Question: So my question is this:
$$V=\frac{4}{3}\pi r^3$$ 
And, 
$$\frac{dV}{dr}=4\pi r^2=SA$$
Is this a coincidence or are there some mathematical hoodoos that I'm unaware of?

P.S. are there any more tags that I should use?

Answer: 
How do you obtain the volume of a sphere? You just calculate a volume integral on the sphere

$$V=\int\_{\mathbb{R}^3}\chi(x,y,z)\;d\mathbf{x}$$

where $\chi$ is a function that equals $1$ inside the sphere and $0$ outside. Of course is comfortable to switch to spherical coordinates. The determinat of the Jacobian is $|J|=r^2\sin\theta$, s0

$$ V=\int\_0^Rr^2dr\int\_0^\pi \sin\theta d\theta\int\_0^{2\pi}d\varphi $$

calculating the two rightmost integral you obtain

$$ V=\int\_0^R4\pi r^2dr =\int\_0^R \frac{dV}{dr}dr \tag{1}$$

How do you calculate the surface area of a sphere? Through a surface integral

$$SA=\int\_{\mathbb{R}^3}\sigma(x,y,z)\;d\mathbf{x}$$

where $\sigma(x,y,z)=\chi(x,y,z)\delta (r-R)$. In spherical coordinates:

$$ SA=\int\_0^\pi r^2\sin\theta d\theta\int\_0^{2\pi}d\varphi = 4\pi r^2 \tag{2}$$

So, confronting equation $(1)$ and $(2)$ is possible to prove that

$$SA=\frac{dV}{dr}$$

This, of course, means that

$dV=SAdr$

i.e., the infinitesimal increment of the Volume $dV$ is obtained through the product of the surface area $SA$ and the infinitesimal increment of the radius $dr$.
I think I have something but I'm not sure what. 

The volume of an object is the integral $\int 1 d\tau$ where $\tau$ is the volume element and the boundaries are the boundary of the surface. 

The divergence of $\vec{r}/3$ is 1 in any co-ordinate system. So the above integral can be expressed as the volume integral of a divergence. Where $\vec{r}$ is the position vector. 

By Gauss' Law $\int \nabla\cdot (\vec{r}/3) d\tau=\int \frac{\vec{r}}{3}\cdot \hat{n} dA$ 

Where $\hat{n}dA$ is the vector form of the infinitesimal surface area. 

For a figure of constant radius, $\hat{n}=\hat{r}$, so the integrand on the right becomes (r/3). 

I'm not sure what comes next, but there does seem to be a tie in there between the entire volume and only considering geometric features of the boundary.