My unclarity, my bad. I wanted the volume of the intersection of a sphere with a pyramid whose axis of symmetry intersects the sphere's center. In the special case of an apex solid angle of 4π/6, a correct h will give 1/6 of the "fat cube" intersection of six spheres centered at the vertices of an octahedron, (not to be confused with two colliding alpha particles) gosper.org/fatcube.png , analogous to the area problem gosper.org/areaansa.png whose answer is clearly some π - some algebraic + some rational. --rwg On 2016-03-08 10:20, Brent Meeker wrote:
Volume of "round bottomed pyramid" = (s-1)[/]6 + (h-r+1/2)/3
where s is the volume of the sphere and r is it's radius.
That assumes that "at height h above the center of one such square" refers to the center of a "spherical square", not the center of the face of the cube.
Brent
On 3/8/2016 2:00 AM, rwg wrote:
--rwg Problem: Project a (unit, say) cube onto its circumsphere, producing six "spherical squares". What is the volume of the round-bottomed "pyramid" whose apex lies at height h above the center of one such square?