# Rigorous Geometric Proof That dA=rdrdθ?

"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 $$d\theta$$

• at small enough $\Delta r$ and $\Delta\theta$ (i.e. $dr,d\theta$) we can visualise the area as a rectangle of side length $dr$ and $rd\theta$. Hence the area is $rdrd\theta$ – Henry Lee Feb 19 at 15:33
• You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner. – Brevan Ellefsen Feb 19 at 22:58

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 $$1/n^2$$).
$$\lim_{d\theta,dr\to0}\frac{r\,dr\,d\theta+\frac12dr^2d\theta}{dr\,d\theta}=\lim_{d\theta,dr\to0}(r+\tfrac12dr)=r.$$