# Name of the intersections of a ball with the octants?

The eight regions of space defined by the eight possible combinations of signs for $(+/- , +/-, +/-)$ for $x$, $y$, $z$ are called octants.

Given a ball of radius 1 centered in the origin $(0, 0, 0)$. How are the eight sections called obtained by the intersection of the ball with the octants?

More generally, given a (regular) triangulation of the sphere (I hope such a thing exists). Now connecting such a spherical triangle with the origin how are the pieces cut out of the (unit) ball this way called?

Pointer to literature on these topics are appreciated.

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For your first question, I would think the easiest way to refer to it is simply as the portion of the sphere in the first octant, (or whatever octant you are referring to). This does invite the question of which octant should be considered the first, etc. There may be a better way to describe this, however. Secondly if you draw a triangle $T$ on a sphere, and then connect the three vertices of the triangle to the origin $\vec{0}$, the resulting solid could be called the convex hull of $T$ and $\vec{0}$ (which could be written as $Conv(T,\vec{0})$. – JavaMan Aug 25 '11 at 15:00
Given that "hemisphere" means half a sphere, from the Greek "hemi," I would like to coin the ogdoosphere, pronounced og-do-o-sphere, using the Greek "ogdoos" for eighth. Google reports no matching documents, so I think this is a safe new coinage. :-) – Joseph O'Rourke Aug 25 '11 at 16:48
@Joseph: If you want people to pronounce it og-do-o-sphere, perhaps a diaeresis would be helpful: ogdoösphere? – Rahul Aug 25 '11 at 17:55
@Rahul: Great idea! Makes it seem more exotic. :-) But perhaps a Greek scholar should be consulted... – Joseph O'Rourke Aug 25 '11 at 18:18