the radius EC; we say that the zone with one base, described by the revolution of the arc EF about EC, will have for its measure EG x circ. EC. For let us suppose, in the first place, that this zone has a less measure, and, if it be possible, let this measure be equal to EG × circ. AC. Inscribe in the arc EF a portion of a regular polygon EMNOPF, the sides of which shall not touch the circumference described with the radius CA, and let fall upon EM the perpendicular CI, the surface described by the polygon EMF, turning about EC, will have for its measure EG x circ. CI (533). This quantity is greater than EG × circ. AC, which, by hypothesis, is the measure of the zone described by the arc EF. Consequently the surface described by the polygon EMNOPF would be greater than the surface described by the circumscribed. arc EF; but on the contrary this last surface is greater than the first, since it encloses it on all sides; therefore the measure of any spherical zone with one base cannot be less than the altitude of this zone multiplied by the circumference of a great circle. We say, in the second place, that the measure of the same zone cannot be greater than the altitude of this zone multiplied by the circumference of a great circle. For, let us suppose that the zone in question is the one described by the arc AB about AC, and, if it be possible, let the zone AB be greater than ADX circ. AC. The entire surface of the sphere composed of the two zones AB, ADx circ. AC + DH × circ. AC; if then the zone AB be greater than AD × circ. AC, the zone BH must be less than DH x circ. AD, which is contrary to the first part already demonstrated. Therefore the measure of a spherical zone with one base cannot be greater than the altitude of this zone multiplied by the circumference of a great circle. It follows then that every spherical zone with one base has for its measure the altitude of this zone multiplied by the circumference of a great circle. Let us now consider any zone of two bases described by the Fig. 220. revolution of the arc FH (fig. 220) about the diameter DE, and let FO, HQ, be drawn perpendicular to this diameter. The zone described by the arc FH is the difference of the two zones described by the arcs DH and DF; these have for their measure respectively DQ x circ. CD and DO x circ. CD; therefore the zone described by FH has for its measure (DQ— DO) × circ. CD or OQ × circ. CD. We conclude then that every spherical zone with one or two bases has for its measure the altitude of this zone multiplied by the circumference of a great circle. 539. Corollary. Two zones are to each other as their altitudes, and any zone whatever is to the surface of the sphere as the altitude of this zone is to the diameter. THEOREM. 265. 540. If the triangle BAC (fig. 264, 265) and the rectangle Fig. 264, BCEF of the same base and same altitude turn simultaneously about the common base BC, the solid generated by the revolution of the triangle will be a third of the cylinder generated by the revolution of the rectangle. Demonstration. Let fall upon the axis the perpendicular AD (fig. 264); the cone generated by the triangle ABD is a third of Fig. 264. the cylinder generated by the rectangle AFBD (524); also the cone generated by the triangle ADC is a third of the cylinder generated by the rectangle ADCE; therefore the sum of the two cones, or the solid generated by ABC, is a third of the sum of the two cylinders, or of the cylinder generated by the rectangle BCEF. If the perpendicular AD (fig. 265) fall without the triangle, Fig. 165. the solid generated by ABC will be the difference of the cones generated by ABD and ACD; but, at the same time, the cylinder generated by BCEF will be the difference of the cylinders generated by AFBD, AECD. Therefore the solid generated by the revolution of the triangle will be always the third of the cylinder generated by the revolution of the rectangle of the same base and same altitude. 541. Scholium. The circle of which AD is the radius has for -2 its surface × AD; consequently л × AD × BC is the measure of the cylinder generated by BCEF, and x ADX BC is the measure of the solid generated by the triangle ABC. Fig. 266. Fig. 267. PROBLEM. 