When differentiated with respect to $r$, the derivative of $\pi r^2$ is $2 \pi r$, which is the circumference of a circle.

Similarly, when the formula for a sphere's volume $\frac{4}{3} \pi r^3$ is differentiated with respect to $r$, we get $4 \pi r^2$.

Is this just a coincidence, or is there some deep explanation for why we should expect this?

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    $\begingroup$ (I realise that it might not be clear what the $n$-dimensional generalisation is of this, but perhaps this would happen even in different geometries or metric spaces?). $\endgroup$
    – bryn
    Jul 24, 2010 at 3:01
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    $\begingroup$ Its deep. Look at the most general version of the fundamental theorem of calculus en.wikipedia.org/wiki/… $\endgroup$ Jul 24, 2010 at 3:23
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    $\begingroup$ And next explain why it fails for the square... or the ellipse... $\endgroup$
    – GEdgar
    Dec 6, 2011 at 14:22
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    $\begingroup$ You mentioned that it's true for the 2-sphere, and for the 3-sphere, but it should be noted that it is also true for the 1-sphere, which is the interval from -r to r, which has 1-volume of 2r. The derivative of 2r wrt r is 2, which is the measure of its "surface", measure for 0-dimensional items being the same as cardinality. $\endgroup$ Jan 7, 2012 at 11:47
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    $\begingroup$ @GEdgar : I make out the area of a square of 'radius' $r$ as $4r^2$ and the perimeter as $8r$; the idea continues to work there (for essentially the same uniformity reason that it does on the sphere). Of course, it doesn't work on rectangles for the same reason it doesn't work on ellipses... $\endgroup$ Apr 8, 2012 at 18:11

8 Answers 8


Consider increasing the radius of a circle by an infinitesimally small amount, $dr$. This increases the area by an annulus (or ring) with inner radius $2 \pi r$ and outer radius $2\pi(r+dr)$. As this ring is extremely thin, we can imagine cutting the ring and then flattening it out to form a rectangle with width $2\pi r$ and height $dr$ (the side of length $2\pi(r+dr)$ is close enough to $2\pi r$ that we can ignore that). So the area gain is $2\pi r\cdot dr$ and to determine the rate of change with respect to $r$, we divide by $dr$ and so we get $2\pi r$. Please note that this is just an informative, intuitive explanation as opposed to a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.

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    $\begingroup$ Inner radius of the annulus or inner circumference? $\endgroup$
    – Huey
    Sep 4, 2015 at 10:31

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Bd}{\partial}\DeclareMathOperator{\vol}{vol}$The formulas are no accident, but not especially deep. The explanation comes down to a couple of geometric observations.

  1. If $X$ is the closure of a bounded open set in the Euclidean space $\Reals^{n}$ (such as a solid ball, or a bounded polytope, or an ellipsoid) and if $a > 0$ is real, then the image $aX$ of $X$ under the mapping $x \mapsto ax$ (uniform scaling by a factor of $a$ about the origin) satisfies $$ \vol_{n}(aX) = a^{n} \vol_{n}(X). $$ More generally, if $X$ is a closed, bounded, piecewise-smooth $k$-dimensional manifold in $\Reals^{n}$, then scaling $X$ by a factor of $a$ multiplies the volume by $a^{k}$.

  2. If $X \subset \Reals^{n}$ is a bounded, $n$-dimensional intersection of closed half-spaces whose boundaries lie at unit distance from the origin, then scaling $X$ by $a = (1 + h)$ "adds a shell of uniform thickness $h$ to $X$ (modulo behavior along intersections of hyperplanes)". The volume of this shell is equal to $h$ times the $(n - 1)$-dimensional measure of the boundary of $X$, up to added terms of higher order in $h$ (i.e., terms whose total contribution to the $n$-dimensional volume of the shell is negligible as $h \to 0$).

The change in area of a triangle under scaling about its center

If $X$ satisfies Property 2. (e.g., $X$ is a ball or cube or simplex of "unit radius" centered at the origin), then $$ h \vol_{n-1}(\Bd X) \approx \vol_{n}\bigl[(1 + h)X \setminus X\bigr], $$ or $$ \vol_{n-1}(\Bd X) \approx \frac{(1 + h)^{n} - 1}{h}\, \vol_{n}(X). \tag{1} $$ The approximation becomes exact in the limit as $h \to 0$: $$ \vol_{n-1}(\Bd X) = \lim_{h \to 0} \frac{(1 + h)^{n} - 1}{h}\, \vol_{n}(X) = \frac{d}{dt}\bigg|_{t = 1} \vol_{n}(tX). \tag{2} $$ By Property 1., if $r > 0$, then $$ \vol_{n-1}\bigl(\Bd (rX)\bigr) = r^{n-1}\vol_{n-1}(\Bd X) = \lim_{h \to 0} \frac{(1 + h)^{n}r^{n} - r^{n}}{rh}\, \vol_{n}(X) = \frac{d}{dt}\bigg|_{t = r} \vol_{n}(tX). \tag{3} $$ In words, the $(n - 1)$-dimensional volume of $\Bd(rX)$ is the derivative with respect to $r$ of the $n$-dimensional volume of $rX$.

