Is this a 2D problem? Because integrating over a closed curve (ie the "surface" of an area) is the 2D equivalent of integrating over the surface of a volume. By Green's Theorem, [tex]
\begin{split}<br />
\int_{\Omega} \nabla \cdot \mathbf{E}\,dA &= \oint_{\partial\Omega} (-E_y, E_x, 0)\cdot \mathbf{t}\,dl \\<br />
&= \oint_{\partial \Omega} (\mathbf{k} \times \mathbf{E}) \cdot \mathbf{t}\,dl \\<br />
&= \oint_{\partial \Omega} \mathbf{E} \cdot (\mathbf{t} \times \mathbf{k})\,dl \\<br />
&= \oint_{\partial \Omega} \mathbf{E} \cdot \mathbf{n}\,dl<br />
\end{split}[/tex] since [itex]\mathbf{k} = \mathbf{n} \times \mathbf{t}[/itex] where [itex]\mathbf{t}[/itex] is the unit tangent of the curve traversed anticlockwise and [itex]\mathbf{n}[/itex] is the outward unit norm