Calculus 8th Edition

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

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 2

Answer

$\sqrt{2} \frac{\pi}{4}$

Work Step by Step

$y = \sqrt{2-x^2}$ then $\frac{dy}{dx} = -\frac{x}{\sqrt{2-x^2}}$ $L = \int^{1}_{0} \sqrt{1+(\frac{dy}{dx})^2}dx = \int^{1}_{0}\sqrt{1+\frac{x^2}{2-x^2}}dx = \int^{1}_{0} \frac{\sqrt{2}dx}{\sqrt{2-x^2}} = \sqrt{2} \int^{1}_{0}\frac{dx}{\sqrt{(\sqrt{2})^2 -x^2}}$ $ = \sqrt{2}[\sin^{-1}(\frac{x}{\sqrt{2}})]^{1}_{0} = \sqrt{2}[\sin^{-1}(\frac{1}{\sqrt{2}})-\sin^{-1}(0)] = \sqrt{2}[\frac{\pi}{4} - 0] = \sqrt{2} \frac{\pi}{4}$
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