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: 16

Answer

$\mathop \smallint \limits_C^{} \left( {6xz - 2{y^2}} \right){\rm{d}}s = \frac{{864}}{{35}}$

Work Step by Step

We have $f\left( {x,y,z} \right) = 6xz - 2{y^2}$ and the parametrization ${\bf{r}}\left( t \right) = \left( {t,\frac{{{t^2}}}{{\sqrt 2 }},\frac{{{t^3}}}{3}} \right)$ for $0 \le t \le 2$. So, $f\left( {{\bf{r}}\left( t \right)} \right) = 2{t^4} - {t^4} = {t^4}$ $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,\sqrt 2 t,{t^2}} \right)\cdot\left( {1,\sqrt 2 t,{t^2}} \right)} = \sqrt {1 + 2{t^2} + {t^4}} $ $ = \sqrt {{{\left( {{t^2} + 1} \right)}^2}} = {t^2} + 1$ By Eq. (4): $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$ $\mathop \smallint \limits_C^{} \left( {6xz - 2{y^2}} \right){\rm{d}}s = \mathop \smallint \limits_0^2 {t^4}\left( {{t^2} + 1} \right){\rm{d}}t = \mathop \smallint \limits_0^2 \left( {{t^6} + {t^4}} \right){\rm{d}}t$ $ = \left( {\frac{1}{7}{t^7} + \frac{1}{5}{t^5}} \right)|_0^2 = \frac{{128}}{7} + \frac{{32}}{5}$ So, $\mathop \smallint \limits_C^{} \left( {6xz - 2{y^2}} \right){\rm{d}}s = \frac{{864}}{{35}}$.
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