Is it simply a coincidence that if you differentiate the formula for the volume of sphere you get the formula for the surface area of sphere? 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?
 A: 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 $dr$.
A: 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.
A: 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. 
