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: 26


$$\frac{ A }{2} \sqrt{1+ A ^{2}}+\frac{1}{2} \ln ( A+\sqrt{1 + A^{2}})$$

Work Step by Step

Since $r= \theta $ , then $f'(\theta )= 1$ and hence: \begin{aligned} \text { Length } &=\int_{0}^{A} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta \\ &=\int_{0}^{A} \sqrt{1+ \theta^{2}} d \theta \end{aligned} Use the formula $$ \int \sqrt{a^{2}+u^{2}} d u=\frac{u}{2} \sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2} \ln (u+\sqrt{a^{2}+u^{2}})+C$$ Then \begin{aligned} \text { Length } &=\int_{0}^{A} \sqrt{1+ \theta^{2}} d \theta \\ &= \frac{ \theta }{2} \sqrt{1+ \theta ^{2}}+\frac{1}{2} \ln ( \theta +\sqrt{1 + \theta^{2}})\bigg|_{0}^{A}\\ &= \frac{ A }{2} \sqrt{1+ A ^{2}}+\frac{1}{2} \ln ( A+\sqrt{1 + A^{2}}) \end{aligned}
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