Perfectly Round Sphere on Perfectly Flat Floor I know that in practice it may be practically impossible to create the following situation, but suppose I did place a perfectly round ball of radius r (not that I think radius r is relevant) on a perfectly flat floor. Also suppose there are no forces acting upon the ball other than Gravity and the reaction of the floor. Finally suppose the ball and floor can not be compressed in any way.
Then what surface area of the ball will be in contact with the floor?
This is not a homework question but just something I was curious about. My thoughts are that an infinitely small area (tends to zero) of the ball touches the floor but I am not sure.
 A: The sphere only contacts the floor at one point, so there's zero surface area. 
Here's an argument that hopefully blends the physical intuition and mathematical rigor you want: For the sake of contradiction, imagine what would happen if there were two or more points of intersection: Viewed in the floor, these two points are connected by a straight line segment. The sphere and the region it bounds form a ball, which is convex and thus contains the whole line segment. But the rigid sphere and floor cannot pass through eachother, so the line segment must lie entirely on the sphere (and not in the interior of the ball). This is a contradiction because a perfectly round sphere contains no straight line segments (since no line segment can have all of its points be the same distance $r$ from another fixed point).
For another perspective, you can model this situation with equations: View the perfectly round sphere of radius $r$ as being defined by $x^2+y^2+z^2=r^2$ and the perfectly flat floor being the plane defined by $z=-r$ (for all $x,y$). The only point which satisfies both these equations (hence lies in both the sphere and the floor) is $(0,0,-r)$. The figure below depicts the case where $r=1$.

