University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 594: 52


$\sqrt 2 \pi$

Work Step by Step

Given: $r=2 \sin \theta+2\cos \theta$ $r'=2 \cos \theta-2 \sin \theta$ The length is given as follows: $L= \int_{0}^{\pi/2} \sqrt{8(\cos^2 \theta+\sin^2 \theta)} d\theta$ Then, $L=[2\sqrt 2 \theta]_{0}^{\pi/2} $ After solving, we get $L=\sqrt 2 \pi$
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