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: 57


$\dfrac{32 \pi}{105}$

Work Step by Step

The volume of a region can be calculated as: Now, $V=\pi \int_{-1}^{1} [(1-x^{2/3})^{3/2}]^2 \ dx \\ =\pi \int_{-1}^1 (1-x^{2/3})^3 \ dx \\=\pi \int_{-1}^1 (-x^2+3x^{4/3}-3x^{2/3}+1) \ dx \\=\pi [-\dfrac{x^3}{3}+\dfrac{9x^{7/3}}{7}- \dfrac{9x^{5/3}}{5}+x]_{-1}^1 \\=\dfrac{32 \pi}{105}$
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