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 ¦π × Aм3× CD (540); the solid generated by the triangle --2 CBD has for its measure BN x CD; therefore the difference of these solids, or the solid generated by ABC, will have for its measure × (AM— BÑ) × CD. 2 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 AM+BN = 2IK (178), and AM-BN=AO; consequently (AM + BN) × (AM—BN), or AM — BN (184), is equal to 2IK × 40. Accordingly the 40 × CD = CP x 4B; moreover CP x AB is double of the area of the triangle ABC; thus we have 4O × CD = 2ABC; consequently the solid generated by the triangle ABC has also for its measure or, since circ. KI is equal to 2л × 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 л × ABC × IK will become л× AВ × IK x CI. But the triangles ABO, CIK, are similar, and give the proportion AB : BO or MN :: CI: IK; consequently AB X IK = MN × CI; therefore the solid generated by the isosceles triangle ABC will have for its measure × MN × -2 544. 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 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 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 × Oi × MQ, MQ being the portion of the axis terminated by the extreme perpendiculars AM, DQ. Demonstration. Since the polygon is regular, all the triangles AOB, BOC, &c., are equal and isosceles. Now, by the corollary of the preceding proposition, the solid generated by the isosceles triangle AOB has for ita measure × OI × MN, the solid generated by the triangle BOC has for its measure × OỈ × NP, and the solid generated by the triangle COD has -2 for its measureл × OI × PQ; therefore the sum of these solids, or the entire solid generated by the polygonal sector AOD, has for its measure × OỈ (MN + NP + PQ), or & π × бỈ× 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 2л× ACX AD (538), we say that the spherical sector will have for its measure this -2 zone multiplied by AC, or § 7 × ÃÑ3× AD. 1. Let us suppose, if it be possible, that this quantity 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 × 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 × CI × EG greater than 2 × CAX 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 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 л × 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 л × CI× EG. But CI is less than CE; consequently the -2 solid is less than л × CE × 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 π R2, and its solidity 4л R2 × R, oг л R3. If we call D the diameter, we shall have R = D, and R3 = ¦ D3 ; therefore the solidity will also be expressed by π × ¦ D3, or π 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 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 |