University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 39


$$\dfrac{2 \pi}{3}$$

Work Step by Step

We need to integrate the integral to compute the volume. We have: Area $=R^2\pi- r^2 \pi=\pi (1-x^2)$ $$Volume = \pi \times \int_{0}^{1} (1-x^2) \space dx \\= \pi [x-\dfrac{x^3}{3}]_{0}^{1} \\= \pi (1-\dfrac{1}{3} -0) \\=\dfrac{2 \pi}{3}$$
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