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: 35

Answer

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

Work Step by Step

We have ${\bf{F}}\left( {x,y,z} \right) = \left( {{{\rm{e}}^z},{{\rm{e}}^{x - y}},{{\rm{e}}^y}} \right)$ and the blue path as is shown in Figure 14. 1. Let the line segment ${C_1}$ from the origin $\left( {0,0,0} \right)$ to $\left( {0,0,1} \right)$ be represented by ${{\bf{r}}_1}$. It can parametrized by ${{\bf{r}}_1}\left( t \right) = \left( {0,0,t} \right)$ for $0 \le t \le 1$. So, ${{\bf{r}}_1}'\left( t \right) = \left( {0,0,1} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right) = \left( {{{\rm{e}}^t},1,1} \right)$. 2. Let the line segment ${C_2}$ from $\left( {0,0,1} \right)$ to $\left( {0,1,1} \right)$ be represented by ${{\bf{r}}_2}$. It can parametrized by ${{\bf{r}}_2}\left( t \right) = \left( {0,0,1} \right) + t\left( {\left( {0,1,1} \right) - \left( {0,0,1} \right)} \right)$ $ = \left( {0,0,1} \right) + t\left( {0,1,0} \right)$ $ = \left( {0,t,1} \right)$, for $0 \le t \le 1$ So, ${{\bf{r}}_2}'\left( t \right) = \left( {0,1,0} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right) = \left( {{\rm{e}},{{\rm{e}}^{ - t}},{{\rm{e}}^t}} \right)$. 3. Let the line segment ${C_3}$ from $\left( {0,1,1} \right)$ to $\left( { - 1,1,1} \right)$ be represented by ${{\bf{r}}_3}$. It can parametrized by ${{\bf{r}}_3}\left( t \right) = \left( {0,1,1} \right) + t\left( {\left( { - 1,1,1} \right) - \left( {0,1,1} \right)} \right)$ $ = \left( {0,1,1} \right) + t\left( { - 1,0,0} \right)$ $ = \left( { - t,1,1} \right)$, for $0 \le t \le 1$ So, ${{\bf{r}}_3}'\left( t \right) = \left( { - 1,0,0} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_3}\left( t \right)} \right) = \left( {{\rm{e}},{{\rm{e}}^{ - t - 1}},{\rm{e}}} \right)$. Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8): $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_3}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ $ = \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_a^b {\bf{F}}\left( {{{\bf{r}}_3}\left( t \right)} \right)\cdot{{\bf{r}}_3}'\left( t \right){\rm{d}}t$ $ = \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^t},1,1} \right)\cdot\left( {0,0,1} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {{\rm{e}},{{\rm{e}}^{ - t}},{{\rm{e}}^t}} \right)\cdot\left( {0,1,0} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {{\rm{e}},{{\rm{e}}^{ - t - 1}},{\rm{e}}} \right)\cdot\left( { - 1,0,0} \right){\rm{d}}t$ $ = \mathop \smallint \limits_0^1 {\rm{d}}t + \mathop \smallint \limits_0^1 {{\rm{e}}^{ - t}}{\rm{d}}t - {\rm{e}}\mathop \smallint \limits_0^1 {\rm{d}}t$ $ = t|_0^1 - \left( {{{\rm{e}}^{ - t}}|_0^1} \right) - {\rm{e}}\left( {t|_0^1} \right)$ $ = 1 - \left( {{{\rm{e}}^{ - 1}} - 1} \right) - {\rm{e}}$ $ = 2 - {{\rm{e}}^{ - 1}} - {\rm{e}}$ So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2 - {{\rm{e}}^{ - 1}} - {\rm{e}}$.
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