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 305: 47


$\dfrac{16 \pi}{35} $

Work Step by Step

The volume of a revolution can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=y^{1/3} $ and $ R_{inside}=y^3$ Now, $V=\pi \int_0^1 (y^{1/3})^2 -(y^3)^2 \ dy =\pi \int_0^1 (y^{7/3} -y^6) \ dy =\pi [\dfrac{3 y^{5/3}}{5}-\dfrac{y^7}{7}]_0^1 \\=\dfrac{16 \pi}{35} $
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