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: 43

Answer

$8\pi$

Work Step by Step

Step 1. Draw a diagram as shown in the figure. The limits of integration with respect to $x$ can be found as from $0$ to $2$. Step 2. Approximate using cylinders; we have $dV=2\pi x(x^2-0) dx=2\pi x^3\ dx$ Step 3. We have $V=\int_0^2 2\pi x^3\ dx=\frac{\pi}{2}x^4|_0^2=8\pi$
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