Integrating the Geometric Mean of the Differentials I'm curious if anyone has seen any literature on the following.  I was playing around with differentials and integration and couldn't help but notice the interesting property of the following integral:

$$
\int{\sqrt{dxdy}}=\int{\sqrt{\frac{dy}{dx}}dx}
$$

A similar integral for functions of n-variables integrating the geometric mean of all the differentials can be written as well.

So the integral of the geometric mean of the two differentials is the integral over $x$ of the square root of the derivative of $y$ with respect to $x$ (or you can also flip $x$ and $y$ around.

In the general case,

$$
\int{(dx_1dx_2\cdots dx_n)^{\frac{1}{n}}}=\int{(\frac{dx_2}{dx_1}\cdots \frac{dx_n}{dx_1})^{\frac{1}{n}}dx_1}
$$

It just seems too elegant to not have an application. Any thoughts greatly appreciated!
I highly suspect this is just misleading notation. the right hand side of your first equation assumes there is a function y(x), such that you can compute it's square root, and proceed Riemann integration. I don't know any definition for the left hand side, so this would need to be done with some rigor before the equation you wrote could be proved, and any application found. You might have implicitly considered x=x(t) and y=y(t), such that:

$$
\sqrt{dxdy}=\sqrt{x'(t) dt y'(t) dt}=\sqrt{x'(t)y'(t)}\sqrt{dt dt} 
$$

But even if it seems trivial, I wouldn't even claim:
$$
\sqrt{dt dt} = dt
$$
Because at the very least:
$$
\sqrt{dt dt} = |dt|
$$

But note that $dt$ is not a number, so the expression itself lacks rigor.
I highly suspect this is just misleading notation. the right hand side of your first equation assumes there is a function y(x), such that you can compute it's square root, and proceed Riemann integration. I don't know any definition for the left hand side, so this would need to be done with some rigor before the equation you wrote could be proved, and any application found. You might have implicitly considered x=x(t) and y=y(t), such that:

$$
\sqrt{dxdy}=\sqrt{x'(t) dt y'(t) dt}=\sqrt{x'(t)y'(t)}\sqrt{dt dt} 
$$

But even if it seems trivial, I wouldn't even claim:
$$
\sqrt{dt dt} = dt
$$
Because at the very least:
$$
\sqrt{dt dt} = |dt|
$$

But note that $dt$ is not a number, so the expression itself lacks rigor.