Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 44

$\pi \sqrt 3$

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