Let $\Omega \subseteq \mathbb R^2$ be compact and with a smooth boundary, such that $(0, 0) \in \mathring\Omega$.
Let $u: \mathbb R^2 \to \mathbb R$ be defined by $u(x, y) = \log(x^2+y^2)$.
We want to show that $\int_{\partial \Omega}\langle\nabla u, N\rangle = 4\pi $ where $N$ is the outwards facing normal to the integration region.
What I tried:
First I calculated the gradient and the divergence -
$\nabla u = \begin{pmatrix} \frac{2x}{x^2+y^2} \\ \frac{2y}{x^2+y^2}\end{pmatrix}$
$\Delta u = \frac{\partial}{\partial x}\frac{2x}{x^2+y^2} + \frac{\partial}{\partial y}\frac{2y}{x^2+y^2} = 0$
Let $\epsilon > 0$ such that $B(0, \epsilon) \subset \mathring \Omega$, and define $\Omega'= \Omega \setminus B(0, \epsilon)$. Inside $\Omega'$ we have that $\nabla u$ is a continuously differentiable vector field, and so divergence theorem can be applied.
Using divergence theorem:
$$0 = \int_{\Omega'}\Delta u = \int_{\partial \Omega'}\langle\nabla u, N\rangle = \int_{\partial \Omega}\langle\nabla u, N\rangle + \int_{\partial B(0, \epsilon)}\langle\nabla u, N\rangle$$
And so overall, $\int_{\partial \Omega}\langle\nabla u, N\rangle = - \int_{\partial B(0, \epsilon)}\langle\nabla u, N\rangle$
So all we need to do is calculate the integral over the surface of the sphere.
The outwards facing normal to the surface of the sphere centered at zero is just the radius, so we should have that $N = \begin{pmatrix}\frac{x}{\epsilon} \\ \frac{y}{\epsilon}\end{pmatrix}$, where we divided so it's normalized length 1.
And now:
$$\int_{\partial B(0, \epsilon)}\langle\nabla u, N\rangle = \int_{\partial B(0, \epsilon)}\langle\begin{pmatrix}\frac{2x}{x^2+y^2} \\ \frac{2y}{x^2+y^2}\end{pmatrix}, \begin{pmatrix}\frac{x}{\epsilon} \\ \frac{y}{\epsilon}\end{pmatrix}\rangle = \int_{\partial B(0, \epsilon)}\langle\begin{pmatrix}\frac{2x}{\epsilon^2} \\ \frac{2y}{\epsilon^2}\end{pmatrix}, \begin{pmatrix}\frac{x}{\epsilon} \\ \frac{y}{\epsilon}\end{pmatrix}\rangle = \int_{\partial B(0, \epsilon)}\frac{2x^2}{\epsilon^3} + \frac{2y^2}{\epsilon^3} = \frac{2}{\epsilon^3}\int_{\partial B(0, \epsilon)}\epsilon^2 = \frac{2}{\epsilon}\int_{\partial B(0, \epsilon)}1 = \frac{2}{\epsilon} \times 2 \pi \epsilon = 4 \pi$$
So the answer should be that $\int_{\partial \Omega}\langle\nabla u, N\rangle = -4\pi$ which is not what we wanted to show.
Where is the mistake? Direction of the normal maybe? why??