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?

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    $\begingroup$ See this note of Dorff and Hall, which has been proposed as the answer to a question very similar to yours here on Math.SE. Unfortunately, I cannot find the precise reference. $\endgroup$ – Giuseppe Negro Jan 30 '19 at 11:22
  • $\begingroup$ You might find the explanations here to be helpful $\endgroup$ – Ben Grossmann Jan 30 '19 at 11:22
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    $\begingroup$ Just saying the area of a circle is $S=\pi r^2$ while the perimeter is $P=2\pi r=\frac{dS}{dr}$ $\endgroup$ – abc... Jan 30 '19 at 11:23
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    $\begingroup$ Intuitively you have $V(r+\varepsilon) \approx V(r) + \varepsilon A(r)$, as the difference in volume is about $\varepsilon$ times the area of the sphere. To make this rigorous you need to make sure that the error is of order $o(\varepsilon)$. So definitely not a coincidence. :-) $\endgroup$ – Klaus Jan 30 '19 at 11:23
  • $\begingroup$ For people's interest: here is a similar but I think rather less obvious question. $\endgroup$ – David Jan 30 '19 at 23:13

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.

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    $\begingroup$ As you can see, nothing here uses the fact that we have a sphere. This argument is equally valid for any family of objects parametrised by a single parameter $r$, such as a cube, or a rectangular prism of constant width and height (and length $r$), or in other dimensions like coupling area to circumference, and so on. As long as things are continuous and differentiable, this approach tells you that the derivative of the volume is the surface area. $\endgroup$ – Arthur Jan 30 '19 at 11:40
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    $\begingroup$ Just to add some keywords: What you describe is essentially the measure-theoretic version of Fubini's theorem, also knwon as Cavalieri's principle. $\endgroup$ – MaoWao Jan 30 '19 at 12:07
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    $\begingroup$ @Arthur what you say in the comment is not exact. Think about a cube: the volume is $V=l^2\times l=A\times l$, where $l$ is the length of a side. $l^2$ is therefore the area of a single face, not the total surface area as it happens with a sphere. $\endgroup$ – Alessandro Zunino Jan 30 '19 at 14:25
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    $\begingroup$ @Arthur. Sorry, I accidentally deleted my previous comment. Yes, I gave the example of a $r \times r \times 2r$ cuboid. But actually example of a cube of $r\times r \times r$ also does not work, because increasing $r$ puts a layer over the cube of thickness $\Delta r/2$. It only really works if $r$ is like a radius, so a $2r\times 2r \times 2r$ cube. $\endgroup$ – Jaap Scherphuis Jan 30 '19 at 14:30
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    $\begingroup$ @AlessandroZunino You're right. I didn't think that one entirely through. The thickness must increase with $r$ everywhere. $\endgroup$ – Arthur Jan 30 '19 at 14:33

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


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


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


This, of course, means that


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 $dr$.

  • $\begingroup$ The integral $V=\int_{\mathbb{R}^3}\sigma(x,y,z)\;d\mathbf{x}$ where $\sigma$ is $1$ only on the surface of a sphere does not and should not give you a surface area: it should give you $0$, because any 2D surface in $\mathbb R^3$ has measure $0$. And if you do exactly the same thing for surface area as you did for volume, you would write down $$V=\int_R^Rr^2dr\int_0^\pi \sin\theta d\theta\int_0^{2\pi}d\varphi$$ and the $\int_R^R$ would make things cancel to $0$. The integral you actually did write down is correct, but it is not justified through the integral of $\sigma(x,y,z)$. $\endgroup$ – Misha Lavrov Jan 30 '19 at 16:06
  • $\begingroup$ @MishaLavrov I disagree. I did not wanted to get the things too much complicated introducing generalized functions, but just write down $\sigma(x,y,z)=\chi(x,y,z)\delta (r-R)$ and see that everything will work exactly as I wrote. $\endgroup$ – Alessandro Zunino Jan 30 '19 at 16:09
  • $\begingroup$ Only if you have a "function" $\delta$ such that $\int_0^\infty \delta(r-R)\,dr = 1$, in which case $\sigma$ is not a real function either, and certainly does not have value $1$ on the surface of the sphere. (But other than the having-value-$1$ thing, I agree that integrating $\sigma(x,y,z)=\chi(x,y,z)\delta (r-R)$ over $\mathbb R^3$ will give you the surface area.) $\endgroup$ – Misha Lavrov Jan 30 '19 at 16:11
  • $\begingroup$ @MishaLavrov it is actually true that any 2D surface has measure zero on $\mathbb R^3$, because these are parametrized by proper functions. Dirac's delta is generalized function (or a distribution) that in this context is just a tool to explain something else, but you are very correct saying that it has not value $1$ on the surface of the sphere, is the integral over $r$ that has value $1$. I'll correct this. $\endgroup$ – Alessandro Zunino Jan 30 '19 at 16:19
  • $\begingroup$ @MishaLavrov I admit this is not the most formal way to define surface integrals, but I just did not wanted to get things too much complicated. $\endgroup$ – Alessandro Zunino Jan 30 '19 at 16:25

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.

  • $\begingroup$ The problem is, if the volume is not a sphere, $\hat r \cdot \hat n$ is not constant. So this only applies to spheres (or their $n$-dimensional equivalents for any $n$). For general shapes, there is no proportionality between change of volume and area. $\endgroup$ – Paul Sinclair Jan 30 '19 at 17:36
  • $\begingroup$ I think the particular relationship i mention doesn't apply, but I think something similar does apply in general, the part before I introduce a figure of constant radius. Consider a solid in 3D with equation $f(x,y,z)=0$. The gradient of f is normal to the surface. Divide by its magnitude to get the normal and let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ We get $d\tau=\frac{x\partial f/\partial x + y \partial f/ \partial y + z \partial f/\partial z}{3 norm} dA$ $\endgroup$ – TurlocTheRed Jan 30 '19 at 19:10
  • $\begingroup$ That is true, but it doesn't give you the $\dfrac {dV}{dr} = A$ formula the OP is asking for, $\endgroup$ – Paul Sinclair Jan 31 '19 at 0:06

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