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: 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, $L=\dfrac{\pi+3}{8}$
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