with which we set out cannot be maintained; therefore the zone or base of a spherical sector multiplied by a third of the radius cannot be the measure of a greater spherical sector. 2. We say that this same product cannot be the measure of a less spherical sector. For, let CEF be the circular sector which by its revolution generates the given spherical sector, and let us suppose, if it be possible, that × CE× EG is the measure of a less spherical sector, of that, for example, generated by the circular sector ACB. The preceding construction remaining the same, the solid generated by the polygonal sector will always have for its measure & « × CỈ × EG. But CI is less than CE; consequently the solid is less than × CE3× EG, which, by hypothesis, is the measure of the spherical sector generated by the circular sector ACB. Therefore the solid generated by the polygonal sector would be less than the solid generated by the spherical sector; but, on the contrary, it is greater, since it contains it. Therefore it is impossible that the zone of a spherical sector multiplied by a third of the radius should be the measure of a less spherical sector. We conclude, then, that every spherical sector has for its measure the zone which answers as a base multiplied by a third of the radius. A circular sector ACB may be increased till it becomes equal to a semicircle; then the spherical sector generated by its revolution is an entire sphere. Therefore the solidity of a sphere is equal to its surface multiplied by a third of the radius. 547. Corollary. The surfaces of spheres being as the squares of their radii, these surfaces multiplied by the radii are as the cubes of the radii. Therefore the solidities of two spheres are as the cubes of their radii, or as the cubes of their diameters. 548. Scholium. Let R be the radius of a sphere, its surface will be 4 R2, and its solidity 4 π R2 × 1 R, or « R3. If we call D the diameter, we shall have R = D, and R3 = ¦ D3 ; therefore the solidity will also be expressed by XD3, or fπ D3. Fig. 270. 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. 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 circumscribed 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 1⁄2 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 polye dron. 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 = { «× CB3× AE (546); the solid generated by the sector DCA = 3 « × CB× AF; DCA={«× consequently the difference of these two solids, or the solid generated by the sector DCB, will be equal to ÷ «× CB3× (AF—AE) = { « × CB3× EF. But the solid generated by the isosceles triangles DCB has for 552. Scholium. The solid generated by the segment BMD is to the sphere whose diameter is BD, as × BD × EF is to ¿ * × BD, or : : EF ; BD, 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 will be equal to ¦a × BD × EF (552), the frustum of a cone generated by the trapezoid BDFE will be equal to -2 ÷ « × EF × (BE2+DFa2+ BE × DF) (527) ; consequently the segment of the sphere which is the sum of these two solids=× EF× (2BE+2DF2+2BE× DF+BĎ). But, by drawing BO parallel to EF, we shall have DO — DF — BE, 2 -2 -2 DỔ = DF — 2DF × BE + BE (182), and consequently an expression which may be decomposed into two parts; the one ↓ « × EF × (3BE+3DF), or EF × (2 πΧ ΒΕ +πχ DF 2 is the half sum of the bases multiplied by the altitude; the other 3 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 segment 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, H its 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 R2 × ¦ H, or ¦ « R2 H. T Let A, B, be the radii of the bases of the frustum of a cone, H its 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 π R2 H. ༔ Let P, Q, be the two bases of a spherical segment, H its altitude, the solidity of this segment will be (P+2) × H+ ÷ «H. If the spherical segment have only one base P, its solidity will be PH+ 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+HA+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. 24% 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 |