On account of the similar triangles SAO, SDC, AO: DC: SA: SD; and, on account of the similar triangles SAF, SDH, AF: DH:: SA: SD; whence AF: DH:: AO: DC :: circ. AO: circ. DC (287). But, by construction, AF = circ. AO; consequently This being premised, the triangle SAF, which has for its measure AF × SA, is equal to the surface of a cone SAB, which has for its measure circ. AO × 1 SA. For a similar reason, the triangle SDH is equal to the surface of the cone SDE. Whence the surface of the frustum ADEB is equal to that of the trapezoid AF+ ADHF. This has for its measure AD (FDH) (178). Therefore the surface of the frustum of a cone ADEB is equal to its side AD multiplied by the half sum of the circumferences of the two bases. 531. Corollary. Through the point I, the middle of AD, draw IKL parallel to AB, and IM parallel to AF; it may be shown as above that IM circ. IK. But the trapezoid (179) = ADHF=AD × IM = AD × circ. IK. Hence we conclude further that the surface of the frustum of a cone is equal to its side multiplied by the circumference of a section made at equal distances from the two bases. 532. Scholium. If a line AD, situated entirely on the same side of the line OC, and in the same plane, make a revolution about OC, the surface described by AD will have for its measure + circ. DC' 2 AD × (circ. 10 c. DC), ᎯᎠ X or AD x circ. IK; the lines AO, DC, IK, being perpendiculars let fall from the extremities and from the middle of the line AD upon the axis OC. For, if we produce AD and OC till they meet in S, it is evident that the surface described by AD is that of the frustum of a cone, of which OA and DC are the radii of the bases, the entire cone having for its vertex the point S. Therefore this surface will have the measure stated. This measure would always be correct, although the point D should fall upon S, which would give an entire cone, and also when the line AD is parallel to the axis, which would give a cylinder. In the first case DC would be nothing; in the second DC would be equal to AO and to IK. LEMMA. 533. Let AB, BC, CD (fig. 262), be several successive sides of Fig. 252. a regular polygon, O its centre, and OI the radius of the inscribed circle; if we suppose the portion of the polygon ABCD, situated entirely on the same side of the diameter FG, to make a revolution about this diameter, the surface described by ABCD will have for its measure MQ × circ. OI, MQ being the altitude of this surface, or the part of the axis comprehended between the extreme perpendiculars AM, DQ. Demonstration. The point I being the middle of AB, and IK being a perpendicular to the axis let fall from the point I, the surface described by AB will have for its measure AB × circ. IK (532). Draw AX parallel to the axis, the triangles ABX, OIK, will have their sides perpendicular each to each, namely, OI to AB, IK to AX, and OK to BX; consequently these triangles will be similar, and will give the proportion AB: AX or MN:: OI: IK:: circ. OI: circ. IK, × therefore AB x circ. IK MN × circ. OI. Whence it will be perceived that the surface described by AB is equal to its altitude MN multiplied by the circumference of the inscribed circle. Likewise the surface described by BC = NP × circ. OI, the surface described by CDPQ× circ. OI. Accordingly the surface described by the portion of the polygon ABCD has for its measure (MN+NP+PQ) × circ. OI, or MQ × circ. OI; therefore this surface is equal to its altitude multiplied by the circumference of the inscribed circle. 534. Corollary. If the entire polygon has an even number of sides, and the axis FG passes through two opposite vertices F and G, the entire surface described by the revolution of the semipolygon FACG will be equal to its axis FG multiplied by the circumference of the inscribed circle. This axis FG will be at the same time the diameter of the circumscribed circle. THEOREM. 535. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Demonstration. 1. We say that the diameter of a sphere multiplied by the circumference of a great circle cannot be the measure of the surface of a greater sphere. For, if it be possiFig. 263. ble, let AB × circ. AC (fig. 263) be the surface of a sphere whose radius is CD. 1 About the circle, whose radius is CA, circumscribe a regular polygon of an even number of sides, which shall not meet the circumference of the circle whose radius is CD; let M and Sbe two opposite vertices of this polygon; and about the diameter MS let the semipolygon MPS be made to revolve. The surface described by this polygon will have for its measure MSX circ. AC (534); but MS is greater than AB; therefore the surface described by the polygon is greater than AB × circ. AC, and consequently greater than the surface of the sphere whose radius is CD. On the contrary the surface of the sphere is greater than the surface described by the polygon, since the first encloses the second on all sides. Therefore the diameter of a sphere multiplied by the circumference of a great circle cannot be the measure of the surface of a greater sphere. 2. We say, also, that this same product cannot be the measure of the surface of a less sphere. For, if it be possible, let DE × circ. CD be the surface of a sphere whose radius is CA. The same construction being supposed as in the first case, the surface of the solid gener ted by the polygon will always be equal to MS X circ. AC. But MS is less than DE, and circ. AC less than circ. CD; therefore, for these two reasons, the surface of the solid generated by the polygon would be less than DE × circ. CD, and consequently less than the surface of the sphere whose radius is AC. But, on the contrary, the surface described by the polygon is greater than the surface of the sphere whose radius is AC, since the first surface encloses the second; therefore the diameter of a sphere multiplied by the circumference of a great circle cannot be the measure of the surface of a less sphere. We conclude, then, that the surface of a sphere is equal to the diameter multiplied by the circumference of a great circle. 536. Corollary. The surface of a great circle is measured by multiplying its circumference by half of the radius, or a fourth of the diameter; therefore the surface of a sphere is four times that of a great circle. 537. Scholium. The surface of a sphere being thus measured and compared with plane surfaces, it will be easy to obtain the absolute value of lunary surfaces and spherical triangles, the ratio of which to the entire surface of the sphere has already been determined. In the first place, the lunary surface, whose angle is A (fig. 276), Fig 276. is to the surface of the sphere as the angle A is to four right angles (493), or as the arc of a great circle, which measures the angle A, is to the circumference of this same great circle. But the surface of the sphere is equal to this circumference multiplied by the diameter; therefore the lunary surface is equal to the arc, which measures the angle of this surface, multiplied by the diameter. In the second place, every spherical triangle is equivalent to a lunary surface whose angle is equal to half of the excess of the sum of its three angles over two right angles (503). Let P, Q, R, be the arcs of a great circle which measure the three angles of a spherical triangle; let C be the circumference of a great circle, and D its diameter; the spherical triangle will be equivalent to the lunary surface whose angle has for its measure P+Q+RC and consequently its surface will be Thus, in the case of the triangle of three right angles, each of the arcs P, Q, R, is equal to C, and their sum is C; the ex94 cess of this sum over C is C, and the half of this excess is C; therefore the surface of a triangle of three right angles CX D, = ¦ which is the eighth part of the whole surface of the sphere. The measure of spherical polygons follows immediately from that of triangles, and it is moreover entirely determined by the proposition of art. 505, since the unit of measure, which is the triangle of three right angles, has just been estimated on a plane surface. THEOREM. 538. The surface of any spherical zone is equal to the altitude of this zone multiplied by the circumference of a great circle. Demonstration. Let EF (fig. 269) be any arc, either less or Fig. 269. greater than a quadrant, and let FG be drawn perpendicular to 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 × 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 X 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 × 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 |