Answer
$$\dfrac{7\pi}{6}$$
Work Step by Step
We need to integrate the integral to compute the volume.
We have:
$$Volume = \int_{0}^{1} (3-2\sqrt y-y) \space dy \times \pi \\= \pi [3y-\dfrac{4 y^{3/2}}{3}-\dfrac{y^2}{2}]_{0}^{1} \\= \pi (\dfrac{18-8-3}{3}) \\=\dfrac{7\pi}{6}$$
