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.
Best Answer
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.