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How do I draw with Mathematica a function like this?enter image description here

I would use a series of circles and find the surface.

Thanks!

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3 Answers 3

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ParametricPlot3D[{r Cos[θ], r Sin[θ], .2 RandomReal[]}, {r, 1, 2}, {θ, 0, 2 π},
  ColorFunction -> Function[y, Blend[{Red, Blue}, y]]]
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ParametricPlot3D[{r Cos[θ], r Sin[θ], Sin[20 θ]/15}, 
 {r, 1, 2}, {θ, 0, 2 π}, 
 ColorFunction -> Function[{x, y, z, u}, Blend[{Red, Pink, LightBlue, Blue}, y]], 
 Mesh -> None, Boxed -> False, Axes -> False, PlotPoints -> 30,
 Background -> Black]

enter image description here

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  • $\begingroup$ Very nice, +1, but I think there should be a certain amount of randomness somewhere $\endgroup$
    – eldo
    Dec 1, 2015 at 22:57
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Just play with it, using functions with periods that divide 2 Pi:

ParametricPlot3D[{r Cos[θ], r Sin[θ], 
  0.015 (Sin[7 θ + 10 r Cos[2 θ]] + Sin[5 θ + 10 r Sin[2 θ]]) + 
   0.025 Sin[30 θ + Sin[θ]] + (r - 2)^2 Cos[3 θ + 0.73] - 0.15 r},
 {r, 2, 3}, {θ, 0, 2 π},
 Mesh -> None, 
 ColorFunction -> Function[{x, y, z, r, θ}, 
   ColorData["ThermometerColors"][y]],
 PlotPoints -> {15, 60}, PlotRangePadding -> {0, 0, 0.5}, 
 Background -> Black, Boxed -> False, Axes -> False]

Mathematica graphics

(I mean, it's not like an anatomy lesson.)

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  • $\begingroup$ Nicely done. (+1) $\endgroup$
    – Edmund
    Dec 2, 2015 at 0:28

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