Re: Integrating over area of intersection

*To*: mathgroup at smc.vnet.net*Subject*: [mg44142] Re: Integrating over area of intersection*From*: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>*Date*: Fri, 24 Oct 2003 04:24:10 -0400 (EDT)*References*: <bn8esp$nle$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Here is an example of the sort of thing you can do. First of all read in the Calculus`Integration` package which gives you the Boole function for doing integrals over regions defined by inequalities: << Calculus`Integration` Now integrate the function x^2*y^2 (as an example) over the region of interest (as an example fJ = 3, fH = 2, f1 = 1/2, f2 = -4^(-1)}): With[{fJ = 3, fH = 2, f1 = 1/2, f2 = -4^(-1)}, Integrate[Boole[x^2 + y^2 < fJ^2 && (x - f1)^2 + y^2 < fH^2 && (x - f2)^2 + y^2 < fH^2]*x^2*y^2, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]] which gives the result: -((56359*Sqrt[247])/163840) + (11/3)*ArcCos[3/16] + (35/6)*ArcSin[Sqrt[13/2]/4] -- Steve Luttrell West Malvern, UK "Toni Danza" <nospam at yoohoo.com> wrote in message news:bn8esp$nle$1 at smc.vnet.net... > OK, I have three functions that are defined within their respective circles. > I would like to integrate over the intersection of the three circles. > > Here's what I have done: > define region of integration: > region = x^2 + y^2 < fJ^2 && (x - f1)^2 + y^2 < fH^2 && (x - f2)^2 + y^2 > < fH^2 > > Then I try to solve for the intersection using > > Reduce[region,{x,y}] > > and the result is something like (only works with numerical parameters...) > > -0.4<x<0.3 && sqrt(....)< y <sqrt(...) || -0.3<x<-0.2 && sqrt(....)< y > <sqrt(...) > > How do I use this result to do integration over the region? >