Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.2 Area of a Surface of Revolution - 8.2 Exercises: 33

Answer

$\int^{b}_{a} 2\pi[c-f(x)] \sqrt{1+[f'(x)]^{2}}dx$

Work Step by Step

The analogue of $f(x_{i}^{*})$in the derivation of (4) is now $c - f(x_{i}^{*})$, so $S = \lim_{n-> \infty} \Sigma^{n}_{i=1} 2\pi [c-f(x_{i}^{*})] \sqrt{1+[f'(x_{i}^{*})]^{2}}\Delta x = \int^{b}_{a} 2\pi[c-f(x)] \sqrt{1+[f'(x)]^{2}}dx$
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