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: 46


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

Work Step by Step

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