RE: graphing in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg54435] RE: [mg54344] graphing in Mathematica*From*: "David Park" <djmp at earthlink.net>*Date*: Sun, 20 Feb 2005 00:10:05 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

I have to admit that getting ContourPlot3D to work relatively efficiently, or to work at all takes a certain amount of trial and error. It helps if you have a good idea of the actual surface. In any case, the following works fairly well. ContourPlot3D[ x^2 + y^2 + 4z^2 - 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -0.6, 0.6}, PlotPoints -> {4, 6}, PlotRange -> {{-1.1, 1.1}, {-1.1, 1.1}, {-0.6, 0.6}}]; If we solve for z we could try to parametrize the surface. Solve[x^2 + y^2 + 4z^2 == 1, z] {{z -> (-(1/2))*Sqrt[1 - x^2 - y^2]}, {z -> (1/2)*Sqrt[1 - x^2 - y^2]}} The following does not work because of the warning messages and the ragged edges. ParametricPlot3D[{{x, y, (-(1/2))*Sqrt[1 - x^2 - y^2]}, {x, y, (1/2)*Sqrt[1 - x^2 - y^2]}}, {x, -2, 2}, {y, -2, 2}]; Using the DrawGraphics package at my web site we can plot in polar coordinates and then transform to Cartesian coordinates. This will plot faster. Needs["DrawGraphics`DrawingMaster`"] Draw3DItems[ {{ParametricDraw3D[{{r, \[Theta], (1/2)*Sqrt[1 - r^2]}, {r, \[Theta], (-2^(-1))*Sqrt[1 - r^2]}}, {r, 0, 1}, {\[Theta], 0, 2*Pi}]} /. DrawingTransform3D[#1*Cos[#2] & , #1*Sin[#2] & , #3 & ]}, BoxRatios -> {1, 1, 0.5}, Background -> Linen, ImageSize -> 450]; As usual we obtain less than a smooth rendering near the 'equator' because of the steepness of the z function there. The following does a somewhat better job by a nonlinear scaling of the r parameter. Draw3DItems[ {{ParametricDraw3D[{{s, \[Theta], (1/2)*Sqrt[1 - s^(2/3)]}, {s, \[Theta], (-2^(-1))*Sqrt[1 - s^(2/3)]}}, {s, 0, 1}, {\[Theta], 0, 2*Pi}]} /. DrawingTransform3D[#1^(1/3)*Cos[#2] & , #1^(1/3)*Sin[#2] & , #3 & ]}, BoxRatios -> {1, 1, 0.5}, Background -> Linen, ImageSize -> 450]; David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Nick [mailto:nch2198 at louisiana.edu] To: mathgroup at smc.vnet.net How do I graph the function x^2+y^2+4z^2=1? I am using Mathematica version 4 and using ContourPlot3D doesn't work.