Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{{{\rm{e}}^6}}}{3} + \frac{{3{{\rm{e}}^4}}}{2} + \frac{{2{{\rm{e}}^2}}}{3} + \frac{1}{{3{{\rm{e}}^4}}} - \frac{{65}}{6}$
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 path as is shown in Figure 14. Let the orientation be directed from $A$ to $B$, $B$ to $C$, and $C$ to $A$.
1. Let the line segment ${C_1}$ from $A = \left( {2,0,0} \right)$ to $B = \left( {0,4,0} \right)$ be represented by ${{\bf{r}}_1}$.
It can parametrized by
${{\bf{r}}_1}\left( t \right) = \left( {2,0,0} \right) + t\left( {\left( {0,4,0} \right) - \left( {2,0,0} \right)} \right)$ ${\ \ \ }$ for $0 \le t \le 1$
${{\bf{r}}_1}\left( t \right) = \left( {2,0,0} \right) + t\left( { - 2,4,0} \right)$
${{\bf{r}}_1}\left( t \right) = \left( { - 2t + 2,4t,0} \right)$
So, ${{\bf{r}}_1}'\left( t \right) = \left( { - 2,4,0} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right) = \left( {1,{{\rm{e}}^{ - 6t + 2}},{{\rm{e}}^{4t}}} \right)$.
2. Let the line segment ${C_2}$ from $B = \left( {0,4,0} \right)$ to $C = \left( {0,0,6} \right)$ be represented by ${{\bf{r}}_2}$.
It can parametrized by
${{\bf{r}}_2}\left( t \right) = \left( {0,4,0} \right) + t\left( {\left( {0,0,6} \right) - \left( {0,4,0} \right)} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
${{\bf{r}}_2}\left( t \right) = \left( {0,4,0} \right) + t\left( {0, - 4,6} \right)$
${{\bf{r}}_2}\left( t \right) = \left( {0, - 4t + 4,6t} \right)$
So, ${{\bf{r}}_2}'\left( t \right) = \left( {0, - 4,6} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right) = \left( {{{\rm{e}}^{6t}},{{\rm{e}}^{4t - 4}},{{\rm{e}}^{ - 4t + 4}}} \right)$.
3. Let the line segment ${C_3}$ from $C = \left( {0,0,6} \right)$ to $A = \left( {2,0,0} \right)$ be represented by ${{\bf{r}}_3}$.
It can parametrized by
${{\bf{r}}_3}\left( t \right) = \left( {0,0,6} \right) + t\left( {\left( {2,0,0} \right) - \left( {0,0,6} \right)} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
${{\bf{r}}_3}\left( t \right) = \left( {0,0,6} \right) + t\left( {2,0, - 6} \right)$
${{\bf{r}}_3}\left( t \right) = \left( {2t,0, - 6t + 6} \right)$
So, ${{\bf{r}}_3}'\left( t \right) = \left( {2,0, - 6} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_3}\left( t \right)} \right) = \left( {{{\rm{e}}^{ - 6t + 6}},{{\rm{e}}^{2t}},1} \right)$.
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( {1,{{\rm{e}}^{ - 6t + 2}},{{\rm{e}}^{4t}}} \right)\cdot\left( { - 2,4,0} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{6t}},{{\rm{e}}^{4t - 4}},{{\rm{e}}^{ - 4t + 4}}} \right)\cdot\left( {0, - 4,6} \right){\rm{d}}t$
${\ \ \ \ }$ $ + \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{ - 6t + 6}},{{\rm{e}}^{2t}},1} \right)\cdot\left( {2,0, - 6} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( { - 2 + 4{{\rm{e}}^{ - 6t + 2}}} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( { - 4{{\rm{e}}^{4t - 4}} + 6{{\rm{e}}^{ - 4t + 4}}} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {2{{\rm{e}}^{ - 6t + 6}} - 6} \right){\rm{d}}t$
$ = \left( { - 2t - \frac{2}{3}{{\rm{e}}^{ - 6t + 2}}} \right)|_0^1 + \left( { - {{\rm{e}}^{4t - 4}} - \frac{3}{2}{{\rm{e}}^{ - 4t + 4}}} \right)|_0^1 + \left( { - \frac{1}{3}{{\rm{e}}^{ - 6t + 6}} - 6t} \right)|_0^1$
$ = - 2 - \frac{2}{3}{{\rm{e}}^{ - 4}} + \frac{2}{3}{{\rm{e}}^2} + \left( { - 1 - \frac{3}{2} + {{\rm{e}}^{ - 4}} + \frac{3}{2}{{\rm{e}}^4}} \right) + \left( { - \frac{1}{3} - 6 + \frac{1}{3}{{\rm{e}}^6}} \right)$
$ = - 2 - \frac{2}{{3{{\rm{e}}^4}}} + \frac{2}{3}{{\rm{e}}^2} - \frac{5}{2} + \frac{1}{{{{\rm{e}}^4}}} + \frac{3}{2}{{\rm{e}}^4} - \frac{{19}}{3} + \frac{1}{3}{{\rm{e}}^6}$
$ = \frac{{{{\rm{e}}^6}}}{3} + \frac{{3{{\rm{e}}^4}}}{2} + \frac{{2{{\rm{e}}^2}}}{3} + \frac{1}{{3{{\rm{e}}^4}}} - \frac{{65}}{6}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{{{\rm{e}}^6}}}{3} + \frac{{3{{\rm{e}}^4}}}{2} + \frac{{2{{\rm{e}}^2}}}{3} + \frac{1}{{3{{\rm{e}}^4}}} - \frac{{65}}{6}$.
