Answer
a) $\dfrac{11\pi}{48}$ and b) $\dfrac{11\pi}{48}$
Work Step by Step
a) $V= \pi \int_{1}^{2} (y^{-2})^2)-(1/4)^2 dy=\dfrac{- \pi(3y^4+16)}{48y^3}=\dfrac{11\pi}{48}$
b) We need to use the shell model as follows:
$V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx \\= 2\pi \int_{1/4}^{1} (x) \cdot (x^{-1/2})^2)-1) dy=\dfrac{- x^{3/2}\pi(3\sqrt x-4)}{48y^3}=\dfrac{11\pi}{48}$
