Question: Defining on $R^3$, $V = \iiint\_S dx \, dy \, dz $ as the volume of surface $S$, with $S$ closed, bounded and arc-connected. Which is the $S$ of minimal area, that contains $V$. I know it's a bit general, so maybe you could think of another restriction without losing too much generality

Answer: 
The link given by Brad is the isoperimetric inequality, which states that of all surfaces with area $A$, the one enclosing the most volume is a sphere. This also implies the answer to your question: of all surfaces enclosing a given volume $V$, the one with least surface area is a sphere.

To see this, fix a volume $V$ and let $S$ be any surface enclosing volume $V$. Let $S\_0$ be a sphere having the same surface area as $S$. By the isoperimetric inequality, $S\_0$ encloses more volume than $S$ (or the same amount). So if we scaled down $S\_0$ to get a smaller sphere $S\_1$ whose volume is $V$, it would have smaller surface area than $S\_0$, and hence smaller surface area than $S$. 

So we have shown that the sphere $S\_1$ whose volume is $V$ has smaller surface area than any other surface $S$ enclosing volume $V$.

The relationship between statements like this is called *duality*.
The surface of minimum area for given volume is a Delaunay Unduloid.

The meridians have constant mean curvature with respect to the axis of rotational symmetry.

Among the unduloids, the sphere is the smallest.

It is straight forward to set up the isoperimeric problem, and then solve it by Euler-Lagrange Equations.

In this connection it is to be recalled that the locus of an ellipse focus as it rolls on a straight line without slip corresponds to this meridian. I read about it first from Blashke's book on differential geometry ( German ).