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 933: 38

Answer

$\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 44$

Work Step by Step

First, we find $\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$, where ${C_2}$ is the path in counterclockwise direction as is shown in Figure 16. We have $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 5 = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_3}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ Since $\mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 8$, $\mathop \smallint \limits_{{C_3}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 8$, so $5 = 8 + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + 8$ $\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 11$ Thus, the value of $\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$, where $C'$ is the path that traverses the loop ${C_2}$ four times in the clockwise direction is $\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\cdot\mathop \smallint \limits_{ - {C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\cdot\left( { - \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}} \right) = 4\cdot11 = 44$ So, $\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 44$.
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