       RE: circular plot question

• To: mathgroup at smc.vnet.net
• Subject: [mg28449] RE: [mg28433] circular plot question
• From: "David Park" <djmp at earthlink.net>
• Date: Thu, 19 Apr 2001 03:26:38 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Julian,

There are probably several methods, but I use my DrawingCube package
available at my web site for this kind of graphics problem. Here is a sample
function.

f[r_, \[Theta]_] := (r - r^2)*Cos[\[Theta]]

Needs["Graphics`DrawingCube`"]

The following draws the 3D surface in the r, theta plane (which is not what
you want.) Draw3D is just like Plot3D except that it extracts the primitive
graphics in Graphics3D format. That is, it extracts the list of all the
polygons that make up the surface.

Show[Graphics3D[{Draw3D[f[r, \[Theta]], {r, 0, 2}, {\[Theta], 0, 2*Pi},
PlotPoints -> {15, 31}]}], AspectRatio -> Automatic,
PlotRange -> All];

The following converts the cylindrical coordinates in the graphics
primitives to Cartesian coordinates and produces the desired circular plot
with Cartesian axes. The conversion is done by the DrawingTransform3D
package routine. The fact that Draw3D produces the primitive graphics makes
it easy to operate on them with routines like DrawingTransform3D.

Show[Graphics3D[{Draw3D[f[r, \[Theta]], {r, 0, 2}, {\[Theta], 0, 2*Pi},
PlotPoints -> {15, 31}] /. DrawingTransform3D[
#1*Cos[#2] & , #1*Sin[#2] & , #3 & ]}],
AspectRatio -> Automatic, PlotRange -> All];

David Park

> From: Julian Sweet [mailto:jsweet at engineering.ucsb.edu]
To: mathgroup at smc.vnet.net
> 	How can I go about getting Mathematica to create a 3D plot of
> a function with z and phi dependence (cylindrical coordinates) such
> that it will show it's cylindrical symmetry.
> 	Right now I am able to make a 3D plot using z and phi, but I
> would like it to appear in a circle.
>
>
> Thanks
>
>
> E-mail response preferred:  jsweet at engineering.ucsb.edu
>
>

```

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