E A If the perpendicular AD falls without the triangle; the solid described by ABC will, in that case, be the difference of the two cones described by ABD and ACD ; but at the same time, the cylinder described by BCEF will be the difference of the two cylinders described by AFBD and AECD. Hence the solid, described by the revolution of the triangle, will still be a third part of the cylinder described by the revolution of the rectangle having the same base and the same altitude. B Scholium. The circle of which AD is radius, has for its neasure × AD2; hence × AD2 × BC measures the cylinder described by BCEF, and × AD2x BC measures the solid described by the triangle ABC. PROPOSITION XII. LEMMA. If a triangle be revolved about a line drawn at pleasure through its vertex, the solid described by the triangle will have for its measure, the area of the triangle multiplied by two thirds of the circumference traced by the middle point of the base. Let CAB be the triangle, and CD the line about which it revolves. Produce the side AB till it meets the axis CD in D; from the points A and B, draw AM, BN, perpendicular to the axis, and CP perpendicular to DA produced. The solid described by the tri- C angle CAD is measured by X Р B MKN AM2 CD (Prop. XI. Sch.); the solid described by the triangle CBD is measured by x BN2x CD; hence the difference of those solids, or the solid described by ABC, will have for its measure (AM2-BN2) × CD. To this expression another form may be given. From I, the middle point of AB, draw IK perpendicular to CD; and through B, draw BO parallel to CD: we shall have AM+BN=2ÏK (Book IV. Prop. VII.); and AM-BN=AO; hence (AM+ BN) × (AM-NB), or AM2-BN2=2IK × AO (Book IV. Prop X.). Hence the measure of the solid in question is ex pressed by X × IK× AO × CD. But CP being drawn perpendicular to AB, the triangles ABO DCP will be similar, and give the proportion hence AO : CP :: AB : CD; AOX CD=CPX AB; but CPX AB is double the area of the triangle ABC; hence we have π AOX CD=2ABC; hence the solid described by the Р B MKN for its measure the area of this triangle multiplied by two thirds of the circumference traced by I, the middle point of the base. Cor. If the side AC=CB, the line CI will be perpendicular to AB, the area ABC will be equal to AB×CI, and the solidity IK will become > ABC× × AB× X C MK N IK CI. But the triangles ABO, CIK, are similar, and D give the proportion AB : BO or MN CI: IK; hence ABX IK=MNx CI; hence the solid described by the isosceles triangle ABC will have for its measure ×CI3× MN: that is, equal to two thirds of п into the square of the perpendicular let fall on the base, into the distance between the two perpendiculars let fall on the axis. Scholium. The general solution appears to include the supposition that AB produced will meet the axis; but the results would he equally true, though AB were parallel to the axis. Thus, the cylinder described by AMNB P A B is equal to ".AM2.MN; the cone described by ACM is equal to 7.AM2.CM, and the cone described by BCN to AM2 CN. Add the first two solids and take away the third; we shall have the solid described by ABC equal to π.AM2. (MN+CM-CN): and since CN-CM-MN, this expression is reducible to 7.AM2. MN, or 7.CP2.MN: which agrees with the conclusion found above. C M N PROPOSITION XIII. LEMMA. If a regular semi-polygon be revolved about a line passing through the centre and the vertices of two opposite angles, the solid described will be equivalent to a cone, having for its base the inscribed circle, and for its altitude twice the axis about which the semi-polygon is revolved. Let the semi-polygon FABG be revolved about FG: then, if OI be the radius of the inscribed circle, the solid described will be B measured by area OI × 2FG. For, since the polygon is regular, the triangles OFA, OAB, OBC, &c. are equal C and isosceles, and all the perpendiculars let fall from O on the bases FA, AB, &c. will be equal to OI, the radius of the inscribed circle. D G Now, the solid described by OAB is measured by O12+MN (Prop. XII. Cor.) ; the solid described by the triangle OFA has for its measure OI2 FM, the solid described by the triangle OBC, has for its measure OI2 × NO, and since the same may be shown for the solid described by each of the other triangles, it follows that the entire solid described by the semi-polygon is measured by 7012.