Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 45

$\dfrac{7\pi}{6}$

Area $=R^2\pi- r^2 \pi=\pi (2^2-(1+\sqrt y)^2)=\pi (3-2\sqrt y-y)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{1} (3-2\sqrt y-y) dy$ Now, $V= \pi [3y-\dfrac{4 y^{3/2}}{3}-\dfrac{y^2}{2}]_{0}^{1}$ or, $= \pi (\dfrac{18-8-3}{3})$ or, $=\dfrac{7\pi}{6}$