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Re: Re: 3d plots in mathematica 5.0

 Hi Fiz,
> 
> Here is the function
> 
> w=exp(-2(x^2+y^2/r0^2))*exp(-2(z^2/z0^2))
> 
> r0=1.4 and z0=1.4

The function is Exp, not exp, and it needs square brackets not parentheses.

r0 = 1.4;
z0 = 1.4;
w = Exp[(-2(x^2 + y^2/r0^2))]*Exp[(-2(z^2/z0^2))] // FullSimplify

The function you're after is ContourPlot3D, and you use it like

Needs["Graphics`ContourPlot3D`"]
ContourPlot3D[w, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, Contours -> {.75}, \
PlotRange -> All, Axes -> True, BoxRatios -> {1, 1, 1}];

This produces "something", but it's not (IMHO) particularly useful or
informative.

You might get better value from either animating the 3D plot or another
approach.

Dave.

0
7/30/2005 5:23:46 AM
comp.soft-sys.math.mathematica 28821 articles. 0 followers. Follow

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