Answer
We verify that ${\bf{F}} = \nabla f$.
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{9}{4}$
Work Step by Step
1. We have ${\bf{F}}\left( {x,y} \right) = \left( {3,6y} \right)$ and $f\left( {x,y} \right) = 3x + 3{y^2}$.
Evaluate
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) = \left( {3,6y} \right)$
Since ${\bf{F}}\left( {x,y} \right) = \left( {3,6y} \right)$, so ${\bf{F}} = \nabla f$.
2. Evaluate the line integral of ${\bf{F}}$ over ${\bf{r}}\left( t \right) = \left( {t,2{t^{ - 1}}} \right)$ for $1 \le t \le 4$.
Since ${\bf{F}} = \nabla f$, by definition ${\bf{F}}$ is conservative. By Theorem 1, the line integral of ${\bf{F}}$ over the given path is
$\mathop \smallint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{\bf{r}}\left( 4 \right)} \right) - f\left( {{\bf{r}}\left( 1 \right)} \right) = f\left( {4,\frac{1}{2}} \right) - f\left( {1,2} \right) = 12 + \frac{3}{4} - \left( {3 + 12} \right) = - \frac{9}{4}$
