# Divergence Theorem on a surface without boundary

1. Nov 24, 2012

### lmedin02

Reading through Spivak's Calculus on Manifolds and some basic books in Analysis I notice that the divergence theorem is derived for surfaces or manifolds with boundary. I am trying to understand the case where I can apply the divergence theorem on a surface without boundary.

2. Nov 25, 2012

### lurflurf

There is always a boundary, it is an important part. The are times when the boundary (in whole or part) is at infinity, cancels due to symmetry, the function is constant, the integral is known, or some other situation; then the integral over the boundary might be replaced by some other term.

3. Nov 26, 2012

### lavinia

In 3 space every Surface that has no edges -e.g. a sphere or a torus - is the boundary of a solid. This is difficult to prove though.

On a manifold, one can have surfaces that are not boudaries of any solid but a divergence free fluid flows across them with non-zero total flux. For a one dimensional example take a circle on a torus that loops through the ring. The perpendicular flow has non-zero flux and is divergence free.

Last edited: Nov 26, 2012
4. Nov 28, 2012

### mathwonk

i would answer your question by saying that every manifold has a boundary, but the boundary may be the empty set. hence on a manifold without boundary, i.e. with empty boundary, the theorem still holds, but one side, the integral over the boundary, is zero.

this yields the "residue" theorem on a compact riemann surface, i.e. for every meromorphic one form on a compact riemann surface, the sum of its residues, i.e. the integral over the boundary, is zero.

that is avery useful application of the theorem. (the divergence theorem is a version of green's theorem used here, or stokes thorem....)

5. Nov 28, 2012

### lavinia

This confuses me. To calculate the residues one needs to excise disks around the singularities of the meromorphic 1 form. One then integrates over a surface with a finite number if circles as its boundary. No?

6. Nov 28, 2012

### mathwonk

it may be that the proof involves surfaces with boundary, but the resulting statement applies to surfaces without boundary.