Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 35

$\frac{2 \pi}{3}$

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