What is the meaning of $\int_\omega f\, \partial \Omega$? In a paper I am reading, I have come across the following formula:
$$\int_\omega f\, \partial \Omega$$
in which $\Omega$ is a bounded region in $\mathbb{R}^n$, $f:\Omega \to \mathbb{R}$ a function, and $\omega \subset \Omega$.
What is the meaning of $\partial \Omega$ in the integral? Is this standard notation?
EDIT:
I am not actually so much interested in the concrete problem in which the above formula occurs. I just did not understand the use of the "curvy delta" symbol in the integrand, and presumed it would be some kind of standard notation I did not know. FWIW, here is a link to the paper, the formula in question is number (4) on page 268.
 A: If you compare the paragraph before Equation (5) in that paper, which states that for a function $\phi$ such that $\phi$ is positive on $\omega$ and negative outside $\omega$, you have
$$ \int_{\omega} \partial\Omega = \int_{\Omega} H(\phi) dx $$
where $H(\cdot)$ is the Heaviside function, one may infer that by $\partial\Omega$ the authors meant to express some sort of induced volume element on $\omega\subset \Omega$. 
But this certainly is not mainstream/standard notation; I for one have never seen its like before. 
A: Assuming what you originally had was
$$
  \int_{\partial\Omega}f\,\omega
$$
this means simply: integrate $f\omega$ over the boundary of $\Omega$.
$\omega$ is a differential form here, an $n-1$-form if $\Omega\subset\mathbb{R}^n$. It can then be written as
$$
  \omega = \omega_1\cdot\mathrm{d}x_2\wedge\ldots\wedge\mathrm{d}x_n
  + \omega_2\cdot\mathrm{d}x_1\wedge\mathrm{d}x_3\wedge\ldots\wedge\mathrm{d}x_n
  + \ldots\ldots
$$
where the $\omega_i$ are functions $\mathbb{R}^n\to\mathbb{R}$ (as $f$ also is).
(Of course, you could swap the meaning of $\omega$ and $f$, but this is the more common variant.)
As a simple example, let $\Omega$ be the unit square in $\mathbb{R}^2$. Then, $\omega$ is a 1-form, let it be simply
$$
  \omega := \mathrm{d}x.
$$
Let furthermore
$$
  f(x,y) := x\cdot y
$$
then your integral is simply
$$
  \int_{\partial\Omega}f\omega
  = \int_{\partial\Omega}xy\;\mathrm{d}x
$$
We yet have to figure out what $\partial\Omega$ is, but in this case it's simply a path consisting of four straight segments, two in $x$- and two in $y$-direction. As it is an integral over $\mathrm{d}x$, we only need the parts in $x$-direction, for one of these, $y=0$, for the other $y=1$. So
$$
  \int_{\partial\Omega}f\omega
  = \int\limits_0^1x\cdot 0\;\mathrm{d}x
    + \int\limits_0^1x\cdot 1\;\mathrm{d}x
  = \frac12.
$$
