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: 6

Answer

We verify that ${\bf{F}} = \nabla f$. $\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$

Work Step by Step

1. We have ${\bf{F}}\left( {x,y,z} \right) = \frac{z}{x}{\bf{i}} + {\bf{j}} + \ln x{\bf{k}}$ and $f\left( {x,y,z} \right) = y + z\ln x$. Evaluate $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {\frac{z}{x},1,\ln x} \right)$ Since ${\bf{F}}\left( {x,y,z} \right) = \frac{z}{x}{\bf{i}} + {\bf{j}} + \ln x{\bf{k}}$, so ${\bf{F}} = \nabla f$. 2. Since ${\bf{F}} = \nabla f$, so ${\bf{F}}$ is conservative. By Theorem 1, the circulation of ${\bf{F}}$ around a closed circle is zero: $\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
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