than LC, one point of division, at least, will fall between L and C. Let I be such a point, and pass the arc of a great circle through A and I. Now, surface of sphere: lune AIEB :: BCmDn: BI, (2) since the arc BI is commensurable with the circumference. In (1) and (2), the antecedents being equal, the consequents should be proportional, hence we should have lune ACEB: lune AIEB :: BL: BI. But this is absurd, since lune ACEB > lune AIEB, whereas BL < BI. In a similar manner we can show that surface of sphere is not to lune ACEB:: BCmDn any arc greater than BC. Hence, as the fourth term can neither be less nor greater than BC, it must be equal to BC, and we have surface of sphere: lune ACEB :: BCmDn: BC, i. e., as 4 right angles, to the angle of the lune. 610. SCH. 2.—To obtain the area of a lune whose angle is known, on a given sphere, find the area of the sphere, and multiply it by the ratio of the angle of the lune (in degrees) to 360°. Thus, R being the radius of the sphere, 47R2 is the surface of the sphere; and the lune whose angle is 30° is or the surface of the sphere, i. e., 1⁄2 of 4πR2 = πR2. PROPOSITION XXIX. 611. Theorem.-If two semicircumferences of great circles intersect on the surface of a hemisphere, the sum of the two opposite triangles thus formed is equivalent to a lune whose angle is that included by the semicircumferences. DEM.-Let the semicircumferences CEB and DEA intersect at E on the surface of the hemisphere whose base is CABD; then the sum of the triangles CED and AEB is equivalent to a lune whose angle is AEB. For, let the semicircumferences CEB and DEA be produced around the sphere, intersecting on the opposite hemisphere, at the extremity F of the diameter through E. Now, FBEA is a lune whose angle is AEB. Moreover, the triangle AFB is equivalent to the triangle DEC: since angle AFB = AEB = DEC, side AF = side ED, each being the supplement of AE; and BF CE, each being the supplement of EB. Hence, the sum of the triangles GED and AEB is equivalent to the lune FBEA, Q. E. D. = FIG. 350. PROPOSITION XXX. 612. Theorem.-The area of a spherical triangle is to the area of the surface of the hemisphere in which it is situated, as its spherical excess is to four right angles, or 360°. DEM.-Let ABC be a spherical triangle whose angles are represented by A, B, and C; then is surf. of hemisphere :: A + B + C 180° : 4 right angles, or 360°. FIG. 351. But, by (607), ...... Let lune A represent the lune whose angle is the an gle A of the triangle, i. e., angle CAB, and in like manner understand lune B and lune C. Now, triangle AHG+ AED = lune А (611), BHI + BEF = lune B, CGF+ CDI = - lune C. Adding, 2ABC + hemisphere = lune (A + B + C)*, (1) since the six triangles AHG, AED, BHI, BEF, CGF, and CDI, make the whole hemisphere and 2ABC besides, ABC being reckoned three times. From (1), we have by transposing and remembering that a hemisphere is a lune whose angle is 180°, and dividing ABC = lune (A + B + C — 180°). lune (A + B + C — 180°) : surf. of hemisph. : : A + B + C — 180° : 4 right angles. Therefore, ABC: surf. of hemisph. :: A + B + C - 180° : 4 right angles. 613. SCH. 1.-To find the area of a spherical triangle on a given sphere, the angles of the triangle being given, we have simply to multiply the area of the hemisphere, i. e., 2πR2, by the ratio of the spherical excess to 360°. Thus, if the angles are A = = 110°, B = 80°, and C = 50°, we have A+B+C-180° 360° area ABC = 2πR2 × 60 = 2πR2 × 360 614. SCH. 2.—This proposition is usually stated thus: The area of a spherical triangle is equal to its spherical excess multiplied by the trirectangular triangle. When so stated the spherical excess is to be estimated in terms of the right angle; i. e., having subtracted 180° from the sum of its angles, we are to divide the remainder by 90°, thus getting the spherical excess in right angles. In the example in the preceding scholium, the spherical excess estimated in this 110° +80° 50° — 180° way would be = 3; and the area of the triangle would 90° *This signifies the lune whose angle is A + B + C, which is of course the sum of the three lunes whose angles are A, B, and C. be of the trirectangular triangle. Now, the trirectangular triangle being of the surface of the sphere (577) is of 4πR', or R'. This multiplied by gives R2, the same as above, The proportion, ABC surf. of hemisph. :: A + B + C 180° : 360°, -- is readily put into a form which agrees with the enunciation as given in this scholium. Thus, surf. of hemisph. =2πR2, whence 615. Theorem.-The volume of a sphere is equal to the area of its surface multiplied by of the radius, that is, πR, R being the radius. = DEM.