## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 17

#### Answer

$$6$$

#### Work Step by Step

By implicit differentiation \begin{aligned} &\frac{2}{3} x^{-1 / 3}+\frac{2}{3} y^{-1 / 3} y^{\prime}=0\\ &y^{\prime}=-\frac{x^{-1 / 3}}{y^{-1 / 3}}=-\frac{y^{1 / 3}}{x^{1 / 3}} \end{aligned} and \begin{aligned} s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x &=\int_{0}^{1} \sqrt{\frac{1}{x^{2 / 3}}} d x \\ &=\int_{0}^{1} \frac{1}{x^{1 / 3}} d x \\ &=\left[\frac{3}{2} x^{2 / 3}\right]_{0}^{1} \\ &=\frac{3}{2} \end{aligned} The total arc length is therefore $4 · \frac{3}{2} = 6.$

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