542. The triangle CAB (fig. 266) being supposed to make a revolution about the line CD, drawn at pleasure without the triangle through the vertex C, to find the measure of the solid thus generated. Solution. Produce the side AB until it meet the axis CD in D, and from the points A, B, let fall upon the axis the perpendiculars AM, BN. The solid generated by the triangle CAD has for its measure л × АM × CD (540); the solid generated by the triangle CBD has for its measure BN × CD; therefore the difference of these solids, or the solid generated by ABC, will have for its measure 7 × (AM2— BÑ) × CD. This expression will admit of another form. From the point I, the middle of AB, draw IK perpendicular to CD, and through the point B draw BO parallel to CD, we shall have -2 AM+BN = 2IK (178), and AM-BN=AO; consequently (AM + BN) × (AM—BN), or AM — BN (184), is equal to 2IK × AO. Accordingly the measure of the solid under consideration will also be expressed by ¦ π × IK × AO × CD. But, if the perpendicular CP be let fall upon AB, the triangles ABO, DCP will be similar, and will give the proportion AO: CP :: AB: CD; whence 40 × CD = CP x AB; moreover CP × AB is double of the area of the triangle ABC; thus we have AO × CD=2ABC; consequently the solid generated by the triangle ABC has also for its measure or, since circ. KI is equal to 2π × KI, this same measure will be ABC circ. KI. Therefore, the solid generated by the revolution of the triangle ABC has for its measure the area of this triangle multiplied by two thirds of the circumference described by the point I the middle of the base. 543. Corollary. If the side AC = CB (fig. 267), the line CI will be perpendicular to AB, the area ABC will be equal to ABCI, and the solidity π × ABC × IK will become × AB × IK × CI. But the triangles ABO, CIK, are similar and give the proportion AB: BO or MN:: CI: IK; consequently f ABX IK = MN x CI; therefore the solid generated by the isosceles triangle ABC will -2 have for its measure л × MN × CI. 544. Scholium. The general solution seems to suppose that the line AB produced would meet the axis, but the results would not be the less true, if the line AB were parallel to the axis. Indeed the cylinder generated by AMNB (fig. 268) has for its Fig. 268. measure л × AM × MN, the cone generated by ACM is equal to -2 л × AM × CM, -2 and the cone generated by BCN = ¦ π × AM × CN. Adding the two first solids together and subtracting the third from the sum, we have for the solid generated by ABC -2 л × AM × (MN + ¦ CM — ¦ CN); and, since CM- CN=-(} CN — } CM) = — MN, the 18 -2 above expression reduces itself to л× AM × MN, or 545. Let AB, BC, CD (fig. 262), be several successive sides of a Fig. 262, regular polygon, O its centre, Ol the radius of the inscribed circle; if we suppose the polygonal sector AOD, situated on the same side of the diameter FG, to make a revolution about this diameter, the solid -2 generated will have for its measure π × Ol × MQ, MQ being the -2 -2 -2 for its measure 1⁄2 л × O1 × PQ; therefore the sum of these solids, or the entire solid generated by the polygonal sector AOD, has for its measurex OỈ (MN + NP + PQ), or 3 7 × OỈ × MQ. Fig. 269. THEOREM. 546. Every spherical sector has for its measure the zone which serves as a base multiplied by a third of the radius, and the entire sphere has for its measure its surface multiplied by a third of the radius. Demonstration. Let ABC (fig. 269) be the circular sector, which, by its revolution about AC, generates the spherical sector; the zone described by AB being AD × circ. AC, or 2 × Ꭿ Ꮯ × ᎯᎠ (538), we say that the spherical sector will have for its measure this -2 Zone multiplied by AC, or л × АC × AD. 1. Let us suppose, if it be possible, that this quantity -2 zxx is the measure of a greater spherical sector, of the spherical sector, for example, generated by the circular sector ECF similar to ACB. Inscribe in the arc EF a portion of a regular polygon EMNF the sides of which shall not meet the arc AB, then suppose the polygonal sector ENFC to turn about EC at the same time with the circular sector ECF. Let CI be the radius of a circle inscribed in the polygon, and let FG be drawn perpendicular to EC. The solid generated by the polygonal sector will have for its measure 7 × CI× EG (545); now CI is greater than AC, by construction, and EG is greater than AD; for, if we join AB, EF, the triangles EFG, ABD, which are similar, give the proportion EG: AD:: FG: BD:: CF: CB; therefore EG>AD. For this double reason 7 × CIX EG greater than -2 л × CA × AD; the first expression is the measure of the solid generated by the polygonal sector, the second is, by hypothesis, that of the spherical sector generated by the circular sector ECF; consequently the solid generated by the polygonal sector would be greater than the spherical sector generated by the circular sector. But on the contrary the solid in question is less than the spherical sector, since it is contained in it; accordingly the hypothesis |