-Let OL R be the radius of a sphere. Conceive a circumscribed cube, that is, a cube whose faces are tangent planes to the sphere. Draw lines from the vertices of each of the polyedral angles of the cube, to the centre of the sphere, as BO, CO, DO, AO, etc. These lines are the edges of six pyramids, having for their bases the faces of the cube, and for a common altitude the radius of the sphere (?). Hence the volume of the circumscribed cube is equal to its surface multiplied by R. a P D B FIG. 352. Again, conceive each of the polyedral angles of the cube truncated by planes tangent to the sphere. A new circumscribed solid will thus be formed, whose volume will be nearer that of the sphere than is that of the circumscribed cube. Let abc represent one of these tangent planes. Draw from the polyedral angles of this new solid, lines to the centre of the sphere, as ao, bo, and co, etc.; these lines will form the edges of a set of pyramids whose bases constitute the surface of the solid, and whose common altitude is the radius of the sphere (?). Hence the volume of this solid is equal to the product of its surface (the sum of the bases of the pyramids) into R. Now, this process of truncating the angles by tangent planes may be conceived as continued indefinitely; and, to whatever extent it is carried, it will always be true that the volume of the solid is equal to its surface multiplied by R. Therefore, as the sphere is the limit of this circumscribed solid, we have the volume of the sphere equal to the surface of the sphere, which is 47R, multiplied by R, i. e., to πR3. Q. E. D. 616. COR.-The surface of the sphere may be conceived as consisting of an infinite number of infinitely small plane faces, and the volume as composed of an infinite number of pyramids having these faces for their bases, and their vertices at the centre of the sphere, the common altitude of the pyramids being the radius of the sphere. 617. A Spherical Sector is a portion of a sphere generated by the revolution of a circular sector about the diameter around which the semicircle which generates the sphere is conceived to revolve. It has a zone for its base; and it may have as its other surfaces one, or two, conical surfaces, or one conical and one plane surface. a ILL. Thus let ab be the diameter around which the semicircle aCb revolves to generate the sphere. The solid generated by the circular sector AOα will be a spherical sector having a zone (AB) for its base; and for its other surface, the conical surface generated by AO. The spherical sector generated by COD, has the zone generated by CD for its base; and for its other surfaces, the concave conical surface generated by DO, and the convex conical surface generated by CO. The spherical sector generated by EOF, has the zone generated by EF for its base, the plane generated by EO for one surface, and the concave conical surface generated by FO for the other. b 618. A Spherical Segment is a portion of the sphere included by two parallel planes, it being understood that one of the planes may become a tangent plane. In the latter case, the segment has but one base; in other cases, it has two. A spherical segment is bounded by a zone and one, or two, plane surfaces. PROPOSITION XXXII. 619. Theorem.-The volume of a spherical sector is equal to the product of the zone which forms its base into one-third the radius of the sphere. DEM.-A spherical sector, like the sphere itself, may be conceived as consisting of an infinite number of pyramids whose bases make up its surface, and whose common altitude is the radius of the sphere. Hence, the volume of the sector is equal to the sum of the bases of these pyramids, that is, the surface of the sector, multiplied by one-third their common altitude, which is one-third the radius of the sphere. Q. E. D. 620. COR.-The volumes of spherical sectors of the same or equal spheres are to each other as the zones which form their bases; and, since these zones are to each other as their altitudes (604), the sectors are to each other as the altitudes of the zones which form their bases. PROPOSITION XXXIII. 621. Theorem.-The volume of a spherical segment of one base is πА (RA), A being the altitude of the segment, and R the radius of the sphere. DEM.-Let CO = R, and CD = A; then is the volume of the spherical segment generated by the revolution of CAD about CO equal to πA (R — }A). For, the volume of the spherical sector generated by AOC is the zone generated by AC, multiplied by R, or 2πAR × }R = }πAR2. From this we must subtract the cone, the radius of whose base is AD, and whose altitude is DO. To obtain this, we have DO = RA: whence, from the right angled triangle ADO, AD = √/R2 - (RA)2 = 1/2AR - A2. Now, the volume of this cone is OD × πAD2, or π(R — A) (2AR — A2) = }π(2AR2 Subtracting this from the volume of the spherical sector, we have πAR2 π(2AR2 3A2R + A3) - = π(A2R — }A3) = πA2(R − ¦À). Q. E. D. 622. SCH.-The volume of a spherical segment with two bases is readily obtained by taking the difference between two segments of one base each. Thus, to obtain the volumes of the segment generated by the revolution of bCAc about aO, take the difference of the segments whose altitudes are ac and ab. A FIG. 354. 3A'R + A3). a FIG. 355. EXERCISES. 1. What is the circumference of a small circle of a sphere whose diameter is 10, the circle being at 3 from the centre? Ans., 25.1328. 2. Construct on the spherical blackboard a spherical angle of 60°. Of 45°. Of 90°. Of 120°. Of 250°. |
