Conference rusure::math

11.0. "Sphere and Tetrahedron" by HARE::STAN () Sun Jan 22 1984 12:27

A recent query in net.math on USENET asked for the volume of the
intersection of a unit sphere and a regular tetrahedron with one
vertex at the center of the sphere.  Here is my reply:

The volume of a regular tetrahedron (with side s) is s^3 sqrt(2)/12.

The volume of a sphere with radius r is 4/3 pi r^3.

If a plane intersects a sphere, it divides it into two pieces called
segments.  The altitude of a segment is the greatest distance from
the spherical part to the plane.  The volume of a segment of a sphere
with radius r and altitude h is 1/3 pi h^2 (3r-h) or equivalently,
1/6 pi h (3q^2+h^2) where q is the radius of the circular base of
the segment.

If you start with a segment of a sphere and join each point on the
circular base to the center of the sphere, the resulting figure
is called a sector of the sphere.  The volume of the sector is
2/3 pi r^2 h where h is the altitude of the segment. This can be
obtained by adding together the volumes of the segment and the
right circular cone with vertex at the center of the sphere.
(The volume of a right circular cone is 1/3 A h, where A is the
area of the base and h is the altitude).

Now, unforunately, the tetrahedron in question does not intersect
the sphere in a plane, so a sector is not formed.

Each face of the tetrahedron (emenating from the center of
the sphere) meets the sphere in an arc of a great circle. Thus a
spherical triangle is formed on the surface of the sphere by these
three arcs.  The area of a spherical triangle is pi r^2 E / 180
where E is the spherical excess of the triangle (in degrees).
The spherical excess is the sum of the angles of the spherical triangle
minus 180. (The angle between two circular arcs is the dihedral angle
between the planes determining these arcs.)

If you join each point on the spherical triangle to the center of the
sphere, you get a figure called a spherical pyramid.  This is the
true intersection of the tetrahedron and the sphere.  The volume of
a spherical pyramid is 1/3 r A where r is the radius of the sphere
ad A is the area of the spherical triangle.  In other words, the
desired volume is 1/540 pi r^3 E where E is the spherical excess of
the spherical triangle.

So we need to find the spherical excess of the spherical triangle formed.
This triangle is equilateral and each angle is equal to the dihedral angle
of the regular tethrahedron.  To get this angle, drop an altitude from one
vertex to a base and then drop a perpendicular in that base to one edge.
You form in this manner a right triangle with sides of length sqrt(6)/3,
sqrt(3)/6, and sqrt(3)/2.  The desired dihedral angle is the angle opposite
the side of length sqrt(6)/3.  Thus this angle has a sine of 2 sqrt(2) / 3.
The angle is thus approximately 70 degrees 32 minutes.  This makes the
spherical excess E=31.586338 degrees (approx). Using this value, we
calculate the volume of the spherical pyramid as approximately 0.183761866.
This is the desired volume.

I am not an expert on spherical trigonometry, so you should check the above
caculations carefully.  Two good references are:

James and James, Mathematics Dictionary. D. Van Nostrand Company, Inc.
	Princeton, N.J.: 1968.

L. Lines, Solid Geometry. Dover Publications, Inc. New York: 1965.

Another approach would be to look up the measure of the trihedral angle
of a regular tetrahedron.  Solid angles are measured in steradians,
which represents the area of the part of the unit sphere (centered at the
vertex of the solid angle) cut off by the solid angle.  The desired volume
should then be 1/3 the measure of this trihedral angle (in steradians).
Unfortunately, I looked in several reference books and couldn't find one
that listed this value.

	- Stanley Rabinowitz -

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