THEOREM. 549. The surface of a sphere is to the whole surface of the circumscribed cylinder, the bases being comprehended, as 2 is to 3; and the solidities of these two bodies are in the same ratio. Fig. 270. Demonstration. Let MPNQ (fig. 270) be a great circle of the sphere, ABCD the circumscribed square; if the semicircle PMQ, and the semisquare PADQ, be made to turn at the same time about the diameter PQ, the semicircle will generate the sphere, and the semisquare will generate the cylinder circum scribed about the sphere. The altitude AD of the cylinder is equal to the diameter PQ, the base of the cylinder is equal to a great circle, since it has for a diameter AB equal to MN; consequently the convex surface of the cylinder is equal to the circumference of a great circle multiplied by its diameter (523). This measure is the same as that of the surface of the sphere (535); whence it follows that the surface of the sphere is equal to the convex surface of the circumscribed cylinder. But the surface of the sphere is equal to four great circles; consequently the convex surface of the circumscribed cylinder is also equal to four great circles. If we add the two bases, which are equal to two great circles, the whole surface of the circumscribed cylinder will be equal to six great circles; therefore the surface of the sphere is to the whole surface of the circumscribed cylinder as 4 is to 6, or as 2 is to 3. This is the first part of the proposition which it was proposed to demonstrate. In the second place, since the base of the circumscribed cylinder is equal to a great circle, and its altitude equal to the diameter, the solidity of the cylinder will be equal to a great circle multiplied by the diameter (516). But the solidity of the sphere is equal to four great circles multiplied by a third of the radius (546), which amounts to a great circle multiplied by of the radius, or of the diameter; therefore the sphere is to the circumscribed cylinder as 2 is to 3, and consequently the solidities of these two bodies are to each other as their surfaces. 550. Scholium. If a polyedron be supposed, all whose faces touch the sphere, this polyedron might be considered as composed of pyramids having the centre of the sphere for their common vertex, the bases being the several faces of the polyedron. Now it is evident that all these pyramids will have for their common altitude the radius of the sphere, so that each pyramid will be equal to a face of the polyedron, which serves as a base, multiplied by a third of the radius; therefore the entire polyedron will be equal to its surface multiplied by a third of the radius of the inscribed sphere. It will be perceived by this that the solidities of polyedrons circumscribed about a sphere are to each other as the surfaces of these same polyedrons. Thus the property which we have demonstrated for the circumscribed cylinder is common to an infinite number of other bodies. We might have remarked, also, that the surfaces of polygons circumscribed about a circle are to each other as their perimeters. PROBLEM. 551. The circular segment BMD (fig. 271) being supposed to Fig. 271 revolve about a diameter exterior to this segment, to find the value of the solid generated. Solution. Let fall upon the axis the perpendiculars BE, DF, and upon the chord BD the perpendicular CI, and draw the radii CB, CD. The solid generated by the sector BCA = × CBX ЛЕ (546); the solid generated by the sector DCA=«× CB × AF; consequently the difference of these two solids, or the solid generated by the sector DCB, will be equal to -2 π 2 × CBX (AF-AE) = a × CB × EF. But the solid generated by the isosceles triangles DCB has for its measure 2 2 × CIX EF (543); consequently the solid generated by the segment BMD = × EFX (CB-CỈ). Now in the right-angled triangle CBI we have CB-CỈ=BỈ=¦BD; therefore the solid generated by the segment BMD has for its measure × EF × BD, or BD × EF. 2 2 552. Scholium. The solid generated by the segment BMD is to the sphere whose diameter is BD, × BD, or :: EF : BD. as X BDX EF is to Fig. 271. THEOREM. 