Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{1}{2}\pi $
Work Step by Step
We have ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{2y}}{{{x^2} + 4{y^2}}},\frac{x}{{{x^2} + 4{y^2}}}} \right)$ and the path ${\bf{r}}\left( t \right) = \left( {\cos t,\frac{1}{2}\sin t} \right)$ for $0 \le t \le 2\pi $
Evaluate:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{\sin t}}{{{{\cos }^2}t + {{\sin }^2}t}},\frac{{\cos t}}{{{{\cos }^2}t + {{\sin }^2}t}}} \right) = \left( {\sin t,\cos t} \right)$
${\bf{r}}'\left( t \right) = \left( { - \sin t,\frac{1}{2}\cos t} \right)$
By Eq. (8) in Section 17.2:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } \left( {\sin t,\cos t} \right)\cdot\left( { - \sin t,\frac{1}{2}\cos t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } \left( { - {{\sin }^2}t + \frac{1}{2}{{\cos }^2}t} \right){\rm{d}}t$
Recall that $\cos 2x = 1 - 2{\sin ^2}x$ and $\cos 2x = 2{\cos ^2}x - 1$, so the integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } \left[ {\frac{1}{2}\left( {\cos 2t - 1} \right) + \frac{1}{4}\left( {\cos 2t + 1} \right)} \right]{\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{2\pi } \left( {\frac{3}{4}\cos 2t - \frac{1}{4}} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{4}\mathop \smallint \limits_0^{2\pi } \left( {3\cos 2t - 1} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{4}\left( {\frac{3}{2}\sin 2t - t} \right)|_0^{2\pi } = \frac{1}{4}\left( { - 2\pi } \right) = - \frac{1}{2}\pi $
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{1}{2}\pi $.
