Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{{13}}{{18}}$
Work Step by Step
We have ${\bf{F}}\left( {x,y,z} \right) = \left( {{x^2}y,{y^2}z,{z^2}x} \right)$ and the path ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^{ - t}},{{\rm{e}}^{ - 2t}},{{\rm{e}}^{ - 3t}}} \right)$ for $0 \le t < \infty $.
Evaluate:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{{\rm{e}}^{ - 4t}},{{\rm{e}}^{ - 7t}},{{\rm{e}}^{ - 7t}}} \right)$
${\bf{r}}'\left( t \right) = \left( { - {{\rm{e}}^{ - t}}, - 2{{\rm{e}}^{ - 2t}}, - 3{{\rm{e}}^{ - 3t}}} \right)$
By Eq. (8) in Section 17.2:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^\infty \left( {{{\rm{e}}^{ - 4t}},{{\rm{e}}^{ - 7t}},{{\rm{e}}^{ - 7t}}} \right)\cdot\left( { - {{\rm{e}}^{ - t}}, - 2{{\rm{e}}^{ - 2t}}, - 3{{\rm{e}}^{ - 3t}}} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^\infty \left( { - {{\rm{e}}^{ - 5t}} - 2{{\rm{e}}^{ - 9t}} - 3{{\rm{e}}^{ - 10t}}} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \left( {\frac{1}{5}{{\rm{e}}^{ - 5t}} + \frac{2}{9}{{\rm{e}}^{ - 9t}} + \frac{3}{{10}}{{\rm{e}}^{ - 10t}}} \right)|_0^\infty = \left( { - \frac{1}{5} - \frac{2}{9} - \frac{3}{{10}}} \right) = - \frac{{13}}{{18}}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{{13}}{{18}}$.
