University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Practice Exercises - Page 594: 53



Work Step by Step

Given: $r=8 \sin^3 (\dfrac{\theta}{3})$ $r'=8 \sin^2 (\dfrac{\theta}{3}) \cos (\dfrac{\theta}{3})$ The length is given as follows: $L= \int_{0}^{\pi/4} \sqrt{r^2+r'^2} d\theta=\int_{0}^{\pi/4} \sqrt {[64\sin^{4} (\dfrac{\theta}{3})} d\theta$ Then, $L=[4 \theta-6\sin(\dfrac{2\theta}{3})]_{0}^{\pi/4} $ After solving, we get $L=\pi-3$
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