Plot a cone and a cylinder [closed] I 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, RunnyKineApr 27 '16 at 17:34

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• Show us the code you have tried so far. – MarcoB Apr 27 '16 at 14:57
• There is an image of what i have tried – Andrés Bustamante Apr 27 '16 at 15:03
• 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. – MarcoB Apr 27 '16 at 15:08
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• Lose the z = inside Plot. – Michael E2 Apr 27 '16 at 15:21

ContourPlot3D[
{x^2 + y^2 == 1, z == Sqrt[x^2 + y^2]},
{x, -2, 2}, {y, -2, 2}, {z, -.5, 2},
ContourStyle -> Opacity[.65]] 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

Plot3D[
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}] If you are simply interested in drawing the solid object, rather than plotting it from an equation, you can also use graphics primitives:

Graphics3D[
Cone[{{0, 0, 4}, {0, 0, 0}}, 4],
Axes -> True
] • Thank you all this is what i was looking for. – Andrés Bustamante Apr 27 '16 at 15:36