Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 304: 9



Work Step by Step

The volume is given by $$ \pi \int_{1}^{3}\left(\frac{2}{x+1}\right)^{2} d x=4 \pi \int_{1}^{3}(x+1)^{-2} d x=-\left.4 \pi(x+1)^{-1}\right|_{1} ^{3}=\pi. $$
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