This argument fails for non-cubical boxes and ellipsoids (to name two) because for these objects, uniform scaling about an arbitrary point does not add a shell of uniform thickness (i.e., Property 2. fails). Equivalently, adding a shell of uniform thickness does not yield a new region similar to (i.e., obtained by uniform scaling from) the original.

(The argument also fails for cubes (etc.) not centered at the origin, again because "off-center" scaling does not add a shell of uniform thickness.)

In more detail:

  • Scaling a non-square rectangle adds "thicker area" to the pair of short sides than to the long pair. Equivalently, adding a shell of uniform thickness around a non-square rectangle yields a rectangle having different proportions than the original rectangle.

  • Scaling a non-circular ellipse adds thicker area near the ends of the major axis. Equivalently, adding a uniform shell around a non-circular ellipse yields a non-elliptical region. (The principle that "the derivative of area is length" fails drastically for ellipses: The area of an ellipse is proportional to the product of the axes, while the arc length is a non-elementary function of the axes.)

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    $\begingroup$ +1 A fantastic answer! It is a shame that this answer is not getting as much attention as it deserves. The standard answer to this question is effectively "Well, it works by geometry, but let's not worry about why $S^n$ is a very special case." Instead, this answer really focuses on why it works, and how it generalizes. A very good answer indeed. $\endgroup$ Nov 1, 2015 at 19:17
  • $\begingroup$ Thanks; glad you found it useful! It was pleasant to be able to add to a "first-thousand" post, not to mention getting to ponder a fun mathematical issue. $\endgroup$ Nov 1, 2015 at 21:45
  • $\begingroup$ Could this be related to a special case of Reynolds transport theorem or generalized Differentiation under the integration rule? en.m.wikipedia.org/wiki/… $\endgroup$
    – Mann
    May 29, 2020 at 21:57

The explanation is very simple. Take a sphere of radius $r$, volume $V$, and surface area $A$. Now paint it, with a layer of thickness $\delta r$. The volume of paint required is (to first order in $\delta r$) $A\delta r$, which gives you straight away: $$\delta V = A \delta r$$ Hence, in the limit:

$$\frac{dV}{dr} = A$$

  • $\begingroup$ Most intuitive answer here. $\endgroup$
    – AleksandrH
    Aug 23, 2019 at 18:29

There is an article on the web that deals, in depth, with this question. Here is a quote from it:

“We were intrigued by the students' work, and this paper is the result of our attempt to answer the question, “When is surface area equal to the derivative of volume?"”

Here is the link:


  • $\begingroup$ Thank you, I've visited that article three times in the last couple years, it seems to be the definitive word on the matter. I'd like to add another article, one that takes a less formal route (I figured here was the best place.) It also examines when the volume-area-circumference relationships apply, and generalizes them to 2D polygons and 3D polyhedra. I hope others will find this article as helpful as I have. apcentral.collegeboard.com/apc/members/courses/teachers_corner/… $\endgroup$ Apr 3, 2017 at 4:48

The circle (and sphere) is not really that special. It also works for the square if you measure it using not the side length $s$, but half that, $h=s/2$. Then its area is $A=(2h)^2=4h^2$ with derivative $dA/dh=8h$ which is its perimeter.

  • $\begingroup$ It works for the cube as well. $\endgroup$
    – Lucian
    Apr 24, 2014 at 0:18
  • $\begingroup$ ... and Andrew D. Hwang's answer explains why. $\endgroup$ Oct 10, 2018 at 23:05

The size of the boundary times the rate at which the boundary moves equals the rate at which the size of the bounded region changes.

There appears to be no conventional name for this fact. I've called it the boundary rule sometimes.


I recommend the article by J. Tong, Area and perimeter, volume and surface area, College Math. J. 28 (1) (1997) 57. He shows that for any region where the area can be written as $A(s)=c s^2$ and the perimeter as $L(s)= k s$, you can set $x=(2c/k) s$, and you will get $A'(x)=L(x)$. That means that by careful parametrization, the above holds for rectangles and ellipses, too.


How does one set up the integral to find the area of a circle? An area was defined for a square or rectangle to be the width times the length. It is the equivalent for all geometries. For a circle working in polar coordinates the differential area equivalent is $dr$ while the differential width would be $r \,d\theta$.

So... $$dA = r\, d \theta\, dr.$$ Here $r \,d\theta$ is the differential arc (width) times the differential length $dr$. You can see that by inspecting the form of this differential equation the fundamental form for finding the area of a circle is in the form of what we know to be the circumference of a circle. If we divide through by $dr$. So the connection is implicit in the basic geometry. Because we are working in a polar system.

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    $\begingroup$ You can use LaTeX pretty much as usual. Just enclose your formulas in dollar signs. For example, $\theta$ gives $\theta$. $\endgroup$
    – t.b.
    Dec 6, 2011 at 14:52

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