PROPOSITION XII. LEMMA. If a triangle be revolved about any line drawn through its vertex in the same plane, the solid generated 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. then Let CAB be a triangle, I the middle point of the base, and CD the line about which it is to be revolved: will the solid generated be measured by solid generated by CAD=3«×AM3× CD, solid generated by CBD=3«×BN3× CD: hence, the difference of these solids, which is the solid generated by the triangle CAB, has for its measure (AM-BN)× 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 then have (B. IV., P. 7, s.), AM+BN=2IK, and AM-BN=AO; hence, (AM+BN)×(AM-BN)=AM2-BN2=2IK×AO:hence, the measure of the solid is also equal to TXIK×AOX CD. But CP being perpendicular to AB produced, the triangles AOB and CPD are similar; hence, and, AO : CP :: AB : CD. AOX CD=CPX AB. But CPXAB is double the area of the triangle CAB; therefore, AOX CD=2CAB: hence, the solid generated by the triangle CAB is measured by × CAB×IK=CAB×3«×IK; and since 2TXIK=circ. IK, we have, solid=CABX circ. IK. CAB in the measure of the solid before found, viz.: solid=CAB×TM×IK, gives, solid=3>AB × IK × CI. But the triangles AOB, CKI, are similar (B. IV., P. 21); : BO or MN :: CI : IK, hence, AB which gives, AB×IK=MN × CI. Substituting for ABXIK, we have, solid=CIX MN: that is, the solid generated by the revolution of an isosceles triangle about any line drawn through its vertex, is measured by two-thirds of into the square of the perpendicular let fall on the base, into the distance between the two perpendic ulars let fall from the extremities of the base on the axis. Scholium. The demonstration appears to involve the supposition that AB prolonged will meet the axis: but the results are equally true if AB is parallel to the axis. Thus, the cylinder generated by Р MNBA is measured by × AM3× MN: the cone generated by CAM is measured by ×AM3× CM; and the cone generated by CBN is measured by XAM × CN. A M N and since MN+}CM=}CN, we have But AM CP and MN=AB; hence, M solid by CAB=AB×CP×3«×CP=CAB×3circ. CP. But the circumference traced by P is equal to the circumference traced by the middle point of the base: hence, the result agrees with the general enunciation. PROPOSITION XIII. LEMMA. If a regular semi-polygon be revolved about a line passing through its centre and the vertices of two opposite angles, the solid generated will be measured by two-thirds the area of the inscribed circle multiplied by the axis. Let GDBF be a regular semi-polygon and OI the radius of the inscribed circle: then, if this semi-polygon be revolved about GF, the solid generated will have for its measure, area OIX GF. For, since the polygon is regular, the triangles, OFA, OAB, OBC, &c., are isosceles and equal; then, all the perpendiculars let fall from O on their bases, will be equal to OI, the radius of the inscribed circle. Now, we have the following measures for the solids generated by these triangles (P. 12, c.): viz., R G hence, the entire solid generated by the semi-polygon is measured by 3π×ÕI2 (FM+MN+NO+ OQ+QR+RG): The solidity of a sphere is equal to its surface multiplied by a third of its radius. Let O be the centre of a sphere and OA its radius: then its solidity is equal to its surface into one-third of OA. semi-circle For, inscribe in the ABCDE a regular semi-polygon, having any number of sides, and let OI be the radius of the circle inscribed in the polygon. If the semicircle and semi-polygon be revolved about EA, the semicircle will generate a sphere, and the semipolygon a solid which has for its measure OIX EA (P. 13); and this is true whatever be the number of sides A B D E of the semi-polygon. But if the number of sides of the polygon be continually doubled, the limit of the solids generated by the polygons will be the sphere; and when we pass to the limit the expression for the solidity will become XOA XEA, or by substituting 20A for EA, it becomes XOA3× OA, which is also equal to 4×OA3× OA. But 4×бA is equal to the surface of the sphere (P. X., C. 1): 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. For, the solid described by any portion of the regular polygon, as the isosceles triangle OAB, is measured by OTXAF (P. 12, c.); and when we pass to the limit which is the spherical sector, the expression for this measure becomes XAOXAF, which is equal to 2 XA0XAFX3A0. But 2,XA0 is the circumference of a great circle of the sphere (B. V., P. 16), which being multiplied by AF gives the surface B D A E of the zone which forms the base of the sector (P. X., c. 2); 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 4XR (P. X., C. 1), 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 4 XR, and its solidity by 4≈×R3×3R, or ̃×R3. If the diameter be denoted by D, we shall have R={D, and R3={D3: hence, the solidity of the sphere may be expressed by 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 |