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