• Oct 4th 2011, 02:03 AM
agentmulder
The volume of a sphere is

V = 4/3(pi)r^3

If you differentiate with respect to r you get

dV/dr = 4(pi)r^2

This is the surface area of the sphere.

How come this works so well for the sphere? It doesn't work for the cube...

volume of cube is

V = s^3

differentiate with respect to s

dV/ds = 3s^2

this is only half the true surface area 6s^2

Is there something special about the sphere, within the calculus, that makes it work out perfectly or is it just a coincidence?

If you differentiate the surface area of the sphere with respect to r

d(SA)/dr = 8(pi)r

you get 4 times the circumference of a great circle on the sphere.

It would be nice if you got 2(pi)r but you don't, you get 8(pi)r.

On the other hand, differentiate the area of a circle with respect to r you get the circumference. If you differentiate the area of the square you get half the perimeter, or twice the side, either way, not as nice as the circle. You get half the surface area for the cube, you get half the perimeter for the square...coincidence?

Are there other geometrical objects that work out nicely by applying differentiation or are the sphere, circle the only ones?(Thinking)

 I found that this nice relationship between volume and surface are also works for hyperspheres.

http://en.m.wikipedia.org/wiki/N-sphere

[end edit]
• Oct 4th 2011, 07:35 AM
Opalg
The procedure does work for a cube, if you apply it in the right way. In the case of a sphere, you measure the volume and surface area in terms of the radius, which is the distance from a point on the sphere to the centre. You need to do the same thing for the cube. More precisely, if you have a cube with side s, then the distance x from the centre of the cube to a midpoint of a face of the cube is given by $x=\tfrac12 s$, or $s=2x.$ The volume of the cube is $s^3 = 8x^3.$ Differentiate that with respect to x and you get $24x^2 = 6s^2,$ which correctly tells you the surface area.