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: 24



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, $L=\int_{0}^{\pi} \sqrt{a^2(\dfrac{1-\cos \theta}{2})^2+(\dfrac{asin \theta}{2})^2} d \theta$ Then, we have $L=\dfrac{a}{2} \int_{0}^{\pi} \sqrt {2 (1-\cos \theta)} d\theta$ or, $L =a [2\cos (\theta/2)]_{0}^{\pi}$ Thus, $L=2a$
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.