volume calculation..!!?

Question

i want to calculate the volume of the tetrahedron type figure formed in the folowing manner....take a tetrahderon of length r and a sphere of adius r....place the tetrahedron such its one corner lie on the center of sphere and other three corners on the surface of the sphere...from the three corners instead of joining through straight plane (triangular) take the curved surface of the sphere....
hence the figure has three faces of tetrahedron and the fourth one as a part of sphere..!!

Answer 1

The area of a spherical triangle is ((A+B+C)/180 - 1) π r², where r is the radius of the sphere and A, B, C are the angles of the spherical triangle. If A=B=C=60°, then the area would be zero. In this case of a tetrahedron's face projected onto the spherical surface, the angles A=B=C are actually slightly larger, 70.5288°, so that the area comes out to 0.551286 r². The reason for this slight increase over 60° is because the angle is projected onto a plane tilted slighly above the face. Since the total area of the sphere is 4 π r², the volume is simply that fraction of the total volume of the sphere, or

V = ((0.551286 r²)/( 4 π r²)) (4/3) π r³ = 0.183762 r³

The exact answer is actually:

(4/3)ArcTan((2-√3)^(3/2)) r³

As a note, the icosahedron has 20 equilateral triangle faces, but the tetrahedron in this problem covers 1/22.7947 of the spherical surface, so that's one way of proving that the icosahedron isn't composed of regular tetrahedrons clustered at a common center. As a matter of fact, for a sphere of radius 1, the inscribed icosahedron has side length of about 1.05146.