Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/390

300 be the locus of the point K. Let that curve meet the plane of the circle in L. In PA take PH equal to PD, and erect the perpendicular HI meeting that curve in I; and the attraction of the corpucle P towards the circle will be as the area AHIL drawn into the altitude AP. Q. E. I.

For let there be taken in AE a very mall line Ee. Join Pe, and in PE, PA taka PC equal to Pe. And becaue the force with which any point E of the annulus decribed about the centre A with the interval AE in the aforeaid plane, attracts to it elf the body P, is uppoed to be as FK; and therefore the force with which that point attracts the body P towards A is as $$\scriptstyle \frac {AP \times FK}{PE}$$ and the force with which the whole annulus attracts the body P towards A, is as the annulus and $$\scriptstyle \frac {AP \times FK}{PE}$$ conjunctly; and that annulus alo is as the rectangle under the radius AE and the breadth Ee, and this rectangle (becaue PE and AE, Ee and CE are proportional) is equal to the rectangle PE x CE or PE x Ff; the force with which that annulus attracts the body P towards A, will be as Pe x Ff and $$\scriptstyle \frac {AP \times FK}{PE}$$ conjunctly; that is as the content under Ff x FK x AP, or as the area FKkf drawn into AP. And therefore the um of the forces with which all the annuli, in the circle decribed about the centre A with the interval AD, attract the body P towards A, is as the whole area AHIKL drawn into AP. Q. E. D.