Answer
a) $\dfrac{\pi}{6}$ ; b) $\dfrac{\pi}{6}$
Work Step by Step
a) $V= 2 \pi \int_{0}^{1} (x) (x-x^2) dx=\dfrac{-x^3 \pi(3x-4)}{6}=\dfrac{\pi}{6}$
b) We need to use the shell model as follows:
$V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx \\= \int_{0}^{1} (2 \pi) \cdot (1-x) (x-x^2) \\=\dfrac{\pi}{6}$
