Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.4 Area and Arc Length in Polar - Exercises - Page 625: 34


$$ \text { The length }= \int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{5-4\cos \theta} d \theta $$

Work Step by Step

Since $r=f(\theta)=(2-\cos \theta)^{-1}$, then $f'(\theta)=-(2-\cos \theta)^{-2}\sin\theta$ The length is given by $$ \text { The length }=\int_{0}^{2\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ =\int_{0}^{2\pi} \sqrt{(2-\cos \theta)^{-2}+(2-\cos \theta)^{-4}\sin^2\theta} d \theta\\ =\int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{(2-\cos \theta)^{2}+\sin^2\theta} d \theta\\ =\int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{5-4\cos \theta} d \theta $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.