Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - 8.2 Volume and Average Value - 8.2 Exercises - Page 439: 17

Answer

$V=\frac{16\pi}{15} $

Work Step by Step

If y=0, then x=1, x=-1, so that a=1, b=-1. The volume is $V=\int^{1}_{-1}\pi(1-x^{2})^{2}dx = \int^{1}_{-1}\pi(1-2x^{2}+x^{4})dx = \pi(x-\frac{2x^{3}}{3}+\frac{x^{5}}{5})|^{1}_{-1} =\frac{16\pi}{15} $
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