Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 21

$36 \pi$

Area $=\pi r^2=\pi (\sqrt{9-x^2})^2$ We integrate the integral to calculate the volume as follows: $V= (2) \times \int_{0}^{3} \pi (9-x^2) dx$ Now, $V=( 2\pi)\times [9x-\dfrac{x^3}{3}]_0^3$ or, $=(2 \pi)(27-9)$ or, $=36 \pi$