Given a volumen. Which is the suface, that contains it, that has minimal area?

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 1

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.

Answer 2

The minimal surface is a sphere.