Answer
(a) we show that ${\bf{F}}\left( {r,\theta } \right) = {r^{ - 1}}\left( { - \sin \theta ,\cos \theta } \right)$
(b) we show that ${\bf{F}}\cdot{\bf{r}}'\left( \theta \right)d\theta = d\theta $
(c) we show that $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = {\theta _2} - {\theta _1}$
Work Step by Step
(a) We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$.
In polar coordinates: $x = r\cos \theta $, $y = r\sin \theta $. So,
${\bf{F}}\left( {r,\theta } \right) = \left( {\frac{{ - r\sin \theta }}{{{r^2}}},\frac{{r\cos \theta }}{{{r^2}}}} \right)$
Thus, ${\bf{F}}\left( {r,\theta } \right) = {r^{ - 1}}\left( { - \sin \theta ,\cos \theta } \right)$.
(b) We have ${\bf{r}}\left( \theta \right) = \left( {f\left( \theta \right)\cos \theta ,f\left( \theta \right)\sin \theta } \right)$.
Taking the derivative of ${\bf{r}}\left( \theta \right)$ with respect to $\theta $ gives
${\bf{r}}'\left( \theta \right) = \left( {f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta ,f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta } \right)$
So,
${\bf{F}}\cdot{\bf{r}}'\left( \theta \right)d\theta $
$ = {r^{ - 1}}\left( { - \sin \theta ,\cos \theta } \right)\cdot\left( {f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta ,f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta } \right)d\theta $
$ = {r^{ - 1}}\left( { - f'\left( \theta \right)\cos \theta \sin \theta + f\left( \theta \right){{\sin }^2}\theta + f'\left( \theta \right)\sin \theta \cos \theta + f\left( \theta \right){{\cos }^2}\theta } \right)d\theta $
$ = {r^{ - 1}}f\left( \theta \right)d\theta $
Since $r = f\left( \theta \right)$, so ${\bf{F}}\cdot{\bf{r}}'\left( \theta \right)d\theta = d\theta $.
(c) 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_{{\theta _1}}^{{\theta _2}} {\bf{F}}\left( {{\bf{r}}\left( \theta \right)} \right)\cdot{\bf{r}}'\left( \theta \right){\rm{d}}\theta $
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{\theta _1}}^{{\theta _2}} {\rm{d}}\theta = {\theta _2} - {\theta _1}$
Thus, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = {\theta _2} - {\theta _1}$.
