Area of annulus $dA = 2\pi r \ dr$, is it a coincidence that $\frac{dA}{dr} = 2\pi r \implies A = \pi r^2$? When considering how to derive the area of a circle, a physicist approach was to consider an annulus of infinitesimal thickness $dr$, such that with variable radii, a summation of rings could approximate the area of a circle  using a single integral, despite this being classically a double integral - by replacing $dA$ with $dA = 2\pi r \ dr $, as the infinitesimal area element has been approximated in a way that we need only integrate in one "direction". 
Fair enough. However, I spotted that this implies $A = \pi r^2$ if we assume taking $dA/dr$ is the derivative of the area with respect to the radius (which I think is fair to do but my justification for it could be better). This implication looks pretty good. 
My question is, however, two-fold. One, does $dA/dr$ make sense in this context as the derivative of area with respect to radius, since it was kind of an afterthought for me after considering the derivation and not a formal application of a derivative operator, and two, does this approach hold for finding the area of other shapes?
 A: No, not a coincidence.  Your interpretation of $dA$ as the area of an infinitesimal annulus is exactly why the derivative of the area will give the perimeter. 
This will not hold for any shape. However the circle is not the only shape for which this holds although the circle is the only figure for which $\vec{E}$ is normal to the surface everywhere, provided the "radius" is chosen correctly.
For even-sided regular polygons, choose the radius of the inscribed circle as "$r$" and we find that for perimeter $P(r)$ and area $A(r)$, $P(r)=\frac{dA}{dr}$.
Examples: square (n=4). Let $r=\frac{a}{2}$ where $a$ is the side of the square.  Then $A=4r^2$ and $P=8r$ and $P(r)=\frac{dA}{dr}$.
Hexagon(n=6)
$P=4\sqrt{3}r$ and $A=2\sqrt(3)r$ and $P(r)=\frac{dA}{dr}$.
Octagon (n=8)
$P=16(\sqrt{2}-1)r$ and $A=8(\sqrt{2}-1)r^2$ and $P(r)=\frac{dA}{dr}$.
The circle belongs to this class of figures.  It is the limiting case as $n\to \infty$
A: It's valid. $\frac{{\mathrm d}A}{{\mathrm d}r}$ is the ratio of the area change to the infinitesimal change in the radius, since the direction is normal everywhere to the outbounding circle, it follows that $\frac{{\mathrm d}A}{{\mathrm d}r}=2\pi r$.
This also apply to $n$-ball and $(n-1)$-sphere. Take $n=3$ as an example, we get $$\frac{{\mathrm d}V}{{\mathrm d}r}=\frac43\frac{{\mathrm d}(\pi r^3)}{{\mathrm d}r}=4\pi r^2=S.$$
