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Re: 3D graphics domain

• To: mathgroup at smc.vnet.net
• Subject: [mg55762] Re: 3D graphics domain
• From: dh <dh at metrohm.ch>
• Date: Tue, 5 Apr 2005 05:45:12 -0400 (EDT)
• References: <d2tesa$qj2$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Hi Dick,
you may simply plot your function on a rectangular region and color the 
valid region differently. E.g.:

f[x_, y_] = -64*x +  320*(x^2) - 512*(x^3) + 256*(x^4) + 20*y - 64*x*y + 
64*(x^2)*y - 4*(y^2);
s[x_, y_] = If[(y = 4*x*(1 - x)) && (y = 4*x*(1 - 2x)) && (y = 4*(x - 
1)*(1 - 2x)), Hue[1], Hue[0.5]];
Plot3D[{f[x, y], s[x, y]}, {x, 0, 1}, {y, 0, 1}, PlotPoints -> 50]

Sincerely, Daniel


Richard Bedient wrote:
> Thanks to Bob and Dan for helping me get this far. Again, I've exhausted
> my Mathematica knowledge along with anything I can find in the Help
> files.  I now need to take the function they found for me and graph it
> in 3D over a restricted domain. Here's the problem:
> 
> Graph the function
> 
> f(x,y) = -64*x + 320*(x^2) - 512*(x^3) + 256*(x^4) + 20*y - 64*x*y +
> 64*(x^2)*y - 4*(y^2) 
> 
> over the domain:
> 
> y <= 4*x*(1-x)
> y >= 4*x*(1 - 2x)
> y >= 4*(x - 1)*(1 - 2x) 
> 
> Thanks for any help.
> 
> Dick
> 

• Follow-Ups:
  • Re: Re: 3D graphics domain
    • From: DrBob <drbob@bigfoot.com>