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 934: 61

Answer

We obtain the value of $\mathop \smallint \limits_{{C_R}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $, which does not depend on $R$.

Work Step by Step

We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$. The circle ${C_R}$ of radius $R$ centered at the origin can be parametrized by ${\bf{r}}\left( t \right) = \left( {R\cos t,R\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le 2\pi $ So, we obtain ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{ - \sin t}}{R},\frac{{\cos t}}{R}} \right)$ $d{\bf{r}} = {\bf{r}}'\left( t \right)dt$ $d{\bf{r}} = \left( { - R\sin t,R\cos t} \right)dt$ Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$: ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {\frac{{ - \sin t}}{R},\frac{{\cos t}}{R}} \right)\cdot\left( { - R\sin t,Rcost} \right)dt = dt$ Calculate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8) when ${C_R}$ is oriented counterclockwise: $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$ $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } {\rm{d}}t = 2\pi $ Thus, the value of $\mathop \smallint \limits_{{C_R}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $, does not depend on $R$.
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