{30}$$
Work Step by Step
(a) $Volume= \int_{0}^{1} (2 \pi) (y) (y-y^3) dy=\dfrac{-2y^3 \pi(3y^2-5)}{15}=\dfrac{4 \pi}{15}$
b) Consider the shell model to compute the volume:
$$Volume=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\= 2 \pi \times \int_{0}^{1} (1-y) (y-y^3) dy \\=\dfrac{(2y^2 \pi) [12y^3-15y^2-20y+30]}{30} \\=\dfrac{7\pi}{30}$$
