Dec. 5, 1884.] 



♦ KNOWLEDGE ♦ 



457 



the curved arc A P B. You can divide this area by a 

 series of such parallels as are shown in Fig. 1, — say equi- 

 distant parallels, — into spaces such as L P Q N, in which, 

 since P Q is curved, you would seem to have the same 

 difficulty to deal with as in the origical figure. But if you 

 complete the rectangles N /, L )*, you see that the area 

 P Q ^ is very small compared with either of these rect- 

 angles ; and you can easily see that if the strip P N is 

 made very narrow indeed the area P Q Z mi^ht be neglected, 

 compared with Q L or P N, either of which rectangles you 

 might take, instead of P Q L, without appreciable error. 

 Doing this for the whole area you have rectangles to deal 

 with, instead of surfaces bounded by a curved line, or by 

 curved lines. 



A . Yes ! — but — though you thus make each little area as 

 F Q I very small, you make the number of such areas very 

 large, and a very large number of very small things may 

 make an appreciable total. 



J/. True ; but if each such part as P Q i is very small 

 compared with the total area P Q N L, must not the sum 

 of all such parts be very small compared with the sum of 

 all such areas 1 — or in other words must not all the little 

 areas like P I Q, taken together, be insignificant compared 

 with the area A B M O 1 



A. I cannot say I quite see this. 



Fig. 1. 



M. Suppose P ; Q a millionth part of P Q N L, and none 

 of the other small areas like F IQ to be 7nore than a 

 millionth part of the area like P Q N L of which it is a 

 portion. Would not all such areas as P ZQ, together, be 

 less than a millionth part of the area A B M ^ 



A. Yes. I see that. But it does not seem possible to 

 prove that you can so arrange matters that such areas as 

 F IQ may be any given very small proportion of such areas 

 asPQNL. 



J/. Why, regarding P for the moment as fixed, may not 

 Q be brought so close up to P that P I may be as small as 

 we please, and therefore the rectangle I n bear as small a 

 ratio as we please to the rectangle Q L. If this is done, 

 then assuredly the area P / Q will bear a smaller ratio still 

 to Q L. Suppose you wanted F I to have some given value ; 

 well, all you have to do is to measure off P ? of that value, 

 and from I to draw / Q parallel to O X ; then you have 

 your points P, Q, from which to draw parallel pei-pen- 

 diculars to O X. 



A. The proof, I see, depends on making the parallels 

 of Fig. 1 as close together as possible. 



M. Quite so. And we may easily prove the soundness 

 of the conclusion we have just reached. From PQ draw 

 P p m, Q q k, perpendicular to A 0, and let a series of such 

 perpendiculars be drawn from all such points as P, Q, (fee, 



B being the last of them. Then if the parallels are equi- 

 distant the rectangle I n is equal to the rectangle p k, and 

 the like for all such rectangles as I n ; giving, for the sum 

 of them all, the rectangle oa, and therefore for the sum of 

 all such areas as P / Q an area less than the rectangle o a. 

 But o a may be made as small as we please by sufficiently 

 increasing the number of the parallels. Even if the 

 parallels are not equidistant, the proof remains good, so 

 long as the greatest distance between the parallels may be 

 made eventually as small as we please. For obviously the 

 rectangle we then get instead of « o may also be made 

 less than any area which can be assigned. 



A. But now, all this is very indefinita What is the 

 curve A P B ? and how can such a series of parallels be 

 drawn as your argument requires 1 



J/. There is no more occasion for actually drawing such 

 parallels, than for making our straight lines, circles, «fec., in 

 ordinary geometry, be perfectly accordant with the defi- 

 nitions. 



..1. But will you not give an instance of the actual 

 application of this method ! 



J/. Certainly. Let us take my old friend the cycloid, 

 which affords more examples of various ways of measuring 

 areas than any curve I know of. I wUl take one of the 

 simplest proofs of the area of the cycloid from among 

 nearly a score which I have given in my " Geometry of 

 Cycluid.-'." 



A g V li ,■• J 



Fig. 2. 



Let A P D, Fig. 2, be a half cycloid, A B its axis, A p B 

 half the rolling circle. We want to find the area A P D B. 

 Complete the rectangle A B D 6, whose area (since B D is 

 equal to arc A/) B) is equal to twice that of the rolling 

 circle. Let A Q D be an equal half cycloid having D h 

 as axis. Then if ?)iQ/)P be drawn parallel to A 6 we 

 know that MP=arc Aj3-f M^ 



and M Q = arc A^ - Mjj 

 Hence, subtracting, Q P=2 M;j 



Therefore, if we draw a parallel \jlknF. very near 

 indeed to M P, as in Fig. 2, 



rectangle Q n^l rect. M^. 

 Imagine an immense number of such parallels athwart the 

 semicircle and the cycloids. The sum of all the rectangles 

 in the space A P D Q is equal to twice the sum of aU the 

 rectangles in the semicircle Ap B. Wherefore eventually 

 (when the number of parallels becomes indefinitely large) 

 we have 



Area A P D Q = 2 semicircle A /I B (i.) 

 But (since rect B6 = 2 rolling circle), 



AQDB-fAPD6=2 semicircle A/) B 

 or. Area A Q D B=semicircle A ;; B (ii.) 



. •. adding (i.) and (ii.). 



Area A P D B=3 semicircle A^ B 

 or, the area of a cycloid is equal to thrice the area of its 

 generating circle. 



A. That is very curious. Let me see, — I want another 

 figure to make the matter clear to me, and to see some of 

 its consequences. I take the oddly-shaped space A Q D P, 

 and set the rolling circle of either cycloid midway upon it, 



