Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 671: 21

Answer

$\dfrac{19}{3}$

Work Step by Step

The length of the curve can be calculated as: $L=\int_{p}^{q}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$ Here, we have $L=\int_{0}^{\sqrt 5} \sqrt{(\theta^4+(2\theta))^2} d \theta=\dfrac{1}{2}\int_{0}^{\sqrt 5} \sqrt {\theta^2+4}(2\theta d\theta)$ Plug $\theta^2+4 =k $ or, $2\theta d\theta= dk$ $L =\dfrac{1}{2}\int_{4}^{9}(k^{1/2}) dk=(\dfrac{1}{2})[(\dfrac{2}{3}) (k^{3/2})]_{4}^{9} $ $\implies L=\dfrac{27-8}{3}=\dfrac{19}{3}$
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