Expressing the integral over boundary of a domain as an integral over the domain Let $\Omega \subset \mathbb{R}^2$ be a domain which is "well behaved" (has all "wishable" properties), so as its boundary. For every  $u \in C^\infty(\Omega,\mathbb{R})$, I would like to express the following integral
$$\int_{\partial \Omega} \frac{((Du)^\perp \cdot n)^2}{|Du|^3} dS,$$
where $dS$ is the Hausdorff measure of $\partial \Omega$ and $n$ its unit normal,  as an integral over $\Omega$ with the standard Lebesgue measure. Here, $(Du)^\perp=(\partial u/\partial x_2, - \partial u/\partial x_1)$ and $|Du|=\sqrt{(\partial u/\partial x_1)^2 + (\partial u/\partial x_2)^2 }$.
The presence of the normal might indicate that the divergence theorem could be of use, but the fact that the term is quadratic makes it hard to handle. Any ideas are appreciated!
 A: $\newcommand{\Om}{\Omega}$Let
\begin{equation*}
    I_u(\Om):=\int_{\partial\Om} \frac{((Du)^\perp\cdot n)^2}{|Du|^3}\,dS. 
\end{equation*}
Then
\begin{equation*}
    I_u(\Om)=\int_\Om \frac{I_u(\Om)}{|\Om|}\,dz 
\end{equation*}
if $|\Om|:=\int_\Om dz>0$.
This trivial representation of $I_u(\Om)$ as "an integral over $\Omega$ with the standard Lebesgue measure" is probably not what you had in mind when asking your question.
Apparently, you wanted the integrand to not depend on $\Om$ and you wanted the integral representation of $I_u(\Om)$ to hold for all nice enough domains $\Om$. So then, what you wanted seems to be a formula of the form
\begin{equation*}
    I_u(\Om)=\int_\Om F(u)(z)\,dz \tag{1}\label{1}
\end{equation*}
for some operator $F$ acting on nice enough functions $u$ and for all nice enough domains $\Om$.
However, such a representation of $F$ would be additive in the sense that, if $|\Om_1\cap\Om_2|=0$ for some nice enough domains $\Om_1$ and $\Om_2$, then $I_u(\Om_1\cup\Om_2)=I_u(\Om_1)+I_u(\Om_2)$. But this is false for "almost all" nice functions $u$ and, say, $\Om_1=[0,1]\times[0,1]$ and $\Om_1=[1,2]\times[0,1]$ -- because in general the integrand $\frac{((Du)^\perp\cdot n)^2}{|Du|^3}$ is $>0$ almost everywhere on $\{1\}\times[0,1]$.
So, a representation of the form \eqref{1} is impossible.
