C A P UT VIII.

21 I

Solutio.

Sit fpatium MPCZ=S, quod inftar unius funftionis abfciflae xt= APpoteft fpeftari. Si abfciffa x=AP capiat incrementum Ax = PQ, duca-turque ordinata Qm, et M p parallela axi abfciffarum V U, altera vero pa-rellela ml ordinatae MP occurrat in l; erit fpatium PMmQ^AS, mp:=Ay, Mp=Ax; hinc AS >MpQP et iimul AS< ImQP pro qualibetdifferentia A x. Habebimus igitur A S >y Ax et Iimul AS< (y+Ay)Ax,unde ob (84. §.) per (131. §.) obtinetur exponens differentialis eS=yex,cujus integrale aequabitur fpatio S.

489- Problema.

Data aequatione ad curvam AX (16. 17. Fig.) inter ordinatas z=BCFig.iS.iad datum punitum C convergentes et angulos variabiles ACB:=y, quadrarefelior em A C B, /eu invenire ejus aream S.

Solutio.

Angulus yr=ACB capiat incrementum BCD —Ay, et ex punfto Cradiis CB —z, CD defcribantur arcus circulares Bd, Db; erit Dd==Bb= Az differentia ordinatae BCr=z, et area feftoris BCDs= AS differen-tia areae Si=BCA (17. 18- §-): 6 ergo ordinata z fpeftetur inftar certaefunftionis anguli y; poterit poni Azc=«Ay-f-/ 3 Ay 2 -f-etc. ( 84 - §-)-

Iam vero patet, aream AS feftoris BDC pro quovis angulo A y itafe habere ad areas feftorum circularium B d C, b D C, ut debeat eife AS BdCet funul AS<bDC (i6.Fig.), vel AS >bDC et AS< BdC (17.Fig.);ea propter, cum fit arcus circularis Bd~Ay.BC — Ay.z, et b 0 —Ay.OC,feu bD = Ay(z + Az) in (16.Fig.), et bD—-Ay(z —Az) in (17. Fig.) per(5. §. z. Scbol.); debebit eile AS >^z 2 Ay, et AS< \ z 2 A y -f- (z+J (x A y+ , 3 Ay 2 + etc.)) >Ay 2 + / 3 Ay 3 +etc.); vel ASC |:Z 2 Ay, etAb>f.z 2 Ay+ (i(«Ay + / 3 Ay 2 -f etc.) — z) (x A y 2 + /3 A y 3 + etc.).

Eft igitur eS —iz 2 ey (i3i.§.) exponens differentialis areae S, cujusintegratio dabit idcirco ipsam aream S —ABC.

490. Problema.

Data aequatione ad curvam BPC ( 18- Fig-) inter ejus coordinatas rig. ig.x —Ba, y —aP, cujus revolutione circa axem abfcijfarum A E cogitetur ge-nerari corpus rotundum BCD; invenirefoliditatem ejusdem corporis.

Dd 2

Sol u-