Fig.150. ing triangle DEF of the other; and we say-Two symmetrical spherical triangles are equal in area. Fig. 147. Fig.151. 299. The two opposite triangles CMK, LFI, (fig. 147) being symmetrical, are equal in area; if we add to each of these equal areas the triangle CMI, we shall have CMK + CMI = CMI+LFI; but the triangles CMK, CMI, constitute the lunary surface whose angle is CIM; the sum of two triangles CMI, LFI, in the nearer hemisphere, is therefore equivalent to this lunary surface. We have therefore this general rule:-If two semi-circumferences cut each other in a hemisphere, the sum of two opposite triangles will be equal to a lunary surface whose angle is equal to the angle made by the semi-circumferences. 300. Let ABC (fig. 151) be a spherical triangle, and DEFGHI the circumference of a great circle which does not cut this triangle. Produce the sides of the triangle both ways till they meet this circumference; and we shall have IBD + GBF = Lun.B, HAG + DAE= Lun.A, FCE + HCI = Lun. C. It is manifest that the sum of these six triangles is equal to the hemisphere plus twice. the triangle ABC, that is, equal to 2 ABC + Lun. 2 D ; we shall therefore have 2 ABC + Lun. 2 D = Lun. (A+B+C), or ABC= Lun.(A + B + C − 2 D). We say therefore-The area of a spherical triangle is equivalent to half the lunary surface whose angle is equal to the sum of the three angles of the triangle minus two right-angles. 2 Suppose a, b, c, to designate the arcs which measure the angles A, B, C, of any spherical triangle; and let C denote the circumference of a great circle, and R the radius. We shall have, for the area of the spherical triangle, RX (a + b + c — ! C). 2 301. To ascertain the volume of the body generated by the revolution of the polygonal sector abcd O (fig. Fig.149. 149), it will be necessary to examine the body generated Fig.152. by the isosceles triangle ABO (fig. 152) revolving about the straight line OD as an axis. Produce AB till it meet the axis in D; and from the points A, B, draw AM and BN, perpendicular to OD. The volume of the body generated by the triangle AOD, will have for its measure π × (AM)2 × OD, (174). The body generated by the triangle OBD, has for its measure π (BN)2 × OD; therefore the difference of these bodies, or the body generated by ABO, will have for its measure. [(AM2) - (BN2)] × OD. Fig.152. This expression may take another form. From I, the middle of AB, draw IK perpendicular to OD, and through B draw BP parallel to OD; we shall have AM + BN = 2 IK, and AM = AM-BN ᎪᏢ ; consequently (AMBŃ) × (AM — BN) = (AM)2 (BN)2, (176), 2 IK × AP. The volume of the body generated by ABO, may therefore be expressed by л × IK × AР × OD. But if we draw OI, as OAB is an isosceles triangle, OI will be perpendicular to AB; and the two triangles APB, OID, are similar, and give 3 3 3 But AB; and substituting OI × AB for its value in the 302. [If the axis of revolution is parallel to the base of the triangle, MN will equal AB, the surface generated by AB will be a cylindrical surface, and the volume of the body generated by the triangle will be two thirds of the volume of a cylinder the radius of whose base is OI, the height of the triangle, and whose height is AB, the base of the triangle. It follows, therefore, that the volume of a body generated by the revolution of the same triangle about its base as an axis, will be equivalent to one third of the cylinder of the above dimensions.] 303. Returning to figure 149, it will be readily scen that the portion of the polygon abcd, inscribed in the quadrant ad is composed of isosceles triangles whose common height is the radius of the inscribed circle. The volume of the body generated by aOb, will have for its 3 3 Fig.149. measure π × (Oh)2 × αe; that generated by b0c will have 2 × (Оi)2 × eƒ; and that generated by doc will have π × (01)2 ׃O. As Oh, Oi and Ol are equal, the sum of these bodies, or the body generated by the portion of the inscribed polygon abcd, will have for the meaaure of its volume, π × (Oh)2 × (a e + eƒ‡ƒО), or ñ × (О h)2 × a O; and if the portion of the inscribed polygon occupied the entire semicircle, the body generated by it, in a revolution about the diameter ap, will have for its measure π × (0 h)2 × a р. 2 3 2 p. If we apply this result to the body generated by the circumscribed polygon, we shall have for the measure of its volume, × (α O)2 × AP. π (a 3 By increasing the number of sides of the polygons, the inscribed polygon increases continually, and the limit of this increase is the semicircle; but with this increase of the number of sides, the factor Oh