So i need to calculate $$\int \int _{S} F . N d\sigma$$ oriented outside, $F = r/|r|³$ and S is the paraboloid part of $z = 3 - x²-y²$ with $0 \leq z \leq 3$
I thought that the right way to solve it would be $$\int \int _{S} F . N d\sigma + \int \int _{S_{2}} F . N_{2} d\sigma - \int \int _{sphere/2} F . \hat r d\sigma = \int \int div F dv => \int \int _{S} F . N d\sigma + 0 -2\pi = 0 => \int \int _{S} F . N d\sigma = 2\pi$$
S2 is the disk for z=0 and "sphere/2" is a semisphere located at z>0. Fortunatelly, it gives the right answer.
My question is if the expression i wrote is right. That is, i choose a half sphere because in this way it is inside the original surface S of the problem, but is it right to do that? I mean, since F diverges at the origin, generally we need to choose a sphere or a circle to surround the origin, does it need to be inside the original surface/curve? Or it was just coincidence that this expression gives the right answer?
Your working is correct but you should clearly define the surfaces you have chosen. We choose closed region such that it does not contain the origin and the divergence of the vector field anywhere but origin is zero. We can choose a hemisphere surface above $z = 0$ either outside the paraboloid or inside it and close the region with a washer disk, such that the closed region does not contain the origin. We make sure the sphere does not intersect the paraboloid surface.
One such example is a unit hemisphere surface centered at the origin -
$S1: \rho = 1, 0 \leq \phi \leq \frac{\pi}{2}$ where $\phi$ is second polar.
In that case, the washer disk surface is
$S2: 1 \leq r \leq \sqrt3, z= 0$, which closes the region formed by paraboloid and the sphere.
Flux through $S2$ is zero as the $z$ component of vector field at $z = 0$ is zero and outward normal vector to the disk is $(0, 0, -1)$.
$\vec F (S2) \cdot (0, 0, -1) = 0$
Outward flux through $S1$ is
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} \frac{\vec \rho \cdot ( - \hat \rho)}{|\rho|^3} \rho^2 \sin\phi \ d\phi \ d\theta$
$ \displaystyle = \int_0^{2\pi} \int_0^{\pi/2} - \sin\phi \ d\phi \ d\theta = - 2\pi$
Since net flux through the closed surface is zero (as divergence is zero), outward flux through paraboloid surface is $2\pi$.