2D Partial Integration I have a (probably very simple problem): I try to find the variational form of a PDE, at one time we have to partially integrate:
$\int_{\Omega_j} v \frac{\partial}{\partial x}E  d(x,y)$ where v is our testfunction and E ist the function we try to find. We have $v, E: \mathbb{R}^2\rightarrow \mathbb{R}$.
Our approach was the partial integration: $\int_{\Omega_j} v \frac{\partial}{\partial x}E  d(x,y) = - \int_{\Omega_j} E \frac{\partial}{\partial x} v d(x, y) + \int_{\partial\Omega_j} v E ds$
But I don't think that this makes sense, as we usually need the normal for the boundary integral, but how can we introduce a normal, when our functions are no vector functions?
I hope you understand my problem

For the following, I refer to Holland, Applied Analysis by the Hilbert Space Method, Sec. 7.2.  Let $D$ be a simply connected region in the plane, and $f$, $P$, and $Q$ be functions that map $D \rightarrow \mathbb{R}$.  Then
$$\iint_D f \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right ) \,dx\,dy = \int_{\partial D} f (P dx + Q dy) - \iint_D \left( \frac{\partial f}{\partial x} Q - \frac{\partial f}{\partial y} P \right )\,dx\,dy $$
This is a form of 2D integration by parts and is derived using differential forms.  This applies to your problem, in that $f=v$, $P=0$, and $Q=E$.  Your integration by parts then takes the form
$$\iint_{\Omega_j} v \frac{\partial E}{\partial x} \,dx\,dy = \int_{\partial \Omega_j} v \, E \, dy - \iint_{\Omega_j} E \frac{\partial v}{\partial x} \,dx\,dy$$
The important observation here is that the lack of a function that has a derivative with respect to $y$ in the original integral produces a boundary integral that is integrated over the $y$ direction only.
