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MathGroup Archive 2005

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Re: Re: grphing in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54462] Re: [mg54374] Re: grphing in Mathematica
  • From: jeffb at wolfram.com
  • Date: Sun, 20 Feb 2005 00:11:44 -0500 (EST)
  • References: <bsire1sg1x3j@legacy> <200502190732.CAA06135@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

  ContourPlot3D works just fine for this problem.  Try the following:

In[1]:= <<Graphics`

In[2]:=
ContourPlot3D[x^2+y^2+4 z^2,{x,-2,2},{y,-2,2},{z,-2,2},Contours->{1},
  PlotPoints->{7,7},Axes->True,AxesLabel->{"x","y","z"}]

Using version 5.0, I get a resulting ellipsoid in just a few seconds.

-Jeff

-------------
Jeff Bryant
Wolfram Research, Inc.

> On 17 Feb 05 12:08:58 -0500 (EST), Nick wrote:
>>How do I graph the function x^2+y^2+4z^2=1?
>>I am using Mathemetica version 4 and
>>using ContourPlot3D doesn't work.
>
>
> Mathematica plots implicit curves using ImplicitPlot, however, it's a
> planar version only and as far as I am aware standard packages of
> Mathematica
> 5.1 implement no 3D version. If you are able to parametrize the curve,
> use ParametricPlot3D. Otherwise you're left to create a suitable
> package yourself.
>
> Jaroslav
>


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