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Related to "volume of sphere", why is the surface area of a cube not equal to the derivative of its volume?

If you think about a sphere, it makes sense that the rate of change of the volume (with respect to $r$) yields the surface area ($SA$) = $\frac{\text{d}}{\text{d}r} \frac{4\pi r^3}{3} = 4\pi r^2 $, because we are "skinning" on "layers" of sphere. The surface area of a sphere is the rate of change of the volume of that sphere at radius R because an infinitely thin skin of sphere laid on top of the sphere is exactly the surface area.

So for a cube, you'd think the same thing. But $V_{\text{cube}} = x^3$ and $SA_{\text{cube}} = 6x^2$, where you'd expect $SA_{\text{cube}}=3x^2$ if surface area were just the derivative of volume.

I think it has to do with the cube-side length growing on two sides (hence $2 \text{d}x$?) but I can't quite put my finger on it.

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up vote 18 down vote accepted

If you express the length of the cube in terms of its half-length $y=x/2$, then the volume is $V=8y^3$ and the surface area is $S = 24y^2$, so you have $S=\mathrm{d}V/\mathrm{d}y$ just like for the sphere.

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To put it another way, if you did your sphere formulas in terms of the diameter instead of the radius, the area would be off by a factor of 2 from the derivative of the volume. – Gerry Myerson Jun 16 '11 at 23:55

If you compare a cube of side length $x$ to a cube of side length $x + \epsilon$, the difference is a bunch of slices corresponding to the faces, but they have width $\frac{\epsilon}{2}$, not $\epsilon$. As Chris Taylor says, to get the expected normalization you work in terms of half of the side length. Then you get

$$(2y + 2 \epsilon)^3 = 8(y^3 + \epsilon (3y^2) + \epsilon^2 (3y) + \epsilon^3)$$

where the Taylor coefficients are, in order: the volume $8y^3$, the surface area $24y^2$ times $\epsilon$, the total length $24y$ of the edges times $\epsilon^2$, and the number $8$ of vertices times $\epsilon^3$.

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Or to put it another way: if you think of your cube with one corner at the origin and edges parallel to the axes, growing the side length by $\epsilon$ adds a layer of thickness $\epsilon$ to the three faces that don't hit the origin, but not to the three that do. If you have your cube centred at the origin, growing the side length by $\epsilon$ adds a layer to all the faces, but it has thickness $\epsilon/2$. – Robert Israel Jun 16 '11 at 18:12

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