Circular domain for Plot3D

Mathematica's Plot3D works with rectangular domains.

In other words, we write:

Plot3D[Function, {x,a,b}, {y,c,d}]

Here the domain is $[a,b]\times[c,d]$. And hence, the surface is cut in a rectangular projection.

But, what if I want my surface to be cut otherwise, say, as a circle?

To illustrate my question, I present two different images of a paraboloid: The first one is drawn with Plot3D and the second one is obtained by revolution of a simple parabola.

RevolutionPlot3D, however, generates surfaces, which have axial symmetry only.

What should I do, if I have a non-symmetrical surface and want to cut its edges in a circle (or, if it's possible, in any other way)?

RegionFunction is what you are looking for.

Plot3D[4 Sin[x] + y, {x, -10, 10}, {y, -10, 10},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 < 100]] Since version 10 you can also specify the plotting domain using a geometric region. In your case, for instance, you could use:

Plot3D[3 x^2 + y^2, {x, y} ∈ Disk[]] This allows for interesting constructions using the full power of geometric regions:

Plot3D[x^2 + y^2,
{x, y} ∈ RegionDifference[
Polygon[CirclePoints],
Polygon[0.5 CirclePoints]
]
] RevolutionPlot3D, however, generates surfaces, which have axial symmetry only.

Actually, RevolutionPlot3D[] can be used to plot in cylindrical coordinates, and not just surfaces of revolution. Not many people seem to be aware of this.

To use paw's example:

RevolutionPlot3D[With[{x = r Cos[θ], y = r Sin[θ]}, 4 Sin[x] + y],
{r, 0, 10}, {θ, -π, π}, MeshFunctions -> {#1 & , #2 &}] • What has RevolutionPlot3D ever done to you to deserve such abuse? – Brett Champion Oct 26 '15 at 2:04
• @Brett, hey, CylindricalPlot3D[] up and left me cold. What else was I to do, but find another? – J. M. will be back soon Oct 26 '15 at 2:35
• I'm only talking about the conversion back to cartesian coordinates. General plotting in cylindrical coordinates is most definitely a RevolutionPlot3D feature. :-) – Brett Champion Oct 26 '15 at 4:01
• Oh, you mean the mesh? Well, I suppose I could've stuck with the default mesh... – J. M. will be back soon Oct 26 '15 at 4:07
• No, I mean using cylindrical coordinates to plot $f(x,y)$ instead of $f(r,t)$. – Brett Champion Oct 26 '15 at 4:58