Can flux be proportional to $r^2$ in divergence theorem?

Question

The motivation for the divergence being interpreted as the flux of stuff used the following:

$$\text{div} F(a) = \lim_{r\to0}\frac{3}{4\pi r^3}\int_{|x-a|=r} F\cdot n dA$$

Without the $r^3$ in the denominator the right hand side would converge to $0$. But in the context of the divergence theorem what happens if it is $r^2$ instead, can the limit be nonzero?

Answer 1

$\mathbf F=1 \mathbf e_r$. Then the integral is just $4\pi r^2$ and the limit is 3.

I'm not clear on what you mean by "in the context of the divergence theorem", though. In three dimensions, the size of the sphere of radius $r$ is always proportional to $r^3$, and there is no way you would get the above formula with $r^2$ instead.

Of course, in two dimensions, this is exactly what one does find.

$\nabla\cdot \mathbf F=\lim \frac 1 {\pi r^2} \int_{\text{circle}} \mathbf {F\cdot dn}$

Try $\mathbf F = r \mathbf e_r$ for a simple example. Both sides should give 2.

Answer 2

The equation $$\text{div} F(a) = \lim_{r\to0}\frac{3}{4\pi r^3}\int_{|x-a|=r} F\cdot n dA$$ assumes that $F$ is continuously differentiable in the region $|x-a|\le r$. That is the hypothesis of the divergence theorem. As Sharkos's example shows, it does not hold for $F=x/|x|$, which is not defined at the origin. Another example is $F(x)=(0,0,1)$ when $x_3\ge0$, and $(0,0,0)$ when $x_3<0$. In that case the integral is $\pi r^2$.

So the short answer to your question is yes, but not when $F$ is continuously differentiable.