increases, and the limit of this increase is O a; and as the other factors remain the same, the limit for the product is π X (Ο α) Χαρ. 2 3 But by increasing the number of sides of the exterior polygon, it becomes continually less, and the limit of this diminution is the semicircle; but with this increase in the number of sides, the factor AP is continually diminished, and the limit of this diminution is a p; and as the other factors remain the same, the limit of the product expressing the volume of the body generated by the portion of the polygon circumscribed about the semicircle, is × (Oa)2 × ap. It appears therefore that 1⁄2 л × (0 а)2 × ap, cannot be the measure of a volume greater or less than that of the body generated by the semicircle whose radius is a O, revolving about the diameter ap; but this body is the sphere whose radius is a O. 2 3 3 2 3 4 3 If we denote a O by R and ap by 2 R, the expression becomes л × R2 × 2R, or ¦ π × R3; and may take the form 4 л . R2 × ¦ R. But 4 π R2 is the area of the sphere whose radius is R; we therefore say—The sphere has for the measure of its volume, the area of its surface, multiplied by one third of its radius; or the area of a great circle multiplied by two thirds of the diameter. 304. If we have a cone and a cylinder whose common Fig.148. height is equal to the diameter of the sphere (fig. 148), and whose common base is equal to a great circle of the Fig.148. sphere; we may express these convex surfaces as fol lows: Remark. The sphere πχ R3 2 π. R2 × R; 2 л. R2 × 2R; л. R2 × 3 R. is therefore two thirds of the circumscribed cylinder. The cone is one third of the circumscribed cylinder. The area of the sphere is equivalent to that of the convex surface of the circumscribed cylinder. 305. The bodies of revolution which have now been discussed, are said to be similar when the figures which generate them are similar. Two cones generated by similar triangles revolving about homologous sides, have their heights in proportion to the radii of their bases. A Let A represent the area of a cone whose height is H and the radius of whose base is R, and whose side is S; we shall have the equation Л=2л.R × 1 S, (272). For another similar cone we shall have a 2л. r × 1 s. If we compare these expressions, cancelling the common R S factors, we have A a RX S ; or because The areas of similar cones are to each other as the second powers of their corresponding dimensions. 306. Two similar cylinders are generated by similar rectangles (fig. 144). Their areas will have for their expressions, A 2л. R× Н, a 2л.r × h. If we compare these neglecting the common factors in the terms of the second ratio, we shall have that is―The areas of similar cylinders are to each other as the second powers of their homologous measures. 307. All spheres are similar, because every sphere has for its generatrix a semicircle, and all semicircles Fig.144. are similar. The areas of two spheres have for their expressions, A=4ñ.R2, a= 4π.r2; comparing these and cancelling the common factors in the second ratio, A R2 ; that is-The areas of two spheres will give α r2 are to each other as the second powers of their radii. 308. If we compare the volumes of two similar cones, which are expressed as follows; V V = x.r2 × h; we shall have R as, h3 r H 138 V π. R2 × H, and R2 × H R3 3 r2 X h 2.3 ; that is-The volumes of similar cones are to each other as the third powers of their homologous lines. 309. The volumes of two similar cylinders have for their expressions, V' =л. R2 × H, v' π . R2 × H, v' = ñ.r2 × h. If we compare these volumes, we have R2 × H volumes of similar cylinders are to each other as the third powers of their homologous measures. 310. Two spheres, being always similar bodies, and having for the expressions of their volumes, V" 4. R3, and v".r3, may be readily compared ; R3 D3 That is-The vol umes of two spheres are to each other as the third powers of their radii, or as the third powers of their diameters. 311. It was observed (227) that a regular polyedron has all its faces equal regular polygons, all its diedral angles equal, and all its polyedral angles equal. Let us inquire into the number of regular polyedrons which may be constructed. In the first place, it is manifest that the faces of these regular bodies must be equilateral triangles, squares or regular pentagons; for the angles of a regular hexagon are equal to four right angles; no polyedral angle therefore, can be formed with regular hexagons, or with regular polygons of a greater number of sides. Of those bodies contained by equilateral triangles, the |