For, supposing the construction in the last lemma and theorem to remain, conceive the axis of the sphere AB to be divided into innumerable equal parts Dd, and the whole sphere to be divided into as many sphe rical concavo-convex laminz EFfe; and let the per pendicular do be erected. By the last theorem, the force, with which the lamina EFfe attracts the par ticle P, is as DE' x Ff, and the force of one particle exerted at the distance PE or PF, jointly. But, by the last lemma, Dd is to Ff as PE to PS ; and there P S x PEDd fore Ff is equal to --; and DE' x Ff is equal to Dd x PEPS x —• and therefore the force of the lamina EFfe is as and the force of a particle exerted at the distance PF, jointly • that is, -from the supposition, as DN x Dd, or as the inde finitely small area DNnd. Therefore the forces of all the laminz, exerted upon the particle P, are as all the areas DNnd ; that is, the whole force of the sphere is as the whole area ANB.
Corot. 1. Hence, if the centripetal force tending to the several particles remains always the same at all distances, and Dn be made as DE' x PS , the whole PE force, with which the particle P is attracted by the sphere, is as the area ANB.
Corot. 2. If the centripetal force of the particles is reciprocally as the distance of the particle attracted by it, and DN is made IspEzPEI X PS ' • the force, with which the particle P is attracted by the whole will be as the area ANB.
Cor. 3. If the centripetal force of the particles is reciprocally as the cube of the distance of the par DE' x tide attracted by it, and DN is made as -- • ' the force, with which the particle P is attracted by the whole sphere, will be as the area ANB.
Cor. 4. And universally, if the centripetal force, tending to the several particles of a sphere, is sup posed to be reciprocally as the quantity V, and DN ' is made as DE x the force, with which a par PE x V ticle is attracted by the whole sphere, will be as the area ANB.
Supposing what has been already established, it is required to measure the area ANB.
From the point P let the right line PH be drawn, touching the sphere in H ; and having let fall HI perpendicular to the axis PAB, let PI be bisected in L ; and PE' will be equal to But, because the triangles SPH, SHI are similar, SE' or SH' is equal to the rectangle PSI. There fore PE' is equal to the rectangle contained under PS and PS -1-SI-1-2SD ; that is, under PS and 2LS ; that is, under PS and 2LD. More over, DE' is equal to SE'—SD', or 2SLD—LD' ; that is, 2SLD—LD'----ALB. For LS'—SE', or LSI—SA', is equal to the rectangle ALB. Let therefore 2SLD —ALB be DE' x PS substituted for DE' ; and the quantity PE x V which, according to the fourth corollary of the ceding proposition, is as the length of the ordinate DN, will resolve itself into three parts, 2SLD x PS