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Author Comment/Response
Bill Simpson
10/25/12 2:23pm

In Response To 'Re: Re: Parametric regionfunction'
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I believe you will have more success if you turn your parametric function for your region into a simpler boolean that only depends directly on your plot variables.

Fortunately your parametric function for your region is simple enough to perhaps allow this.

In[1]:= Simplify[{x1==((1-u2)^2*0+2 (u2) (1-u2)*(150)+(u2)^2*(150)),
y1==((1-u2)^2*-(100)+2 (u2) (1-u2)*-(100)+(u2)^2*0)}]

Out[1]= {x1==300u2-150u2^2, y1== -100+100u2^2}

In[2]:= Eliminate[%, u2]

Out[2]= 4x1^2+x1(1200+12y1)==270000+1800y1-9y1^2

In[3]:= Solve[%,y1]

Out[3]= {{y1->2/3(150-10Sqrt[6]Sqrt[150-x1]-x1)}, {y1->2/3(150+10Sqrt[6]Sqrt[150-x1]-x1)}}

Now one of those solutions is the correct one for your regionfunction, substitute values for x1 to determine which is correct.

Can you now take that much simpler result, change the variables to those used in your original ParametricPlot3D and use that to give a regionfunction that Mathematica can understand and correctly use?

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Subject (listing for 'Parametric regionfunction')
Author Date Posted
Parametric regionfunction Paul 10/23/12 2:23pm
Re: Parametric regionfunction Bill Simpson 10/24/12 00:14am
Re: Re: Parametric regionfunction Paul 10/25/12 09:50am
Re: Re: Re: Parametric regionfunction Bill Simpson 10/25/12 2:23pm
Re: Re: Re: Re: Parametric regionfunction Paul 10/26/12 01:49am
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