Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.2 Line Integrals - Exercises - Page 732: 10

Answer

$\mathop \smallint \limits_C^{} \frac{{{y^3}}}{{{x^7}}}{\rm{d}}s = \frac{1}{{576}}\left( {65\sqrt {65} - 2\sqrt 2 } \right)$

Work Step by Step

We have $f\left( {x,y} \right) = \frac{{{y^3}}}{{{x^7}}}$. The curve $y = \frac{1}{4}{x^4}$ for $1 \le x \le 2$ can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,\frac{1}{4}{t^4}} \right)$ for $1 \le t \le 2$. So, $f\left( {{\bf{r}}\left( t \right)} \right) = \frac{1}{{64}}{t^{12}}/{t^7} = \frac{1}{{64}}{t^5}$ $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,{t^3}} \right)\cdot\left( {1,{t^3}} \right)} = \sqrt {1 + {t^6}} $ By Eq. (4): $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $\mathop \smallint \limits_C^{} \frac{{{y^3}}}{{{x^7}}}{\rm{d}}s = \frac{1}{{64}}\mathop \smallint \limits_1^2 {t^5}\sqrt {1 + {t^6}} {\rm{d}}t$ Write $u = 1 + {t^6}$. So, $du = 6{t^5}dt$. Thus, the integral becomes $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \frac{1}{{64}}\cdot\frac{1}{6}\mathop \smallint \limits_2^{65} {u^{1/2}}{\rm{d}}u$ $ = \frac{1}{{384}}\left( {\left( {\frac{2}{3}{u^{3/2}}} \right)|_2^{65}} \right)$ $ = \frac{1}{{384}}\left( {\frac{2}{3}\cdot65\sqrt {65} - \frac{2}{3}\cdot2\sqrt 2 } \right)$ $ = \frac{1}{{576}}\left( {65\sqrt {65} - 2\sqrt 2 } \right)$ So, $\mathop \smallint \limits_C^{} \frac{{{y^3}}}{{{x^7}}}{\rm{d}}s = \frac{1}{{576}}\left( {65\sqrt {65} - 2\sqrt 2 } \right)$.
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