Calculus (3rd Edition)

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

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 24


$$2\pi r $$

Work Step by Step

Since \begin{aligned} x^{2}+y^{2}&=r^{2} \\ y&=\sqrt{r^{2}-x^{2}} \\ y^{\prime} &=-\frac{x}{\sqrt{r^{2}-x^{2}}} \end{aligned} Then \begin{align*} \text{Circumference} &=2 \int_{-r}^{r} \sqrt{1+\left(-\frac{x}{\sqrt{r^{2}-x^{2}}}\right)^{2}} d x\\ &=2 \int_{-r}^{r} \sqrt{1+\frac{x^{2}}{r^{2}-x^{2}}} d x \\ &=2 \int_{-r}^{r} \sqrt{\frac{x^{2}+r^{2}-x^{2}}{r^{2}-x^{2}}} d x\\ &=2 \int_{-r}^{r} \frac{r}{\sqrt{r^{2}-x^{2}}} d x\\ &=2 r \sin^{-1}\left(\frac{x}{r}\right) \bigg|_{-r}^{r}\\ &=2r\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\ &=2\pi r \end{align*}
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