Rigorous Geometric Proof That dA=rdrdθ? I've seen the same question as above for both the geometric derivation of polar coordinates and the x-y transformation of polar coordinates. The first answer states that:
"In the geometric approach, dr2=0 as it is not only small but also symmetric (see here)." And links to a wikipedia article on exterior algebra. Could someone clarify this for me? To reiterate:
Given the polar coordinates of a 'circular wedge'
Geometrically, the exact area would be
$$\frac{(r+dr)^2dθ}{2}−\frac{r^2dθ}{2}$$
$$=(r+\frac{dr}{2})drdθ$$
$$= r dr d\theta + \frac{dr^2 d\theta}{2}$$
How do we get rid of $\frac{dr^2dθ}{2}$? Why are we allowed to consider it insiginficant to the point where we can ignore it as opposed to keeping the tiny values of $dr$ and $d\theta$
 A: The meaning of $dA=r dr d\theta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d \theta$.
Thus your question (because $d\theta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=\sum_i f(r_i) (\Delta r)_i^2$ should all go to zero as you refine the partition.
For bounded $f$ for example $|S| \leq ML\delta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $\delta$ is the longest of the $(\Delta r)_i$'s. This indeed goes to zero as $\delta \to 0$. 
The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).
A: $$\lim_{d\theta,dr\to0}\frac{r\,dr\,d\theta+\frac12dr^2d\theta}{dr\,d\theta}=\lim_{d\theta,dr\to0}(r+\tfrac12dr)=r.$$
