How to show Integral ds = 0 (see below, please)?

Question

Let S be a closed surface in R^3 and let ds be elementary area vector, that is, normal to S at a given point of S, oriented outwards and with magnitude |ds| equal to the elementary area. How can we show that Integral (over S) ds = 0?

Thank you

Answer 1

Let A be a constant non null vector field. Then, the flux F of A over S is given by Integral (over S) A . ds, where . means the dot product. Since A is constant, F = A. Integral (over S) ds.

According to the divergence theorem, F =Integral (over S) A . ds = Integral (over V) div A . dv, where V is the volume enclosed by the surface S and div is the divergent of A. Since A is constant, div A = 0 and F = Integral (over V) 0 dv = 0.

So, A. Integral (over S) ds = 0. Since this holds for every vector A in R^3, we must have Integral (over S) ds = 0, as stated.

Answer 2

I'm a little rusty on my calculus, but it sounds like you want to use something with the properties of divergence, maybe the divergence theorem for surfaces. Sorry I can't help more, but maybe this is the right direction. I have a feeling though that no matter what the surface actually is, if it's closed (and I suppose smooth, even though it's not specified) the divergence is zero when using unit vectors.