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: 22

Answer

$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$

Work Step by Step

We have ${\bf{F}}\left( {x,y} \right) = \left( {{{\rm{e}}^{y - x}},{{\rm{e}}^{2x}}} \right)$, and the piecewise linear path from $\left( {1,1} \right)$ to $\left( {2,2} \right)$ to $\left( {0,2} \right)$. The path consists of two line segments ${C_1}$ and ${C_2}$: 1. The line from $\left( {1,1} \right)$ to $\left( {2,2} \right)$ is represented by ${{\bf{r}}_1}$. It can parametrized by ${{\bf{r}}_1}\left( t \right) = \left( {1,1} \right) + t\left( {1,1} \right) = \left( {1 + t,1 + t} \right)$ for $0 \le t \le 1$. So, ${{\bf{r}}_1}'\left( t \right) = \left( {1,1} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right) = \left( {1,{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)$. 2. The line from $\left( {2,2} \right)$ to $\left( {0,2} \right)$ is represented by ${{\bf{r}}_2}$. It can parametrized by ${{\bf{r}}_2}\left( t \right) = \left( {2,2} \right) + t\left( {\left( {0,2} \right) - \left( {2,2} \right)} \right)$ $ = \left( {2,2} \right) + t\left( { - 2,0} \right)$ $ = \left( {2 - 2t,2} \right)$, for $0 \le t \le 1$. So, ${{\bf{r}}_2}'\left( t \right) = \left( { - 2,0} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right) = \left( {{{\rm{e}}^{2t}},{{\rm{e}}^{2\left( {2 - 2t} \right)}}} \right)$. Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8): $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ $ = \mathop \smallint \limits_a^b {\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right)\cdot{{\bf{r}}_1}'\left( t \right){\rm{d}}t + \mathop \smallint \limits_a^b {\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right)\cdot{{\bf{r}}_2}'\left( t \right){\rm{d}}t$ $ = \mathop \smallint \limits_0^1 \left( {1,{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)\cdot\left( {1,1} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{2t}},{{\rm{e}}^{2\left( {2 - 2t} \right)}}} \right)\cdot\left( { - 2,0} \right){\rm{d}}t$ $ = \mathop \smallint \limits_0^1 \left( {1 + {{\rm{e}}^{2\left( {1 + t} \right)}}} \right){\rm{d}}t - 2\mathop \smallint \limits_0^1 {{\rm{e}}^{2t}}{\rm{d}}t$ $ = \left( {t + \frac{1}{2}{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)|_0^1 - \left( {{{\rm{e}}^{2t}}|_0^1} \right)$ $ = 1 + \frac{1}{2}{{\rm{e}}^4} - \frac{1}{2}{{\rm{e}}^2} - {{\rm{e}}^2} + 1$ $ = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$ So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$.
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