Show that if $r$ is taken to be the independent variable the functional 6.10 becomes
$$
S[\theta]=\int_{r_a}^{r_b} d r \mathcal{F}\left(r, \theta, \theta^{\prime}\right) \text { where } \mathcal{F}=\left(\cos \theta-r \theta^{\prime} \sin \theta\right) F\left(r \cos \theta, r \sin \theta, \frac{\sin \theta+r \theta^{\prime} \cos \theta}{\cos \theta-r \theta^{\prime} \sin \theta}\right)
$$