Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.3 Conservative Vector Fields - Exercises - Page 945: 3

Answer

We verify that ${\bf{F}} = \nabla f$. $\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{9}{4}$

Work Step by Step

1. We have ${\bf{F}}\left( {x,y} \right) = \left( {3,6y} \right)$ and $f\left( {x,y} \right) = 3x + 3{y^2}$. Evaluate $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) = \left( {3,6y} \right)$ Since ${\bf{F}}\left( {x,y} \right) = \left( {3,6y} \right)$, so ${\bf{F}} = \nabla f$. 2. Evaluate the line integral of ${\bf{F}}$ over ${\bf{r}}\left( t \right) = \left( {t,2{t^{ - 1}}} \right)$ for $1 \le t \le 4$. Since ${\bf{F}} = \nabla f$, by definition ${\bf{F}}$ is conservative. By Theorem 1, the line integral of ${\bf{F}}$ over the given path is $\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{\bf{r}}\left( 4 \right)} \right) - f\left( {{\bf{r}}\left( 1 \right)} \right) = f\left( {4,\frac{1}{2}} \right) - f\left( {1,2} \right) = 12 + \frac{3}{4} - \left( {3 + 12} \right) = - \frac{9}{4}$
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