The area of Marc's construction approaches 6*pi*r^2, while the surface of the sphere has area 4*pi*r^2. The surface is rough, as it has 50% more surface area than a smooth surface. Sketch of proof: 3 circular disks of radius r have area 3*2*pi*r^2 (counting both sides). If you arrange these 3 disks to intersect in 3 mutually perpendicular planes, this arrangement still has the same area. Now if you take Marc's spherical construction out of little cubes, you can remove the cubes, one by one, and each of these cube removal steps _does not change the total surface area_, until you reach the stage where you have 3 mutually perpendicular circular disks exactly one cube thick. (This area-preservation property comes from the fact that each cube removal removes 3 exposed sides of the removed cube, but exposes 3 new sides which were previously covered up.) Finally, if we allow the dimension of the cube to shrink towards zero, each disk's area (counting both sides) approaches 2*pi*r^2. QED. At 06:12 PM 1/7/2013, Henry Baker wrote:
No, that can't be right. What is the surface area of Marc's construction, which stuffs a sphere with cubes in a regular lattice, as the cubes get smaller & smaller ?
At 06:05 PM 1/7/2013, Henry Baker wrote:
Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere.