Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{16}}{3}$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {{x^2},xy} \right)$, and the line segment from $\left( {0,0} \right)$ to $\left( {2,2} \right)$.
The line segment can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,t} \right)$ for $0 \le t \le 2$.
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^2},{t^2}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {1,1} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {{t^2},{t^2}} \right)\cdot\left( {1,1} \right)dt = 2{t^2}dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\mathop \smallint \limits_0^2 {t^2}{\rm{d}}t = \frac{2}{3}{t^3}|_0^2 = \frac{{16}}{3}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{{16}}{3}$.
