## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 27

#### Answer

$\dfrac{\pi+3}{8}$

#### 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, $L=\int_{0}^{\pi/4} \sqrt{cos^6 (\theta/3)+cos^4 (\theta/3) \sin^2 (\theta/3)} d \theta$ Then, we have $L=\int_{0}^{\pi/4} cos^2 (\theta/3) \theta$ or, $L=(1/2) [\theta+(3/2)sin(2\theta/3)]_{0}^{\pi/4}$ Thus, $L=\dfrac{\pi+3}{8}$

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.