Answer
(a) $$\dfrac{\pi}{6}$$
and
(b) $$\dfrac{\pi}{6}$$
Work Step by Step
a) $$Volume = (2 \pi) \int_{0}^{1} (x) (x-x^2) \space dx \\=\dfrac{-x^3 \pi(3x-4)}{6} \\=\dfrac{\pi}{6}$$
b) Consider the shell model to compute the volume:
$$Volume=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\
\\= \int_{0}^{1} (2 \pi) \cdot (1-x) (x-x^2) \\=\dfrac{\pi}{6}$$
