Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.4 Area and Arc Length in Polar - Exercises - Page 625: 25


$\frac{1}{3}(\pi^2+4)^{3/2} -\frac{8}{3}.$

Work Step by Step

Since $r=f(\theta)= \theta^2$, then $f'(\theta)=2\theta$. The length is given by \begin{align*} \text {Length }&=\int_{0}^{\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ &=\int_{0}^{\pi}\sqrt{\theta^4+4\theta^2} d \theta\\ &=\int_{0}^{\pi}\theta\sqrt{\theta^2+4} d \theta \\ &=\frac{1}{2}\int_{0}^{\pi}2\theta(\theta^2+4)^{1/2} d \theta\\ &=\frac{2}{6}(\theta^2+4)^{3/2} |_{0}^{\pi}\\ &=\frac{1}{3}(\pi^2+4)^{3/2} -\frac{8}{3}. \end{align*}
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