Re: Re: 3d plots in mathematica 5.0

*To*: mathgroup at smc.vnet.net*Subject*: [mg59127] Re: [mg59124] Re: 3d plots in mathematica 5.0*From*: "David Annetts" <davidannetts at aapt.net.au>*Date*: Sat, 30 Jul 2005 01:25:01 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

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.