# Disobedient RegionPlot3D

I faced a strange behavior of a RegionPlot3D. I am drawing a piece of a plane bounded by a parabola. This works:

    RegionPlot3D[
z < 1 - V^2 && (-0.001 < x < 0.001), {x, -0.9, 0.9}, {V, 0, 1.2}, {z,
0, 1}]


Now I would like to shift it over -0.1 along x:

 RegionPlot3D[
z < 1 - V^2 && ((-0.001 - 0.1) < x < (0.001 - 0.1)), {x, -0.9,
0.9}, {V, 0, 1.2}, {z, 0, 1}]


with the following result:

Do I miss something? If not, do you see a workaround?

• {x, -1, 0.8} Works, looks like a rendering bug though Jun 17, 2016 at 14:40
• @Feyre Yes, but -1 < x < -0.99 does not work Jun 17, 2016 at 14:44
• I think with the thinness of the slice, you just have to find a sweet spot where it will render. Jun 17, 2016 at 14:48
• add this: PlotPoints -> {100, Automatic, Automatic} Jun 17, 2016 at 15:21

The documentation for RegionPlot3D states

You should realize that since it uses only a finite number of sample points, it is possible for RegionPlot3D to miss regions in which pred is True. To check your results, you should try increasing the settings for PlotPoints and MaxRecursion.

This is what happened in your second example, a result of the region being so thin.

As commented by george2079, you can increase the sampling in the x direction only with the option PlotPoints -> {100, Automatic, Automatic}:

RegionPlot3D[z < 1 - V^2 && ((-0.001 - 0.1) < x < (0.001 - 0.1)),
{x, -0.9, 0.9}, {V, 0, 1.2}, {z, 0, 1},
PlotPoints -> {100, Automatic, Automatic}]


You might also be interested in the contourRegionPlot3D function here:

contourRegionPlot3D[z < 1 - V^2 && ((-0.001 - 0.1) < x < (0.001 - 0.1)),
{x, -0.9, 0.9}, {V, 0, 1.2}, {z, 0, 1}]


• Thank you, the contourRegionPlot3D is certainly a good idea. Jun 20, 2016 at 7:36