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?

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 1

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.

Answer 2

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 $dr$.

Answer 3

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.