553. Every segment of a sphere, comprehended between two parallel planes, has for its measure the half sum of its bases multiplied by its altitude, plus the solidity of the sphere of which this same altitude is the diameter. Demonstration. Let BE, DF (fig. 271), be the radii of the bases of the segment, EF its altitude, so that the segment may be formed by the revolution of the circular space BMDFE about the axis FE. The solid generated by the segment BMD × BD × EF (552), the frustum of a cone 2 will be equal to -2 2 × EF × (BE + DF + BE × DF) (527); consequently the segment of the sphere which is the sum of these two solids = × EF × (2BỂ+2DF+2BE × DF+BD). But, by drawing BO parallel to EF, we shall have DO = DF-ВЕ, 빙 = DF 2 2DF × BE + BE (182), and consequently BD = BỔ + DỔ = EF+ DF-2DF × BE + BE. Putting 2 2 this value in the place of BD in the expression for the segment, and reducing it, we shall have for the solidity of the segment × EF × (3BE+3DF+ EF), an expression which may be decomposed into two parts; the one × EF × (3BE+3DF), 3 , or EF X 2 πΧ ΒΕ +π× DF 2 is the half sum of the bases multiplied by the altitude; the other × EF represents the sphere of which EF is the diameter (548); therefore the segment of the sphere, &c. 554. Corollary. If one of the bases is nothing, the seginent in question becomes a spherical segment having only one base ; therefore every spherical segment having only one base is equivalent to half of the cylinder of the same base and same altitude, plus the sphere of which this altitude is the diameter. General Scholium. 555 Let R be the radius of the base of a cylinder, Hits altitude; the solidity of the cylinder will be R2 × H, or R2 H. Let R be the radius of the base of a cone, H its altitude; the solidity of the cone will be a R2 × H, or R2 H. Let A, B, be the radii of the bases of the frustum of a cone, Hits altitude, the solidity of the frustum will be H(A2 + B2 + AB). Let R be the radius of a sphere; its solidity will be R3. Let R be the radius of a spherical sector, H the altitude of the zone, which answers as a base; the solidity of the sector will be ፮ « R2 H. (+) Let P, Q, be the two bases of a spherical segment, H its altitude, the solidity of this segment will be × H + H3. If the spherical segment have only one base P, its solidity will 2 π bePH + H3. APPENDIX TO THE THIRD SECTION OF THE SECOND PART. Of Spherical Isoperimetrical Polygons. THEOREM. 556. Let S be the number of solid angles in a polyedron, H the number of its faces, A the number of its edges; then in all cases we shall have S + H = A + 2. Demonstration. Within the polyedron, take a point, from which let straight lines be drawn to the vertices of all its angles; conceive next, that from the same point, as a centre, a spherical surface is described, meeting all these straight lines in as many points; join these points by arcs of great circles, so as to form on the surface of the sphere polygons corresponding in position and number with the faces of the polyedron. Let ABCDE be one of these polygons (fig. 240), and n the number of its sides; Fig. 240 its surface will be s - 2n + 4, s being the sum of the angles A, B, C, D, E (506). If the surface of each polygon be estimated in a similar manner, and afterwards the whole added together, we shall find their sum, or the surface of the sphere, represented by 8, to be equal to the sum of all the angles of the GEOM. 26 polygons, minus twice the number of their sides, plus 4, taken as many times as there are faces. Now, since all the angles which meet at any one point A are equal to four right angles, the sum of all the angles of the polygons must be equal to 4, taken as many times as there are solid angles; it is therefore equal to 4S. Also, twice the number of sides AB, BC, CD, &c., is equal to four times the number of edges, or to 4A; since the same edge is in every case a side to two faces. Hence we have 8=4S-4A+4H; or, dividing the whole by 4,2 = S - A+H; therefore S + H = A +2. 557. Corollary. From this it follows, that the sum of all the plane angles, which form the solid angles of a polyedron, is equal to as many times four right angles as there are units in S - 2, S being the number of solid angles of the polyedron. For, if we consider a face, the number of whose sides is n, we shall find that the sum of the angles of this face is equal to 2n-4 right angles (64). But the sum of all these 2n's, or twice the number of sides in all the faces, will be 4.1; and 4, taken as many times as there are faces, will be 4H; hence the sum of the angles in all the faces is 4A-4H. Now, by the theorem just demonstrated, we have A -H = S-2, and consequently 4A - 4H = 4 (S-2). Therefore the sum of all the plane angles, &c. THEOREM. 558. Of all the spherical triangles formed with two given sides Fig. 276. CA, CB (fig. 276), and a third assumed at pleasure, the greatest, ABC, is that in which the angle C, contained by the given sides, is equal to the sum of the two other angles, A and B. Demonstration. Produce the two sides AC, AB, till they meet in D; we shall have a spherical triangle BCD, in which the angle DBC is also equal to the sum of the two other angles BDC, BCD. For, BCD + BCA, being equal to two right angles, and likewise CBA + CBD, we have BCD + BCA = ᏟᏴA + ᏟᏴᎠ ; and adding BDC=BAC, we shall have BCD + BCA + BDC = ᏟᏴA + CBD + BAC. |