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: 4



Work Step by Step

Given $$ y= y=\left(\frac{x}{4} \right)^{3}+\frac{1}{2x^2}$$ Since \begin{aligned} 1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{x^3}{4} -\frac{1}{x^3}\right)^{2} \\ &=1+\frac{x^{6}}{16}-\frac{1}{2}+x^{-6} \\ &=\frac{x^{6}}{16}+\frac{1}{2}+x^{-6} \\ &=\left(\frac{1}{4} x^{3}+x^{-3}\right)^{2} \end{aligned} Then the arc length given by \begin{aligned} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{1}^{4} \sqrt{\left(\frac{1}{4} x^{3}+x^{-3}\right)^{2}} d x \\ &=\int_{1}^{4}\left(\frac{1}{4} x^{3}+x^{-3}\right) d x \\ &=\frac{1}{16} x^{4}-\frac{1}{2} x^{-2}\bigg|_{1}^{4}\\ &=\frac{525}{32} \end{aligned}
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