798. THEOREM. The area of a lune is to the area of the surface of its sphere as the angle of the lune is to 360. Given: Lune ABCDA on sphere 0; L= area of lune; s: = area of sphere; great EB whose pole is A. To Prove: L:S=ZA : 360. Proof: I. If arc BD and the circumference of O EB are commensurable. B There exists a common unit of measure. Suppose this unit contained 5 times in BD; 32 times in the circumference. ..arc BD circumference = 5:32 (?). That is, arc BD: 360 = 5:32. Arc BD measures ▲▲ (726). Pass great circles through the several points of division of circumference EB and vertex 4, dividing the surface of the sphere into 32 equal lunes. Then, L: 85 : 32 (Ax. 3). Hence, L: SA: 360 (Ax. 1). = Q.E.D. II. If the arc and circumference are incommensurable. The proof is similar to that found in 244, 302, 368, 539, etc. 799. THEOREM. The number of spherical degrees in the area of a lune is double the number of degrees in its angle. Proof: Let L° denote the area of the lune, expressed in spherical degrees. Then, L° : s = ▲ ▲ : 360 (?) (798). That is, L°: 720=Z A: 360 (Ax. 6). Hence, L° Q.E.D. 800. FORMULA. =24A. Proof: L: S = ZA : 360 (?) (798), and s = 4 πR2 (?). NOTE. The unit of measure in this formula is the square unit. Q.E.D. 801. THEOREM. Two lunes on the same or equal spheres are to each other as their angles. Proof: L:S=≤ ▲ : 360, and L' : s = Z a' : 360 (?) (798). Dividing, L : L' = Z A : Z 4' (Ax. 3). 802. THEOREM. Q.E.D. Two lunes on unequal spheres, having equal angles, are to each other as the squares of the radii of the spheres. Proof: L:s=▲ ▲ : 360, and L' : s' = ≤ 4 : 360 (?). Hence, L: S = L': S' (Ax. 1). .. L : L' = S : S' = R2 : R'2 (292 and 795). 803. THEOREM. The number of spherical degrees in a spherical triangle is equal to the spherical excess of the triangle. Lune CAC'BC = ▲ ABC+▲ AC'B≈ ▲ ABC +▲ A'B'C (Ax. 6). Now, A ABC+AA'B'Clune CAC'BC, and AABC+▲ A'BC= lune ABA'CA, and ▲ ABC+▲ AB'C= - lune BAB'CB } (Ax. 4). Adding, 2 ▲ ABC +▲ ABC + ▲ A'B'C+▲ A'BC+▲ AB'C lune Alune B+ lune C (Ax. 2). But, ▲ ABC+▲ A'B'C+▲ A'BC + ▲ AB'C = 360 spherical degrees (?) (713). Lune 4 + lune B + lune C = 2A + 2 Z B + 2 Z c (799). : Z A .. 2 ▲ ABC + 360 = 2 ≤ 4 + 2 Z B + 2 Z C (Ax. 6). Hence, ▲ ABC= ZA+ZB+ C-180 = E (Ax. 2, Ax. 3, 781). Q.E.D. There are E spherical degrees in a spherical ▲ (?) (803). .. the area of a spherical triangle = 4 πR2 × E. Q.E.D. NOTES. The unit of measure in 803 is a spherical degree. The area of a spherical triangle is determined by its angles. 805. THEOREM. The number of spherical degrees in a spherical polygon is equal to its spherical excess. Given: A spherical n-gon. To Prove: The number of spherical degrees in this n-gon = the excess of the polygon. Ө Proof: From any vertex draw diagonals, dividing the polygon into (n − 2) ▲; the sums of the of these ▲ are denoted by 8, 81, 82, etc. Now, the number of sph. degrees in one A=8-180° (803); the number of sph. degrees in another A= 81 — 180° (?). Etc., for (n-2) A. Adding, the number of sph. degrees in the n-gon = the sum of its - (n-2) 180° (Ax. 2). Excess of the n-gon = sum of its —(n-2) 180° (788). .. the number of spherical degrees = the excess of the polygon (Ax. 1). in a spherical polygon Q.E.D. Ex. 1. Find the area of a spherical triangle whose angles are 80°, 125°, and 95°, on a sphere whose radius is 6.3 in. Ex. 2. Find the area of a spherical polygon whose angles are 135°, 105°, 85°, 155°, 120°, on a sphere whose radius is 15 ft. 806 THEOREM. The volume of a sphere = 4 TR3 3 Given Sphere 0; radius = R; surfaces; volume = V. are all at the center, and whose common altitude is R. The volume of one such pyramid = R its base (?) (625). .. volume of all the pyramids = R · sum of all their bases (Ax. 2); that is, 'R. S'. v' = Indefinitely increase the number of faces of the polyhedron, thus indefinitely decreasing each face, and 'will approach Vas a limit Hence, RS' will approach R S as a limit (?). 4 TR3 :. V= 3 (Ax. 6). Q.E.D. 807. THEOREM. The volumes of two spheres are to each other as the cubes of their radii or as the cubes of their diameters. 808. THEOREM. The volume of a spherical pyramid is equal to one third the product of the polygon that is its base, by the radius of the sphere. Proof: Similar to the proof of 806. 809. THEOREM. The volume of a spherical wedge is to the volume of the sphere as the angle of its base is to 360. Proof: Similar to the proof of 798. 810. THEOREM. The volume of a spherical sector is equal to one third the product of the zone that is its base by the radius of the sphere. Proof: Similar to the proof of 806. 811. FORMULAS. Vol. of a spherical sector = R· zone (810). But the zone=2πR.h (?). Therefore, 1. The volume of spherical sector R·h (Ax. 6). = 2. The volume of a spherical cone=2πR2.h (811, 1). 812. PROBLEM. To find the volume of a spherical segment. 1. Spherical segment of one base. Given: Spherical segment generated by the figure ACX; semicircle XAY; AC=r; radius of sphere R; altitude = CX=h. = Required: To find the volume of the spherical segment. Computation: Draw chords AX, AY, and radius 40. The right A ACO will generate a cone of revolution (?) (671). C The volume of spherical segment ACX= volume of spherical cone OAX minus volume of cone ACO. Volume of spherical cone OAX= TR2 · h (?) (811, 2); = volume of cone ACO 2 Co (?) (695). = Now r2 CX CY=h (2 R− h) (?) (340, II); and COR-h. .. vol. ACO = }πh (2 R − h) (R − h) (Ax. 6). Hence, volume of spherical segment ACX = TR2h - ( + R2h — π Rh2 + Th3) (Ax. 6). That is, volume of spherical segment of one base = } Th2 (3 R − h). 400 394 355 356 |