Can someone offer a general proof of the square cube law of surface area and volume growth? For any fixed three dimensional shape it's clear how to proceed, but considering a general shape, how can one show the volume scales as the cube of the varying parameter, while the surface area scales as the square? I imagine Stoke may have his name somewhere here.
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5The volume of a body in three-dimensional space is the limit of approximations by small cubes. Scaling all distances by a given factor scales all of those cubes by, well, the cube of said factor. Similarly, the area of a surface is the limit of approximations by (say) triangular meshes, and again we can ask what scaling all lengths by a given factor does to those triangular faces in any approximating mesh. – anon Sep 11 '15 at 00:24
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That makes sense. Still, shouldn't there be some proof that also includes the constant values that fall out? – Bob Sep 11 '15 at 01:20
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"the constant values that fall out"? – anon Sep 11 '15 at 01:25
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@anon What I meant is that while your argument is true, each different shape has some constant of proportionality in the growth. I'm wondering how one could actually calculate that number for an arbitrary shape, i.e. what the formula is. e.g. for a sphere we have $4/3\pi r^3/4\pi r^2 = \frac{1}{3}r$, so the "constant value that falls out" is $\frac{1}{3}$. – Bob Sep 12 '15 at 04:50
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Read my first comment again. Each of the small cubes in an approximation is scaled by the cube of your linear scaling factor. – anon Sep 12 '15 at 04:56