Re: Re: graphic

• To: mathgroup at smc.vnet.net
• Subject: [mg104583] Re: [mg104518] Re: graphic
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Wed, 4 Nov 2009 01:37:09 -0500 (EST)
• References: <hcl3qc\$btt\$1@smc.vnet.net> <200911030752.CAA01048@smc.vnet.net>
• Reply-to: drmajorbob at yahoo.com

```Or, more generally,

f1[x_, y_] = x^2 + y^2 - 2;
f2[x_, y_] = x^2;
cp = ContourPlot[f1[x, y] == 0, {x, -2, 2}, {y, -2, 2}];
Graphics3D[
Replace[First[cp],
GraphicsComplex[points_, data_] :>
GraphicsComplex[{#1, #2, f2[#1, #2]} & @@@ points, data]]]

Correct?

Bobby

On Tue, 03 Nov 2009 01:52:03 -0600, Szabolcs Horvát <szhorvat at gmail.com>
wrote:

> On 2009.11.01. 23:58, Nodar wrote:
>> Dear All,
>>
>>    May I build 3D curve graphic z=F2(x,y), where x and y are the
>> solutions of equation F1(x,y)==0 ?
>>    May I receive list of points from a "ImplicitPlot" graphic ?
>
> The simplest way would be to solve the F1(x,y)==0 equation.  Here's a
> hackish solution for the case when this is not possible:
>
> Suppose F1(x,y) = x^2+y^2-2 and F2(x,y) = x^2
>
> cp = ContourPlot[x^2 + y^2 - 2 == 0, {x, -2, 2}, {y, -2, 2}]
>
> Graphics3D[
>   Replace[First[cp],
>    GraphicsComplex[points_, data_] :>
>     GraphicsComplex[{#1, #2, #1^2} & @@@ points, data]]]
>
>
> This will work in Mathematica 6 or later.
>

--
DrMajorBob at yahoo.com

```

• References:
• Re: graphic
• From: Szabolcs Horvát <szhorvat@gmail.com>
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