如何用mathematica或者matlab画三维图形，例如旋转抛物面 x^2+z^2=60y , (y,0,21.6).希望能给出详细代码。

如题所述

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相关建议 2012-07-28

No intention to get the reward. Just show how powerful the Mathematica is!

The basic idea is from neu_lin.

ContourPlot3D[
x^2 + z^2 == 60 y, {x, -40, 40}, {y, 0, 21.6}, {z, -40, 40},
Mesh -> None, ContourStyle -> Directive[Opacity[0.5], Red],
ColorFunction -> Function[{x, y, z, f}, Hue[z]]]

ContourPlot3D[

x^2 + z^2 == 60 y, {x, -40, 40}, {y, 0, 21.6}, {z, -40, 40},

ColorFunction -> Function[{x, y, z, f}, Hue[Sin[x] Sin[z]]],

ColorFunctionScaling -> False, Mesh -> None]

ContourPlot3D[

x^2 + z^2 == 60 y, {x, -40, 40}, {y, 0, 21.6}, {z, -40, 40},

MeshFunctions -> {#1 - #2 &, #1 + #2 &},

MeshShading -> {{Yellow, Blue}, {Green, Cyan}},

ColorFunction -> Function[{x, y, z, f}, ColorData["Rainbow"][z]],

Mesh -> 4,

ContourStyle ->

Directive[Orange, Opacity[0.6], Specularity[White, 30]]]

ContourPlot3D[

x^2 + z^2 == 60 y, {x, -40, 40}, {y, 0, 21.6}, {z, -40, 40},

MeshFunctions -> {#3 &},

ContourStyle -> Directive[Opacity[0.5], Yellow],

MeshShading -> {Red, Automatic}]

ContourPlot3D[

x^2 + z^2 == 60 y, {x, -40, 40}, {y, 0, 21.6}, {z, -40, 40},

Mesh -> None,

ColorFunction -> Function[{x, y, z, f}, ColorData["Rainbow"][z]],

Mesh -> 4,

ContourStyle ->

Directive[Orange, Opacity[0.6], Specularity[White, 30]]]

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其他看法

第1个回答 2012-07-26

figure;

bound=sqrt(60*21.6);

[x,z]=meshgrid(linspace(-bound,bound));

y=(x.^2+z.^2)/60;

y(y>21.6)=NaN;

mesh(x,y,z);

xlabel('x');ylabel('y');zlabel('z');

效果图如下图：

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第2个回答 2012-07-28

自己看看啊

ContourPlot3D[x^2+z^2==60 y,{x,-40,40},{y,0,21.6},{z,-40,40},Mesh->None]

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