University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 25


$3 \sqrt 2+3 \ln (\sqrt 2+1)$

Work Step by Step

The length of the curve is given as: $L=\int_{p}^{q}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$ Thus, $L=\int_{0}^{\pi/2} \sqrt{(\dfrac{6}{1+\cos \theta})^2+(\dfrac{6 \sin \theta}{(1+\cos \theta)^2})^2} d \theta$ Then, we have $L=(3) \int_{0}^{\pi/2}(\sec^3 \theta/2) d\theta$ or, $L=3 [\tan (\theta/2)\sec (\theta/2)+\ln |\sec (\theta/2)+\tan (\theta/2)]_0^{\pi}$ Thus, $L=3 \sqrt 2+3 \ln (\sqrt 2+1)$
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