Mistake in my proof: what is the normalisation factor of the surface integral of a sphere? I was trying to prove
$$ {1\over \varepsilon} \int_{\partial B(a,\varepsilon)} f dS = {1\over r} \int_{\partial B(a,r)} f dS$$
where $0<\varepsilon < r$  and $f$ is harmonic on $\mathbb R^2$ and $n$ is the normal vector to the sphere. 
Here is what I did:
(0) I use the equation 
$$ \int_{D}(f\Delta g - g \Delta f)dV = \int_{\partial D}\left( f {\partial g \over \partial n} - g {\partial f \over \partial n}\right) dS$$
and let $g(x) = \log \|x-a\|$. (this is a radial harmonic function)
(1) I calculated that ${\partial g \over \partial n} = 1$ (this is probably correct, I suspect my mistake to be later in the calculation)
(2) I note that $\int_{B(a,r) \setminus B(a,\varepsilon)} (f \Delta G - G \Delta f)dV  = 0$ because both $f$ and $g$ are harmonic
(3) I showed that $$ \int_{\partial B(a,r)} g {\partial f \over \partial n} dS = 0$$
(4) I put (1)-(3) together so that
$$ 0 = \int_{B(a,r) \setminus B(a,\varepsilon)} (f \Delta G - G \Delta f)dV  = \int_{\partial B(a,r) \sqcup \partial B(a,\varepsilon)} f dS$$
But this results in
$$ \int_{\partial B(a,\varepsilon)} f dS = \int_{\partial B(a,r)} f dS$$
and there is one problem with that:
I am missing the factors of ${1\over \varepsilon}$ and ${1\over r}$.
Hence my question is:

How can I calculate the normalisation of the surface integral of the sphere?



(4) I put (1)-(3) together so that
  $$
0 = \int_{B(a,r) \setminus B(a,\varepsilon)} (f \Delta G - G \Delta f)\, dV
  = \int_{\partial B(a,r) \sqcup \partial B(a,\varepsilon)} f\, dS
$$

It appears you've lost the normal derivative of $g$. That is, the right-hand integral should be
$$
0 = \int_{B(a,r) \setminus B(a,\varepsilon)} (f \Delta G - G \Delta f)\, dV
  = \int_{\partial B(a,r) \sqcup \partial B(a,\varepsilon)} f \frac{\partial g}{\partial n}\, dS,
$$
and of course $\dfrac{\partial g}{\partial n}(x) = \dfrac{1}{\|x - a\|}$.
