The surface I want to plot is $z^2 + r^2 = 25\, \theta$.

Please tell me how to do it in Mathematica.


closed as off-topic by Henrik Schumacher, Coolwater, bbgodfrey, Pinti, Sumit Nov 22 '18 at 19:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Coolwater, bbgodfrey, Sumit
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ ContourPlot3D? $\endgroup$ – Henrik Schumacher Nov 19 '18 at 17:05
  • $\begingroup$ @Henrik Schumacher How would you write the function? $\endgroup$ – Brandon Nov 19 '18 at 18:17
  • 1
    $\begingroup$ That depends. What is $z$, $r$, $\theta$? I guess you try to express an equation in cylindrical coordinates... but who knows? $\endgroup$ – Henrik Schumacher Nov 19 '18 at 18:20
  • $\begingroup$ See, for instance, the first example in TransformedField. You can then apply ContourPlot3D[Evaluate@TransformedField[..],...] $\endgroup$ – Michael E2 Nov 22 '18 at 19:50

Brandon: You could modify the example given in: How do I make a 3DPlot using cylindrical coordinates? as follows:

cylinderPlot3D[f_, {rMin_, rMax_}, {\[Theta]Min_, \[Theta]Max_},opts___] :=ParametricPlot3D[{r Cos[\[Theta]], r Sin[\[Theta]],f[r, \[Theta]]}, {r, rMin, rMax}, {\[Theta], \[Theta]Min, \[Theta]Max}, opts]

g[r_, \[Theta]_] := Sqrt[25 \[Theta] - r^2];

cylinderPlot3D[g, {0, 1}, {0, 2 Pi}, Mesh -> None, Boxed -> True]

Not the answer you're looking for? Browse other questions tagged or ask your own question.