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: 23



Work Step by Step

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