COMMENTATIO ad QUAESTIONEM MATHEMATICAM. ag 



Haec pauca sufficiant de historia Problematis , de qua integra volumina conscribi 

 possent. Plura hac de re scire volentibus patent opuscula supra notata. 



Potius igitur ipsum Problema aggrediamur, 



Posito diametro =: aa , ordinata = j et abscissa = x, erit aequatio circuli y^ =: 

 lax — j:*, ergo y := ± X/^^iax^-x^) et ycfx = dxy/Qiax — x'^) = dx^^iax — x^^jl 

 = dx.x^ (2«— «)l; qaod si dividatur et multiplicetur per (a«)l erit, 



dxVi^ax — xr-) z= dx^ia^k xl (i ^)^ 



Quantitas haec, cum absolute integrari nequeat, necessario in seriem evolvenda e«; 

 ut sic singuli ejus termini integrentur , et integralia , quam proxime inveniri queant , 

 habeamus, Hoc si fiat erit : 



et dxVC^-ax — i=) = dx {2a)i xl ( r — — V = 



^ 2a J - 



J 1 r" '<^ * '•' Ky ■** 1-1-3 vx X^ I.I.S-S ^y X* ~^ , , 



dx.xi Lx--X- - - X -. - — ^ X 8^ - ^ X IH^-etc.JXl/^« 

 — /^rF:»:!— ^ V *^ — — i^ -if — ^•'■3 V *^ ^•'•3-5 w ^2 _etc~lv</»- 



_^,j^AP.-_x _ - - X — , - ^ X g-3 - ^^^ X ,-6^4 "'•Jxv^»' 



CUJU8 seriei si integrale quaeratur, erit f ydx =z fdxVC^^^^''^') 



— r^J^— i y riLI _ LI v-H.^ _LIl? v -H? ^l±±5 V °^'"' — etc 1 X\/2a 



et cum omnes termini complectantur zxk , erit : 



rydx=ri ~ lX-^-'^X "' -'-^Jx^3-^-^X-^ ,-etc.lx 2.1 V^.« W 

 J -^ 3 2 5.2« 2.4 7.4«* 2.4.6 ij.Sa^ 2.4.0.8 11.10«« j 



I 2 3 , X^ XS Xz ^ I ,„- 



sive etiam = - «t/aa ; y-—, — — ^^o^i./ „ — 6'<^« • • • • (P) 



Habemus igitur ope hujus seriei spatium quodcunque circuli ; in ea si ponamus 

 * = <j, sive abscissam aequalem wdio, erit pio area quadrantis circuli ex serie (/3) 



57 9 



— etc. 



/' - » / "^ "^ "- 



^ * - -aiVia — ^^ — ^^^^^^ — ^a^ 56321/2« ,,3 



D = 



