Why do we calculate a charged conductor sphere's potential with respect of the centre? If potential is indefinite when measuring on a point/distribution of charge, then why isn't the same for a charge distribution lying on a conductor surface?
To better explain what i mean:
let's consider the potential of a point charge:
$$ V(r) = \frac{q}{4\pi\epsilon_0r}$$This formula comes straight up from the integral of $\vec E$ using $\infty$ as reference potential, $r$ as distance from the charge.  
To calculate the potential of a conductor sphere we basically use the same formula besides that $q$ would be : $4\pi r^2\sigma$ ($\sigma$ as surface distribution of charge) and (here is what i don't get) r as the distance from the center of the coordinate system we use.
Why isn't $r$, in the sphere case, the distance from the charge  "$dq$" on the surface?
Read the edit please

Edit: to clear things up, we can consider $r$ as the euclidean distance from the origin of a general point. The equation i wrote has an implied term : $V(r) = \frac{q}{4\pi\epsilon_0\infty} $ , meaning that i'm considering the second point to which i'm referring to, in calculating $\Delta V$, at infinite distance from the origin, thus making the second term equal to $0$.
  ($\Delta V= V(r)-V(\infty)=V(r)-0=V(r))$ 
My perplexity: my electromagnetism prof showed our class that when we have a  distribution of charge (I'll refer to a volumetric one but it should apply to every type of distribution) we use this formula to calculate the potential over the entire space : $$V(r)=\int_\tau \frac{ \rho d\tau}{4\pi \epsilon_0 |r-r'|}$$ with $r'$ pointing to $dq=\rho d\tau$ and $\tau$ as the volume that contains charge (sorry for the abuse of notation).  
Now if i use this formula to calculate the potential of a conductor sphere (i should use the one that considers a surface distribution since a conductor has no volume charge), if the sphere is centered in the origin then : $r'=R$(sphere radius).
  We can now see that when we approach with our vector $r$ our sphere's surface, the potential, with respect of infinty, has indefinite value.
  Where am I wrong?  
P.S. our professor didn't explicited the correlation between the formula i used above and the integral($\int \vec E dl)$ it comes from, can you please explain me... from what integral it comes from?
  I guess there is a change in $\vec E$, it should be used a formula where the charge isn't centered in the origin.  

 A: I'll show why the electric field is considered from the center of the sphere. I suppose that my point is outside of the sphere.
Notice this, wherever your point is, you can imagine a plane going through the center of the sphere and having our point inside it, it will necessarily divide the sphere in half. Now imagine another plane with the same characteristics but perpendicular to the other plane.
For the last step, imagine two charges symmetric with respect to the axis of intersection of these planes; you can clearly see that the electric fields that they generate on our point are opposite in direction when it comes to their components from the axis of symmetry, while their component along the axis of symmetry add up.
Looking at this from the top, we see:

What this whole situation means is that if we sum up the contribution to the electric field of the sphere in our point, then the total electric field will be directed from the center of the sphere towards our point.
Let's calculate the electric field; Gauss' law:

The net electric flux through any hypothetical closed surface is equal to $\frac{1}{\epsilon_0}$ times the net electric charge within that closed surface.

Or, in math:
$$\iint_\text{S}\vec E\cdot d\vec S=\frac{Q_\text{within hypothetical surface}}{\epsilon_0}$$
$S$ is our hypothetical surface, $d\vec S$ is the infinitesimal surface vector which directed radially (if you imagine any small surface on the sphere, then $d\vec S$ is perpendicular to it. Needless to say, it is dependent on the point of application).
Our best choice is a sphere that encloses the original and whose radius is $r$, the distance from the center of the original sphere to our point.
Since that, at each point on the surface of our hypothetical sphere, $\vec E(r)\cdot d\vec S=E(r)dS$ as they are both directed radially (also, for a given $r$, $E$ is the same if you rotate around the sphere in any way, because of what I have explained earlier), if we compute the integral, then we will get:
$$\iint_\text{S}\vec E\cdot d\vec S=4\pi E(r)r^2$$
The same integral is equal to the charge enclosed within our hypothetical surface, which, in this case, is the charge of our sphere, i.e $Q$ (as it is uniformly charged), divided by $\epsilon_0$.
$$4\pi E(r)r^2=\frac{Q}{\epsilon_0}\Leftrightarrow E(r)=\frac{Q}{4\pi\epsilon_0r^2}$$
Now you can see that $r$ is the distance from the center of the charged sphere to our point, and that, therefore, the electric potential is not infinite when $r=R_\text{sphere}$.
If $E=E(r)$, then $E(r)=-\frac{dV(r)}{dr}$

If potential is indefinite when measuring on a point/distribution of charge, then why isn't the same for a charge distribution lying on a conductor surface?

Not true, we suppose that those charges are fixed and we perform the path integral using their combined electric (superposition of electric fields) and the path of our "test charge".

To calculate the potential of a conductor sphere we basically use the same formula besides that $q$ would be : $4πr^2σ$ ($σ$ as surface distribution of charge) and (here is what i don't get) r as the distance from the center of the coordinate system we use.

The same formula as in the same integral, yes. And $r$ is not necessarily the distance from the center of our coordinates:
In the finding out of the electric field due a surface charged sphere that I show above, $r$ is the distance from the center of the sphere, because I have defined the center of the sphere to be the origin of my coordinate system. It is useful to have the center of the sphere as our origin because of the symmetries of the sphere. In any case, if the sphere's center is given by $\vec{OC}$, where $O$ is the origin, then my $r$ is the magnitude of $\vec r=\vec{OP}-\vec{OC}$, where $\vec{OP}$ is the position vector of the point at which I want to find the electric field due to the sphere.

Why isn't $r$, in the sphere case, the distance from the charge "$dq$" on the surface?

I think that my derivation of the electric field shows this. However, if you want to find out the electric field with the usual method, i.e by considering the general formula, then you will write it like that at first, of course, since it is the formula, but the ultimate end result should be in the form I have shown.
About why the electric potential formula is like that, see this:
https://en.wikipedia.org/wiki/Electric_potential#Electric_potential_due_to_a_point_charge
$q_i$ when we consider a continuous distribution becomes $dq=\rho d\tau$, where $\rho$ is a volume charge density.
A: For a conducting surface, the charge is smeared out over the surface and you have to introduce the surface charge density, $\sigma$, which has units $C/m^2$.
Just above the surface the electric field with be perpendicular to the surface with magnitude $\sigma/2 \epsilon_0$.
The easiest case to consider is an infinite plane with uniform surface charge density $\sigma$. In this case the electric field is perpendicular to the surface with magnitude everywhere $\sigma/2 \epsilon_0$ and independent how far above the surface you are.
The potential difference between two points $V$ is related to the electric field $\bf E$ by the relationship $${\bf E} = - \nabla V$$ and this leads to $$V = -\int_{P_1}^{P_2} {\bf E} {\bf \cdot} d {\bf l}.$$ This may be taken to be a generalisation of the result (referred to as the The Fundamental Theorem of Calculus) $$y = \frac{d f}{d x} \implies f = \int y(x') d x'.$$
Griffiths Introduction to Electrodynamics, for example, has a useful introduction to mathematics that explains this for the case in 3 dimensions and it is called The Fundamental Theorem of Gradients. Basically it follows from the result that if you have a scalar function $T(x,y,z)$ then starting at some initial point ${\bf a}$ if you move to the point $\bf{a} + d{\bf a}$ then the function $T$ will change by the amount $\nabla T \cdot d {\bf a}$.
