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: 15


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

Work Step by Step

We need to integrate the integral to compute the volume. $$Volume= (\pi) \times \int_{0}^{2} (1-x+\dfrac{x^2}{4}) dx \\ = (\pi x-\dfrac{\pi x^2}{2}+\dfrac{\pi x^3}{12})]_{0}^{2} \\=\pi (2-2+\dfrac{2}{3}) \\=\dfrac{2 \pi}{3}$$
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