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 468: 20



Work Step by Step

Since \begin{aligned} 1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{(x-3)(x-1)}{6 y}\right)^{2} \\ &=1+\frac{(x-3)^{2}(x-1)^{2}}{36 y^{2}} \\ &=1+\frac{(x-3)^{2}(x-1)^{2}}{4\left(9 y^{2}\right)}\\ &= 1+\frac{(x-1)^{2}}{4 x} \end{aligned} Then \begin{aligned} s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{0}^{3} \sqrt{1+\frac{(x-1)^{2}}{4 x}} d x \\ &=\frac{1}{2} \int_{0}^{3} \sqrt{4+\frac{(x-1)^{2}}{x}} d x \\ &=\frac{1}{2} \int_{0}^{3} \frac{x+1}{\sqrt{x}} d x \\ &=\frac{1}{2} \int_{0}^{3} \sqrt{x}+\frac{1}{\sqrt{x}} d x \\ &=\frac{1}{2}\left[\frac{2}{3} x^{3 / 2}+2 \sqrt{x}\right]_{0}^{3}\\ &=2\sqrt{3} \end{aligned}
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