(FM+MN+NO+OQ+QG), or OI2 × FG ; which is also equal to OI2 × 2FG. But .OI2 is the area of the inscribed circle (Book V. Prop. XII. Cor. 2.): hence the solidity is equivalent to a cone whose base is area OI, and altitude 2FG. PROPOSITION XIV. THEOREM. The solidity of a sphere is equal to its surface multiplied by a third of its radius. Inscribe in the semicircle ABCDE a regular semi-polygon, having any number of sides, and let OI be the radius of the circle inscribed in the polygon. B D A F E If the semicircle and semi-polygon be revolved about EA, the semicircle will C describe a sphere, and the semi-polygon a solid which has for its measure 012× EA (Prop. XIII.); and this will be true whatever be the number of sides of the polygon. But if the number of sides of the polygon be indefinitely increased, the semi-polygon will become the semicircle, OI will become equal to OA, and the solid described by the semi-polygon will become the sphere: hence the solidity of the sphere is equal to OA2x EA, or by substituting 20A for EA, it becomes 7.OA2 × OA, which is also equal to 470A2× OA. But 4π.OA2 is equal to the surface of the sphere (Prop. X. Cor.): hence the solidity of a sphere is equal to its surface multiplied by a third of its radius. Scholium 1. The solidity of every spherical sector is equal to the zone which forms its base, multiplied by a third of the radius. X For, the solid described by any portion of the regular polygon, as the isosceles triangle OAB, is measured by πOI2 × AF (Prop. XII. Cor.); and when the polygon becomes the circle, the portion OAB becomes the sector AOB, OI becomes equal to OA, and the solid described becomes a spherical sector. But its measure then becomes equal to 27.AO2× AF, which is equal to 27.AO × AFxAO. But 27.AO is the circumference of a great circle of the sphere (Book V. Prop. XII. Cor. 2.), which being multiplied by AF gives the surface of the zone which forms the base of the sector (Prop. X. Sch. 1.): and the proof is equally applicable to the spherical sector described by the circular sector BOC: hence, the solidity of the spherical sector is equal to the zone which forms its base, multiplied by a third of the radius. Scholium 2. Since the surface of a sphere whose radius is R. is expressed by 47R2 (Prop. X. Cor.), it follows that the surfaces of spheres are to each other as the squares of their radii; and since their solidities are as their surfaces multiplied by their radii, it follows that the solidities of spheres are to each other as the cubes of their radii, or as the cubes of their diameters. Scholium 3. Let R be the radius of a sphere; its surface will be expressed by 47R2, and its solidity by 4лR2 × R, or R3. If the diameter is called D, we shall have R=D, and RD3: hence the solidity of the sphere may likewise be expressed by = PROPOSITION XV. THEOREM. The surface of a sphere is to the whole surface of the circumscribed cylinder, including its bases, as 2 is to 3: and the solidities of these two bodies are to each other in the same ratio. Let MPNQ be a great circle of the sphere; ABCD the circumscribed D square if the semicircle PMQ and the half square PADQ are at the same time made to revolve about the diameter PQ, the semicircle will gene- M rate the sphere, while the half square will generate the cylinder circumscribed about that sphere. A P The altitude AD of the cylinder is equal to the diameter PQ; the base of the cylinder is equal to the great circle, since its diameter AB is equal to MN; hence, the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter (Prop. 1.). This measure is the same as that of the surface of the sphere (Prop. X.): hence 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; hence the convex surface of the cylinder is also equal to four great circles and adding the two bases, each equal to a great circle, the total surface of the circumscribed cylinder will be equal to six great circles; hence the surface of the sphere is to the total surface of the circumscribed cylinder as 4 is to 6, or as 2 is to 3; which was the first branch of the Proposition. In the next place, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to the diameter, the solidity of the cylinder will be equal to a great circle multiplied by its diameter (Prop. II.). But the solidity of the sphere is equal to four great circles multiplied by a third of the radius (Prop. XIV.); in other terms, to one great circle multiplied by of the radius, or by of the diameter; hence the sphere is to the circumscribed cylinder as 2 to 3, and conse quently the solidities of these two bodies are as their surfacer |