[...] Here's a nice one from Neil: Gee, if the area of a sphere is d/dr of 4/3 Γ° r^3 = 4 Γ° r^2, then the area of a cube must be d/ds of s^3 = 3 s^2 ! This should take care of all sorts of polyhedra!
Yes, it does, but only when used properly. The inradius of a cube is half of the side length, so we have: V = 8 r^3 A = dV/dr = 24 r^2 (A sufficient condition for this method to work is that increasing the 'radius' by h causes all faces to move outwards (i.e. normal to the face) by a distance h. A special case of this is the inradius of a circumscribed polyhedron.)
--rwg Why don't you say sphere area = Γ° d^2? Are you secretly one of those Γ΄ people?
I don't see why people consider pi and tau to be incompatible. Here's a nice formula relating them: Γ΄ = Γ°^(Γ΄/Γ°) Sqrt[Γ°/Γ΄] Sqrt[Γ°/Γ΄ + Γ°/Γ΄ Sqrt[Γ°/Γ΄]] Sqrt[Γ°/Γ΄ + Γ°/Γ΄ Sqrt[Γ°/Γ΄ + Γ°/Γ΄ Sqrt[Γ°/Γ΄]]] ... Or, even better: Γ΄ = Γ°^(Γ΄/Γ°) (Γ°/Γ΄)^(Γ°/Γ΄) ((Γ°/Γ΄) + (Γ°/Γ΄) (Γ°/Γ΄)^(Γ°/Γ΄))^(Γ°/Γ΄) ((Γ°/Γ΄) + (Γ°/Γ΄) ((Γ°/Γ΄) + (Γ°/Γ΄) (Γ°/Γ΄)^(Γ°/Γ΄))^(Γ°/Γ΄))^(Γ°/Γ΄) ... Sincerely, Adam P. Goucher