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 minimal surface [is a sphere](http://en.wikipedia.org/wiki/Isoperimetric_inequality#Isoperimetric_inequality_in_higher_dimensions).