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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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(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?). – bryn Jul 24 '10 at 3:01
Its deep. Look at the most general version of the fundamental theorem of calculus… – Jonathan Fischoff Jul 24 '10 at 3:23
And next explain why it fails for the square... or the ellipse... – GEdgar Dec 6 '11 at 14:22
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. – Hexagon Tiling Jan 7 '12 at 11:47
@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... – Steven Stadnicki Apr 8 '12 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 $2\pi(r+dr)$. At 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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Inner radius of the annulus or inner circumference? – Huey Sep 4 at 10:31

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:

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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:

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

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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.

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It works for the cube as well. – Lucian Apr 24 '14 at 0:18

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.

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Bd}{\partial}\DeclareMathOperator{\vol}{vol}$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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Because you use the integral (read: anti-derivative) to find the area under the curve - even a curve in polar coordinates.

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This doesn't explain why the coefficients match up. – Ben Alpert Jul 24 '10 at 3:35
@Ben: Yes it does. Try reading 'integral' as 'anti-derivative'. – BlueRaja - Danny Pflughoeft Jul 24 '10 at 4:06
This answer is right, but I understand Ben Alpert's confusion. It's slightly unclear when I read it over again. – Justin L. Jul 24 '10 at 11:10
I hadn't thought of this. Does this also explain why it works in 3 dimensions? – bryn Jul 27 '10 at 4:59
I agree it doesn't really explain why the coefficients match up. If this answers the question, then how does this work with a square? Is the derivative of $x^2$ equal to $4x$? Your explanation works for one specific example. How about an explanation that really works in general? At the least, you need more details to explain it. – Graphth Dec 6 '11 at 13:10

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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You can use LaTeX pretty much as usual. Just enclose your formulas in dollar signs. For example, $\theta$ gives $\theta$. – t.b. Dec 6 '11 at 14:52

protected by T. Bongers Mar 5 '14 at 3:50

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