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MathGroup Archive 2007

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area under a parametric curve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74983] area under a parametric curve
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Fri, 13 Apr 2007 02:08:33 -0400 (EDT)

In working with unit square "curves"
I came up with the idea of using a Lemnicape  ( quarter of a curve):

  x = Sqrt[Abs[Cos[t]]]*Sin[t]
  y = Sqrt[Abs[Cos[t]]]*Cos[t]
  x1 = x*Sin[ArcCos[-1/2^((2/3))]] + 1/2
  y1 = y + 1/2
  Solve[1/2 + Sqrt[Abs[Cos[t]]Cos[t] == 0, t]
  ParametricPlot[{x1, 2*y1}, {t, 0, 2*Pi}, PlotRange -> {{0, 1}, {0, 1}}]

My question is how to get the area under such a curve?
Since the curve is symmetrical only {x,0,1/2} is actually necessary.

A second question is how to get rid of the curve "in" part
that comes about because the Lemnicape isn't exactly symmetrical:
what is the maximum x  to y instead of the "2" I've used in this as the 
scaling.
It appears to be about 2.1 instead of 2.
Here a try at a better curve:
  x = Sqrt[Abs[Cos[t]]]*Sin[t]
  y = Sqrt[Abs[Cos[t]]]*Cos[t]
  x1 = x*Sin[2.226604043204] + 1/2
  y1 = y + 1/2
  Solve[1/2 + 2.1 Sqrt[ Abs[Cos[t]] Cos[t]/2 == 0, t]
  ParametricPlot[{x1, 2.1*y + 1}, {t, 0, 2*Pi}, PlotRange -> {{0, 1}, {
  0, 1.0}}]


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