Consider a sphere for example.

It's volume is calculated by the formula: $\frac 4 3 \pi r^3$

The derivative of that is $4\pi r^2$ which represents the sphere's surface area.

The derivative of that is $8\pi r$.

Does that quantity represent something tangible about a sphere? Clearly, it is some sort of linear measurement, but of what, I'm not sure.

Just curious.

  • $\begingroup$ Consider the disc and the circle. $\endgroup$ Aug 3, 2017 at 22:39
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    $\begingroup$ @Salahamam, yes I see that the derivative of the circle's area is it's circumference, but I am having trouble picturing what the Surface Circumference of a sphere is? Why 8πr? $\endgroup$
    – Octopus
    Aug 3, 2017 at 22:42
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    $\begingroup$ My other related question is: integrate the circle's area and you get 1/3⋅πr³, but what is that to the circle? Maybe nothing, but maybe there is a real way to visualize what these quantities represent. $\endgroup$
    – Octopus
    Aug 3, 2017 at 22:45
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    $\begingroup$ If you think of the derivative of the volume as $\lim\limits_{h\to0}\frac{V(r+h)-V(r)}h$, and interpret it geometrically, it becomes easier to see why it should equal $SA(r)$. However, $\lim\limits_{h\to0}\frac{SA(r+h)-SA(r)}h$ has no obvious geometrical significance. $\endgroup$ Aug 3, 2017 at 22:50
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    $\begingroup$ Try thinking of the sphere as covered with lines of latitude and longitude, like a globe, but very finely, so the little areas are almost plane. Now inflate the sphere. The areas will all tear apart like postage stamps and these tears will expose a bit more of the area. It has to be the case that the new area exposed is $8\pi r\;dr$, and it would also be the case if the stamps were defined by arbitrary surface coordinates. Does that line of thinking lead anywhere? $\endgroup$
    – Philip Roe
    Aug 3, 2017 at 23:07

3 Answers 3


Maybe first review why the derivative of the volume is the surface area: It's because $4\pi r^2 \,dr$ is the incremental change in volume when you increase the radius by $dr.$ It makes sense that $4\pi r^2$ is the surface area since the then $4\pi r^2\,dr$ is the volume of a spherical shell, which represents the added volume.

So it's certainly true that $ 8\pi r \,dr $ is the incremental change in surface area when we increase the radius of the sphere by $dr.$ But is there a nice way to picture the difference in surface area in terms of a geometrical object? I can't think of any.


Take a step back and remember what a derivative is: the rate of change.

Imagine a sphere, then imagine painting it. The amount of paint you use is based on the surface area, and becomes part of the volume of a new sphere. The rate of change of the volume of the sphere is equal to the surface area of the sphere. The outside of the paint is the new boundary of the sphere, and the inside of the paint is added to the volume. This explains why the derivative (rate of change) of the volume is the surface area (SA).

In $4$-dimensional space, the SA analogue is the derivative of the Volume analogue of a $4$D sphere.

From wikipedia ($3$-sphere):

The $3$-dimensional cubic hyperarea of a $3$-sphere of radius $r$ is $$2 \pi^2 r^3$$ while the $4$-dimensional quartic hypervolume (the volume of the $4$-dimensional region bounded by the $3$-sphere) is $$\frac{1}{2} \pi^2 r^4$$

If we go back to $3$-space, the surface area of a cube is the derivative of volume of a cube if you use $\frac{s}{2}$ as your basis of measurement (this reflects that we use radius not diameter for most calculations of spheres).

The usual way to write it is:


$$\text{SA} = 6s^2$$

Instead use ($\frac{s}{2})$:

$$V = 8(\frac{s}{2})^3$$

$$\text{SA} = 24(\frac{s}{2})^2 $$

When simplified, the formulas are the same, but in the second example the SA is the derivative of the volume. Similarly, the perimeter of a square is the derivative of its area if you use ($\frac{s}{2}$).

I don't know of any other solids for which this is true. (A regular icosahedron almost works, but this isn't horseshoes.)


A general identity says \begin{align} & \Big(\text{size of boundary}\Big) \times \Big( \text{rate of motion of boundary} \Big) \\[5pt] = {} & \Big( \text{rate of change of size of bounded region} \Big) \tag 1 \end{align} (This equality has no standard name as far as I know; I have sometimes called it the boundary rule.)

It can be looked at this way: $$ \frac{d(\text{size of bounded region})}{d(\text{location of boundary})} = \text{size of boundary} $$ In particular $$ \frac{d(\text{volume of sphere})}{d(\text{radius of sphere})} = \text{surface area of sphere} $$ Can we view the surface of the sphere as growing by virtue of the motion of a boundary, when the radius grows? I don't know how to do that, nor how to find any sort of size of such a supposed boundary, nor what it would mean to speak of the amount by which the location of such a boundary changes (and hence to speak of a rate of motion of the boundary).

Therefore I don't see how such a geometric interpretation of the second derivative can be given. However, the fact that I don't know how to do it doesn't mean it can't be done. But it does make one wonder why I am typing this answer. Part of the reason is that I do know how to do such a thing with a cube.

Consider the cube $[0,s]^n$ in $n$-dimensional Euclidean space. As $s$ changes, there are $n$ moving boundaries, each of which is an $(n-1)$-dimensional cube $$ \underbrace{ \cdots \times [0,s]\times \cdots \times [0,s]} {} \times \{s\} \times {} \underbrace{[0,s]\times \cdots \times [0,s]\times\cdots} {}. $$ The volume of this boundary is $s^{n-1}$ and its rate of motion is the rate of change of $s,$ and thus from $(1)$ we conclude that $\dfrac{ds^n}{ds} = ns^{n-1}.$ The factor $n$ in front of $s^{n-1}$ comes from the fact that that is how many such boundaries there are.

So what about the second derivative? Some of those $n$ components of the whole boundary interface with others and some interface with the motionless $(n-1)$-dimensional coordinate hyperplanes. Those latter interfaces make up the "boundary of the boundary". Each of the $(n-1)$-dimensional components of the boundary of the whole cube has $n-1$ such $(n-2)$-dimensional components of its boundary. Hence $n(n-1)s^{n-2}$ is the size of the "boundary of the (moving part of the) boundary".

How can such an idea be applied to the sphere? At this point I don't know . . .

  • $\begingroup$ I think you are trying to explain that it represent the perimeter(Boundary) of the 2d object required to wrap completely the surface area(Boundary) of a 3d object. 8πr indicate that the Surface Area of a spere can be wrapped by 4 circles of perimeter "r" = 4*2πr : This has been proved in this video youtube.com/watch?v=GNcFjFmqEc8 I sadly don't have enough reputation to provide a long answer so if you could add this yo your answer it would be awesome :) Have an awesome day. Dan $\endgroup$
    – Mr Rubix
    Nov 14, 2023 at 19:26

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