Jtdylktuy

So my question is this:
$$V=frac{4}{3}pi r^3$$
And,
$$frac{dV}{dr}=4pi r^2=SA$$
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?

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
$$
Delta rcdot SA(r)
$$
Thus we have
$$
Delta r cdot V'(r)approx Delta rcdot SA(r)\
V'(r)approx SA(r)
$$

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

$$V=int_{mathbb{R}^3}chi(x,y,z);dmathbf{x}$$

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 $|J|=r^2sintheta$, s0

$$ V=int_0^Rr^2drint_0^pi sintheta dthetaint_0^{2pi}dvarphi $$

calculating the two rightmost integral you obtain

$$ V=int_0^R4pi r^2dr =int_0^R frac{dV}{dr}dr tag{1}$$

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

$$SA=int_{mathbb{R}^3}sigma(x,y,z);dmathbf{x}$$

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

$$ SA=int_0^pi r^2sintheta dthetaint_0^{2pi}dvarphi = 4pi r^2 tag{2}$$

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 $int 1 dtau$ 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 $int nablacdot (vec{r}/3) dtau=int frac{vec{r}}{3}cdot hat{n} dA$

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.