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 355: 10

Answer

$$\dfrac{8}{3}$$

Work Step by Step

We need to integrate the integral to compute the volume. $$ Volume = (2) \int_{-1}^{1} (2) \times (1-y^2) dy \\ =[2y-\dfrac{2y^3}{3}]_{-1}^{1} \\= 2 [1-\dfrac{1}{3}-(-1) + (\dfrac{-1}{3})] \\=\dfrac{8}{3}$$
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