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 306: 58


$\dfrac{(2 a^3+6a ) \pi}{3}$

Work Step by Step

The volume of the hyperboloid can be calculated as: Now, $V=\pi \int_{-a}^{a} (\sqrt {1+x^2})^2 \ dx \\ =\pi \int_{-a}^{a} (1+x^{2}) \ dx \\=\pi (x+\dfrac{1}{3} x^3)_{-a}^a \ dx \\=\pi [a+\dfrac{a^3}{3}+a+\dfrac{a^{3}}{3}] \\=\dfrac{(2 a^3+6a ) \pi}{3}$
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