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

Please tell me how to do it in Mathematica.

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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]
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