enter image description hereI have to plot 0=x^2-2y+y^2 and z=sqrt(x^2+y^2) for a class project but Mathematica does not accept raw input, any ideas? I have tried with "cylinder" in the functions but it does not work.


closed as off-topic by MarcoB, LLlAMnYP, m_goldberg, Michael E2, RunnyKine Apr 27 '16 at 17:34

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  • $\begingroup$ Show us the code you have tried so far. $\endgroup$ – MarcoB Apr 27 '16 at 14:57
  • $\begingroup$ There is an image of what i have tried $\endgroup$ – Andrés Bustamante Apr 27 '16 at 15:03
  • $\begingroup$ Please post code in textual form, rather than images, so people can easily copy / paste it into their own Mathematica notebook and play with it. Here is some help on doing that: copying code; formatting. $\endgroup$ – MarcoB Apr 27 '16 at 15:08
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  • $\begingroup$ Lose the z = inside Plot. $\endgroup$ – Michael E2 Apr 27 '16 at 15:21
 {x^2 + y^2 == 1, z == Sqrt[x^2 + y^2]},
 {x, -2, 2}, {y, -2, 2}, {z, -.5, 2},
 ContourStyle -> Opacity[.65]]

enter image description here


Here are two code samples to get you started. Both snippets achieve the same result:

Plot3D[Sqrt[x^2 + y^2], {x, y} ∈ Disk[{0, 0}, 4]]

or alternatively

   Sqrt[x^2 + y^2], {x, -4, 4}, {y, -4, 4}, 
   RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 4^2]

Either one generates the following:


A different, possibly easier approach, using the fact that a cone is a solid of revolution:

RevolutionPlot3D[t, {t, 0, 4}]

revolution plot

If you are simply interested in drawing the solid object, rather than plotting it from an equation, you can also use graphics primitives:

 Cone[{{0, 0, 4}, {0, 0, 0}}, 4],
 Axes -> True


  • $\begingroup$ Thank you all this is what i was looking for. $\endgroup$ – Andrés Bustamante Apr 27 '16 at 15:36

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