RE: Error?

*To*: mathgroup at smc.vnet.net*Subject*: [mg44717] RE: [mg44680] Error?*From*: "Florian Jaccard" <florian.jaccard at eiaj.ch>*Date*: Sat, 22 Nov 2003 02:17:12 -0500 (EST)*Reply-to*: <florian.jaccard at eiaj.ch>*Sender*: owner-wri-mathgroup at wolfram.com

Hello ! Very interesting question ! As Integrate[Sqrt[r^2 - x^2], x] = (1/2)*(x*Sqrt[r^2 - x^2] + r^2*ArcTan[x/Sqrt[r^2 - x^2]]) you will have to use complex numbers using Newton-Leibnitz's formula with R<d<2R ! As (I suppose) you want to compute a surface under a circle, it is not good ! so I would suggest that you should inetgrate only with d<R ! Indeed : In[34]:= Integrate[Sqrt[r^2 - x^2], {x, -r, -r + d}, Assumptions -> {d > 0, r > 0, d < r}] Out[34]= (1/4)*(2*Sqrt[(-d)*(d - 2*r)]*(d - r) + Pi*r^2 + 2*r^2*ArcSin[(d - r)/r]) So if d>R, just compute for x from 0 to d-R and add Pi r^2/4 ! Meilleures salutations Florian Jaccard -----Message d'origine----- De : Baris Altunkaynak [mailto:altunkai at boun.edu.tr] Envoyé : ven., 21. novembre 2003 11:13 À : mathgroup at smc.vnet.net Objet : [mg44680] Error? This integral below give 0 on Mathematica 5.0 Integrate[Sqrt[R^2 - x^2], {x, -R, -R + d}, Assumptions -> R > 0 && d > 0 && d < 2R] This is a mistake I think, isn't it? Thanks, Baris Altunkaynak