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: 9

Answer

$\mathop \smallint \limits_C^{} \sqrt {1 + 9xy} {\rm{d}}s = \frac{{14}}{5}$

Work Step by Step

We have $f\left( {x,y} \right) = \sqrt {1 + 9xy} $. The curve $y = {x^3}$ for $0 \le x \le 1$ can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,{t^3}} \right)$ for $0 \le t \le 1$. So, $f\left( {{\bf{r}}\left( t \right)} \right) = \sqrt {1 + 9{t^4}} $ $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,3{t^2}} \right)\cdot\left( {1,3{t^2}} \right)} = \sqrt {1 + 9{t^4}} $ 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_0^1 \left( {1 + 9{t^4}} \right){\rm{d}}t = \left( {t + \frac{9}{5}{t^5}} \right)|_0^1 = 1 + \frac{9}{5} = \frac{{14}}{5}$ So, $\mathop \smallint \limits_C^{} \sqrt {1 + 9xy} {\rm{d}}s = \frac{{14}}{5}$.
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