Problem showing graphics in Mathematica

[jorgext]

Hi all,

I solved a problem of electrostatics potencial of a cylinder with special condicions. When I plotted the obtained solution, I used this code:

\[Epsilon] = 8.8541878176 10^-12;
R=1;
maxx=10;
Vin[\[Rho]_, \[Phi]_] := (2 \[Sigma] R)/(\[Pi] \[Epsilon])
Sum[If[OddQ[k], 1/k^2 Sin[k \[Phi]] (\[Rho]/R)^k, 0], {k, 0, maxx}]
Vout[\[Rho]_, \[Phi]_] := (2 \[Sigma] R)/(\[Pi] \[Epsilon])
Sum[If[OddQ[k], 1/k^2 Sin[k \[Phi]] (\[Rho]/R)^-k, 0], {k, 0,
maxx}]
V[\[Rho]_, \[Phi]_] :=
If[R < \[Rho], Vin[\[Rho], \[Phi]], Vout[\[Rho], \[Phi]]]
Show[
RevolutionPlot3D[
V[\[Rho], \[Phi]], {\[Rho], 0, 2}, {\[Phi], 0, 2 \[Pi]}],
Graphics3D[{Blue, Opacity[0.5], Cylinder[]}]
]I got this output:

http://www.alumnos.usm.cl/~jorge.lopezl/problem001.jpg

Do you get the same problem with the image?

I want to see a cylinder of radius R=1, centered in the origin with Z as vertical axis.

Greatings.

PS: Here is the NB file: http://www.alumnos.usm.cl/~jorge.lopezl/Griffiths%203.39.nb

 

[CompuChip]

I get a very nice cylinder of radius 1, as you said one should.
Are you sure you don't have some old definitions from earlier commands hanging around?

(Using Mathematica 6.0 btw)

 

[jorgext]

That all code I used.

Where can I look for definitions?

 

[jorgext]

Can you post the obtained image please?

 

[CompuChip]

I'll attach the notebook with the image in an output cell. If you evaluate the cell with "img = ..." you can do with it what you like (e.g. Export["filename", img, "format"]).

 

[jorgext]

Same problem, I added one constant that wasn't included.

Please check the file I attached here:

http://www.alumnos.usm.cl/~jorge.lopezl/cylinder.nb

 

[Dale]

The cylinder is there in the image you posted.