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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 $d\theta$

user2471881
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    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 '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 '19 at 22:58

2 Answers2

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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 $1/n^2$).

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