I have a curve in the form $\gamma[t_P,t_Q]$ with the endpoints $P=\gamma(t_P)$ and $Q=\gamma(t_Q)$. Also the curve can be written as $r=r(\phi), \phi_P\leq\phi\leq\phi_Q$. The task is to prove these for the surface area of the curve between P and Q: $$A=1/2\int_{P}^{Q} xdy -ydx=1/2\int_{t_P}^{t_Q} x\dot y -y\dot x\ dt=1/2\int_{\phi_P}^{\phi_Q} r(\phi)^2d\phi$$
My idea was to start with the right one in this way: $$\int_{\phi_P}^{\phi_Q}\int_{0}^{r(\phi)} dA=\int_{\phi_P}^{\phi_Q}\int_{0}^{r(\phi)}r\ dr d\phi=1/2\int_{\phi_P}^{\phi_Q}r(\phi)^2 d\phi$$ Do I need a parameterization for $r$ to get to the other two integrals? Normally we use $r(\phi)=(r\cos(\phi),r\sin(\phi))$ and $r^2=x^2+y^2$ but I don't get how I can bring that in the form of $ x\dot y -y\dot x\ $.
I would go the other way around. If $x$ and $y$ are parameterised by $t$, $\mathrm{d}x = \dot{x}\mathrm{d}t$ and $\mathrm{d}y = \dot{y}\mathrm{d}t$ and it's easy to see that $$ {1 \over 2} \int_P^Q x\mathrm{d}y - y\mathrm{d}x = {1 \over 2} \int_{t_P}^{t_Q} x\dot{y} - y\dot{x} \ \mathrm{d}t. \tag{1} \label{1} $$
Supposing that $r$ and $\phi$ can also be parameterised by $t$, we have $$ x = r \cos \phi \quad \dot{x}\mathrm{d}t = \dot{r}\cos{\phi}\mathrm{d}t - r\dot{\phi}\sin{\phi}\mathrm{d}t $$ and $$ y = r\sin{\phi} \quad \dot{y}\mathrm{d}t = \dot{r}\sin{\phi}\mathrm{d}t + r\dot{\phi}\cos{\phi}\mathrm{d}t, $$ so $$ (x\dot{y} - y\dot{x}) \mathrm{d}t = r^2(\cos^2{\phi} + \sin^2{\phi})\dot{\phi}\mathrm{d}t = r^2\dot{\phi}\mathrm{d}t. \tag{2} \label{2} $$ Finally, $$ {1\over2}\int_{t_P}^{t_Q} x\dot{y} - y\dot{x} \ \mathrm{d}t = {1\over2}\int_{t_P}^{t_Q}r^2\dot{\phi}\ \mathrm{d}t = {1\over2}\int_{\phi_P}^{\phi_Q}r^2(\phi)\ \mathrm{d}\phi \tag{3} \label{3} $$ since $\dot{\phi}\mathrm{d}t = \mathrm{d}\phi$, $\phi_P = \phi(t_P)$, $\phi_Q = \phi(t_Q)$ and $r$ can be parameterised by $\phi$.
As an exercise, you could try to work your way backwards from \eqref{3} to \eqref{1}.