Consider Let $\Omega \subset \mathbb{R}^2$ denote the open unit ball in $\mathbb{R}^2$. The unbounded function $$u(x) = log log(1+\frac{1}{|x|})$$ is given. Consider the compactly supported testfunction $\phi$, which vanisches on the boundary of $\Omega$
For $e>0$ it holds using divergence theorem of gauss: $$\int_{\Omega - B_e(0) }u(x) \partial_x \phi(x) + \partial_x u(x) \phi(x)dx = \int_{\partial B_e(0)} u \phi N_i dS + \int_{\partial \Omega} u \phi N_i dS $$
The second integral $\int_{\partial \Omega} u \phi N_i dS =0$ since $\phi$ is $0$ on the boundary.
How can I compute $N_i dS$ It is circle with radius $e$, so I can parameterize it in the following way: For each $\theta \in [0, 2\pi]$ the cirlce is described by $e (\cos(\theta ), (\sin(\theta )) )$ But how can I compute the i-th component of normalvector $N_i$ then?