The area $A$ of metric elements $(dx)^2$ I define a metric:
$$
(d s)^2=(d x)^2+(d y)^2
$$
As per the pythagorean theorem, all terms of the infinitesimal metric $(ds)^2$, $(dx)^2$ and $(dy)^2$ are (infinitesimal) areas. I am trying to connect the terms to their corresponding area, but I am having trouble. For instance, say I define an area as 
$$
A[x]
$$
Then taking its derivative produces $dA[x]=\frac{dA[x]}{dx}dx$. For example with $A[x]=x^2$, I get $dA[x]=2xdx$. In this case $dx$ is not squared. Is there any $dA[x]$ for which $dx$ is squared?
What I am asking is what is the anti-derivative of the infinitesimal area $(dx)^2$? What expression of area $A[x]$ has the differential form of $(dx)^2$?
 A: In fact $dA=dxdy$, so $\frac{dA}{dx}=dy$. This looks like a strange equation unless you accept that infinitesimals in calculus can have different orders. Similarly, in polar coordinates $dA=rdrd\theta$, so $\frac{dA}{dr}=rd\theta,\,\frac{dA}{d\theta}=rdr$.
But quantities such as $dx^2$ aren't meant to be integrated as such. With respect to what would you integrate them? I can write $\int 2x dx=x^2$, but $\int dx^2 dx$ would be... What? Well, it's not $dx^3$ because of the $\int$, and we'd learn nothing new from saying it's $\int dx^3$ (which a reader might interpret to mean something very different, such as integration over a variable called $x^3$ where the superscript isn't an exponent, or as a triple integral $\int d^3 x$).
The way to understand $ds^2=dx^2+dy^2$ is an infinitesimal variant of squares erected on sides of a right-angled triangle, viz. Pythagoras. Note this doesn't require you to integrate the terms in this equation. A more careful way to think about it is $$\int ds=\int\sqrt{\left(\frac{dx}{dp}\right)^2+\left(\frac{dy}{dp}\right)^2}dp$$for a parameter $p$ (provided we can and do see all contributions to the arc length as positive).
