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MathGroup Archive 2003

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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?
>



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