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MathGroup Archive 2005

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Re: Re: 3D graphics domain

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55783] Re: [mg55762] Re: 3D graphics domain
  • From: DrBob <drbob at bigfoot.com>
  • Date: Wed, 6 Apr 2005 03:11:12 -0400 (EDT)
  • References: <d2tesa$qj2$1@smc.vnet.net> <200504050945.FAA00513@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Did you test that code?

You had "=" where you must have meant "<=" or ">=" (three times).

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)), Gray, White];
Plot3D[{f[x, y],
     s[x, y]}, {x, -1, 2}, {y, 0, 3}, PlotPoints -> 50, Mesh -> False]

Notice the If turns out to be false throughout the plotted region.

Bobby

On Tue, 5 Apr 2005 05:45:12 -0400 (EDT), dh <dh at metrohm.ch> wrote:

> 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
>>
>
>
>
>



-- 
DrBob at bigfoot.com


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