Cor. 3. As the ellipsoid HWKV is to the sphere HYKX as the axis VW of the ellipsoid to the axis XY of the sphere, and as the cylinder fhig circumscribing the ellipsoid is to the cylinder, pqrs circumscribing the sphere, as the same axis VW to the same axis XY, or as the length of those cylinders resrespectively; the ellipsoid HWKV bears the same ratio to its circumscribing cylinder fhig, as the sphere HYKX to its circumscribing cylinder pqrs. If S=the sphere, and E=the ellipsoid, and if P=the cylinder circumscribing the ellipsoid, and Q=the cylinder circumscribing the sphere, then, and, XY:S::VW: E XY: Q :: VW : P, hence, (Prop. XIX. B. I. El. Geom.) S: Q:: E: P. Scholium. The segment AJB being equal to the sphere HK, the annexed figure may represent the manner in which they are convertible into each other, as there exists no mathematical reason why the segment AJB may not be changed, as partly represented in the figure. H Cor. 4. It is also evident that an ungula BCDH, whose base BCD is the segment of a circle, similar to a segment AHB of a vertical section through this circular spindle AHBG; if the altitude DH of the ungula is equal to the circumference of the conjugate section HG of the spindle, the solidity of the ungula will be equal to that of the spindle, so also will its curve surface. And, if a cylinder segment ABCDH be so cut that the length AB shall be equal to the circumference of a circle of which HG is the radius, and AG and BG lie in the planes ACE, BFD, then may the A segment be converted into a ring whose outside diameter is equal 2HG; and every section of the ring will be equal to a section through the segment. Cor, 5. Hence, also, if a cylindric segment ABC be cut by two planes meeting in the surface of the cylinder at C, and A D B terminating at A and B on the opposite F A N D 0 B F P G C and at such an angle that the distance FG shall be equal to the circumference of a circle whose radius is CD, and the section AFGB will form a cylindric] ring, whose inner diameter is equal to twice CE. Cor. 6. And because the section ANF, (see diagram above,) if cut from its position and placed in the position BGP, com pletes the cylinder NFPB, which is equal in length to half the sum of sides FG and AB. The solidity of a cylindric ring AB, is equal to that of a cylinder whose base is equal to a radial A section of the ring, and whose altitude is equal to half the sum of its inner and outer circles. And hence, cylindric rings whose sections are equal, are proportional to their inner or outer circumferences. PROPOSITION V. THEOREM. The solidity of a cylinder circumscribing a sphere, is equal to the solidity of a prism circumscribing the cylindrical ungula or ungulas whose solidity is equal to the sphere. Let MNGL be a cylinder circumscribing the sphere PK; and let CRDBKA be a prism circumscribing the two similar ungulas QRKV, SRKV described on the base, KRV=half the base of the cylinder from which they are conceived to be taken; and if QS, the sum of their altitudes, is equal in length to the circumference of the cylinder, then (Prop. IV.) will the ungulas equal the sphere of equal diameter to that of the cylinder, and the cylinder equal MNLG will be equal to the prism CRDBKA. For the altitude AC or KR of the prism, on the base CRD, is equal to the altitude of the cylinder; GL being=the axis, both of the cylinder and the sphere; and the length of the side CD of the prism is equal to the sum of the altitudes VQ, VS, or the length QS of the ungulas taken on the surface of the cylinder; but the length QS is equal to the circumference of the sphere or cylinder by hypothesis. Now, the area of the base LN of the cylinder, is equal to its circumference multiplied by half the radius RE; and the area of the base CDR of the prism, is equal to the line CD, or the circumference of the cylinder multiplied by half the line RE; hence, the base of the prism is equal to that of the cylinder. And the solidity of the cylinder is equal to its base ENL multiplied by its altitude GL; the solidity of the prism is also equal to its base CDR, or the base of the cylinder multiplied by its altitude RK; hence, the solidity of the cylinder is equal to that of the prism. Cor. As the sum of two or more ungulas of equal base, are equal to one greater of the same base, if the sum of their altitudes is equal to the altitude of the greater, (Pr. IX. Cor. 2. B. II.) and because a prism circumscribing an ungula is proportional to the altitude of the ungula, two or more prisms circumscribing ungulas, the sum of which is equal to a sphere, are equal to the cylinder circumscribing that sphere; and also, the several prisms circumscribing ungulas equal to a revoloid, are equal to the prism circumscribing the revoloid composed of those ungulas. PROPOSITION VI. THEOREM. Every right revoloid is equal to two-thirds its circumscribing prism; and every sphere is equal to two-thirds its circumscribing cylinder. Let ABCD, (fig. 1.) be a revoloid, or a sphere circumscribed by the prism or cylinder EFGH; then will the revoloid ABCD be equal to two-thirds the prism EFGH; and the sphere ABCD will be equal to two thirds of the cylinder EFGH. For let the plane EG be a vertical section through the E centre of the revoloid and prism bisecting their opposite sides; in which position, o (Def. 8.) the section of the revoloid is a circle, and the section of the prism through D the same plane is evidently a rectangle. These sections are, also, evidently those of a sphere and its circumscribing cylinder, made by a plane through the common axis, H AC, of the sphere and cylin Fig. 1. A F P S K M N L I B C G der, (Prop. 1, Sph. Geom.) and (Def. 5, B. 1,) through the centre, I, join EI, FI; also let AIC, as an axis, be parallel to EH or FG; and DIB and KL parallel to EF, or HG, the base of the section of the prism or cylinder, the latter line, KL, meeting FI in M, and the circular section of the revoloid or sphere in N; and the plane EIF will represent the vertical section of a pyramid of equal base to that of the prism, and an altitude, IA : or it will represent a vertical section of a cone of equal base to that of the cylinder and of an altitude, IA. Now, if the line KL produced, if necessary, be conceived to revolve on the axis AC, it will cut conjugate sections of those solids in the relation of their magnitudes, viz: KS the section of a prism or cylinder, KN the section of a revoloid or sphere, and KM the section of a pyramid or cone. Now, AF being equal to AI or IB, and KL parallel to AF, then by similar triangles IK = KM, (Prop. XVII, B. IV, El. Geom.,) and since, in the right angled triangle, IKN, IN2=IK* +KN3, (Prop. XXIV, B. IV, El. Geom.) and, because, KL is equal to the radius IB or IN, and KM=IK, therefore, KL2= KM2+KN3; or the longest line forming the section of the prism or cylinder, is equal to the sum of the squares of the other two, forming sections of the revoloid or sphere, the pyramid or cone. Let now the conjugate sections of those solids formed by the revolution of the lines KL, be represented. (Fig. 2.) Fig. 2. A C Thus let the square ÖALB, represent the conjugate section of a prism, described about the quadrangular revoloid, and let the circle OALB be the section of a cylinder circumscribing the sphere; the square FCND will represent the section of the the revoloid, and the circle PCND will represent the sec tion of the sphere; the square SEMF will represent the section of the pyramid, and the circle SEMF a section of the cone. Now, since we have shown that KL, or the square AL described on the line KL is equal to KN3 + KM2, or the square CN + the square EM, being squares described on the lines KN and KM; it results that the square OALB, described on the line OL, which is double the line KL, is equal to the sum of the squares PCND, SEMF, described on the lines PN, SM, being double the lines KN, KM. (Prop. XII, B. I, |