Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 31

Answer

$\frac{16\pi}{35}$

Work Step by Step

Step 1. Draw a diagram as shown in the figure. The limits of integration with respect to $y$ can be found from $0$ to $1$. Step 2. Using the washer approximation, we have $dV=\pi (x_2^2-x_1^2)dy=\pi (y^{2/3}-y^6)dy$ Step 3. We have $V=\int_0^1 \pi (y^{2/3}-y^6)dy=\pi (\frac{3}{5}y^{5/3}-\frac{1}{7}y^7)|_0^1=\frac{16\pi}{35}$
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