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: 21

Answer

$\dfrac{19}{3}$

Work Step by Step

The length of the curve is given as: $L=\int_{0}^{\sqrt 5}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$ Thus, $L=\int_{0}^{\sqrt 5} \sqrt{(\theta^4+(2\theta))^2} d \theta$ Then, we have $L=\dfrac{1}{2}\int_{0}^{\sqrt 5} \sqrt {\theta^2+4}(2\theta d\theta)$ Plug $\theta^2+4 =p \implies dp=2\theta d\theta$ Then,we get $L =\dfrac{1}{2}\int_{4}^{9}p^{1/2} dp=\dfrac{1}{2}[\dfrac{2}{3}p^{3/2}]_{4}^{9} $ Thus, $L=\dfrac{27-8}{3}=\dfrac{19}{3}$
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