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: 
Start with a sphere of radius $r$. Now let the radius of the sphere grow by some tiny amount $\Delta r$. How much has the volume changed? By the definition of the derivative, it has changed by approximately
$$
\Delta r \cdot V'(r)
$$
However, the added volume is basically a thin shell, and its volume is approximately equal to its surface area (the inner one, for convenience), multiplied by its thickness. This is 
$$
\Delta r\cdot SA(r)
$$
Thus we have
$$
\Delta r \cdot V'(r)\approx \Delta r\cdot SA(r)\\
V'(r)\approx SA(r)
$$

Rigorous analysis of this setup will allow you to conclude that the approximation error above is small enough as $\Delta r$ becomes smaller, and thus that $V'(r)$ and $SA(r)$ are indeed equal.
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 $dr$.