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MathGroup Archive 2004

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problem getting the area under a parmetric curve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52689] problem getting the area under a parmetric curve
  • From: Roger Bagula <tftn at earthlink.net>
  • Date: Thu, 9 Dec 2004 20:22:47 -0500 (EST)
  • Reply-to: tftn at earthlink.net
  • Sender: owner-wri-mathgroup at wolfram.com

I worked on this late last night.
I had trouble even having the curve well defined
finding the area under it.
I used the symmetry to integrate the side
that was easiest. There is only one real zero for the x parametric
in t  which helps.
An Infinite integral does appear to exist for the distribution.

(* cubic pair and rotated distribution function*)
Clear[x0,y0,ang,x,y,z]
x0=(1-t^3)/(1+t^3)
y0=2^(1/3)*t*(3+t^6)^(1/3)/(1+t^3)
ParametricPlot[{x0,y0},{t,-2*Pi,2*Pi}]
Simplify[x0^3+y0^3]
ang=4
x=Cos[Pi/ang]*x0-Sin[Pi/ang]*y0
y=Cos[Pi/ang]*y0+Sin[Pi/ang]*x0
NSolve[x==0,t]
N[y/.t->0.486313]
N[x/.t->0.486313]
ParametricPlot[{x,y},{t,-2*Pi,2*Pi}]
f[t_]=x
norm=Integrate[-y*f'[t],{t,0.486313,Infinity}]
a0=N[2*norm]
g1=ParametricPlot[{x,y}/(a0),{t,-2*Pi,2*Pi}]
g2=Plot[Exp[-t^2/2]/Sqrt[2*Pi],{t,-Pi,Pi}]
Show[{g1,g2}]
Respectfully, Roger L. Bagula

tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
alternative email: rlbtftn at netscape.net
URL :  http://home.earthlink.net/~tftn


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