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: 10


$\frac{252 \pi}{5}$

Work Step by Step

The volume is given $$ \pi \int_{1}^{3}(\sqrt{x^{4}+1})^{2} d x=\pi \int_{1}^{3}\left(x^{4}+1\right) d x=\left.\pi\left(\frac{1}{5} x^{5}+x\right)\right|_{1} ^{3}=\frac{252 \pi}{5}. $$
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