Rate at which planet "sweeps out area" I want to show that the rate at which a planet circling around a star in an orbit is "sweeps out area" is $\frac{dA}{dt}=\frac{1}{2}r^2\omega$, where $r$ is the distance from the star to the planet and $\omega$ is the planet's angular velocity ($\dot{\phi}$ in polar coordinates).
It can be seen that the infinitesimal "wedge" area is $dA=\lVert d\vec{A}\rVert=\frac{1}{2}\left\rVert\vec{r}\times(\vec{r}+d\vec{r})\right\rVert = \frac{1}{2}rdr\sin(d\phi)\approx \frac{1}{2}d\phi rdr$ (since $d\phi$ is small).
Dividing by $dt$ gives: $$\frac{dA}{dt} = \frac{1}{2}\frac{d\phi}{dt}rdr=\frac{1}{2}\dot{\phi}rdr = \frac{1}{2}\omega r dr$$
Now, I believe, the next step is to integrate from $r$ to $r+dr$?
I think I'm probably complicating something. Would appreciate some help.

I could be wrong but I believe when finding |r x (r+dr)| , it simplifies to |r x dr| where we usually take the angle between those to be close enough to a right angle ( looks like you've taken it to be phi) to approximate it to rdr. After substituting that and converting dr  in terms of the angle you should get the desired result.
P.s. sorry for the lack of notation, I'm on my phone.
