# Surface integral of an infinitesimal volume = 0?

Cracking open the texts for the first time in a while, I'm prepping for school with a review of vector calc. Question 1 asks to derive the expressions for divergence for cartesian, cylindrical and spherical coordinates, based on the equation

$$\nabla\cdot\vec{f} = \lim_{\Delta V \rightarrow 0} \frac{1}{\Delta V} \left[ \iint \vec{f}\cdot d\vec{s} \right]$$

(Also refreshing myself on latex, \ooint seems to not be a thing in SE?)

So for a given point $$P$$, and the vector $$\vec{f}(P)$$ at that specific point, I calculate the dot product between that vector and the outward facing surface area over the infinitesimal volume surrounding $$P$$.

The problem is, if I'm looking at both sides of that little $$dV$$, wrt each axis, the products should cancel out exactly. So for the surfaces normal to the x-axis, we get

$$f_x(P)\hat{i} \cdot dydz\hat{i} + f_x(P)\hat{i} \cdot dydz \left(\hat{-i}\right) = 0$$

I figure the most likely issues are that I either have something like a sign error, or that integrating $$\vec{f}\cdot d\vec{s}$$ should yield zero, and I'm not doing the calculus that I should be doing. I think it isn't the former, because $$\vec{f}(P)$$ should be the same regardless of which side you're taking the product with. Is it the latter?

• The code is oiint, but I can't get that one to work, either (though I could swear I've seen it before). – Paul Sinclair Jan 24 '19 at 13:09

Your problem is that you are treating $$\vec f$$ as constant over this volume, and the divergence of a constant vector field is $$0$$.
If the vector field is treated as the velocity of some flow, then the divergence at a point is the rate at which the flow comes out of the point (negative when the flow is into the point). If a "fluid" is emerging from a point, then you would expect that on opposite sides of a small volume around the point that the flow would be in opposite directions on opposite sides. This doesn't have to violate continuity, as the vector field of this outward flow can be $$0$$ at the point itself. And if you add in a large locally constant flow past the point, the full vector field is not required to be $$0$$, either. But the divergence of the constant flow is $$0$$, so the divergence only picks up the contribution of the emerging fluid.
So instead of treating $$\vec f$$ as the same everywhere, break it up into its even and odd components about $$P$$: $$\vec f_e(\vec v) = \frac 12(\vec f(P+\vec v)) + \vec f(P -\vec v)\\\vec f_o(\vec v) = \frac 12(\vec f(P+\vec v) - \vec f(P -\vec v))$$
While $$\vec f(Q) = \vec f_e(Q - P) + \vec f_o(Q - P)$$, you will find that the integration for $$\vec f_e$$ is indeed $$0$$, but for $$\vec f_o$$, the vector field has opposite signs in each direction, so it does not cancel out.