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.

It all comes down to evaluating one cross-product, which is best done in the ${\{\hat{r},\hat{\phi},\hat{z}\}}$ basis. 
$$\eqalign{
 r\times (r+dr) &= r\times dr \cr
  &= (\hat{r}\,r)\times (\hat{\phi}\,dr_\phi + \hat{r}\,dr_r) \cr
  &= \hat{z}\,r\,dr_\phi \cr
  &= \hat{z}\,r^2\,d\phi \cr
}$$
Half the magnitude of the cross-product yields the swept area 
$$\eqalign{
 dA &= \frac{1}{2}r^2\,d\phi \cr\cr
 \frac{dA}{dt} &= \frac{1}{2}r^2\,\frac{d\phi}{dt} \cr\cr
}$$
Others have suggested that making the angle really small will make $dr$ nearly perpendicular to $r$, however that is not true.  
The key is that $dr$ has components in both the $\hat{r}$ and $\hat{\phi}$ directions, but the cross product with $r$ annihilates the $\hat{r}$-component.
