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