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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?).2010-07-24
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    Its deep. Look at the most general version of the fundamental theorem of calculus http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Generalizations2010-07-24
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    See [section 3 of TauDay.com](http://tauday.com/) for a rather elegant explanation of this, and the author's reasoning for why the area of a circle should be $\frac{1}{2}\tau r^2$, in parallel to the other famous quadratic forms $\frac{1}{2}m v^2$ and $\frac{1}{2}g t^2$, to make it obvious that the area of a circle is, in fact, an integral of growing rings of circumferences.2010-07-24
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    And next explain why it fails for the square... or the ellipse...2011-12-06
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    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.2012-01-07
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    @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...2012-04-08
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    I'm surprised I hadn't posted an answer here before, but now I have; see below. And I humbly suggest that it's better than the one that's got 20 votes so far, and that one is good.2013-11-12

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