I want to plot the region in the $xy$-plane which satisfies $f(x,y,z) < 0$, where $z$ can take values in the range [-1, 1]. How do I do this in Mathematica?

  • 1
    $\begingroup$ Have you looked at RegionPlot3D? $\endgroup$
    – m_goldberg
    Feb 19, 2019 at 13:03
  • $\begingroup$ This example might be relevant. $\endgroup$
    – Greg Hurst
    Feb 19, 2019 at 19:18
  • $\begingroup$ Thanks, @m_goldberg. As Roman gave an explicit example of how to it. This will solve my problem for now! But I am wondering if I can do it in general for functions of more than 3 variables and I still want a 2D Regionplot. $\endgroup$ Feb 20, 2019 at 10:46

2 Answers 2


In general, for each $z$ value, the projection of the contours of $f(x,y,z)$ onto the $xy$-plane will be different, so there is no single "region in the $xy$-plane which satisfies $f(x,y,z) < 0$". All you can do, I think, is look at contour plots for a sequence values of $z$. Here are two examples of how you might make such a sequence. Since you didn't supply an example of f, I will use this one:

f[x_, y_, z_] := x^2 + y^3 - z^2


  f[x, y, z], {"ZStackedPlanes", 5},
  {x, -3, 3}, {y, -3, 3}, {z, -1, 1},
  RegionFunction -> (f[#1, #2, #3] < 0 &),
  AxesLabel -> {"X", "Y", "Z",}]


Projection onto 2D contour plots

    ContourPlot[f[x, y, z], {x, -3, 3}, {y, -3, 3},
      PlotLabel -> Row[{"z = ", z, "\n"}],
      Contours -> Range[-5, 0],
      ContourShading -> None,
      ImageSize -> Medium],
   {z, Subdivide[-1, 1, 4] // N}]]



As @mgoldberg says, do a RegionPlot3D and view it flat from above (projecting out the $z$-axis):

f[x_, y_, z_] = x^2 + y^3 - z^2;
RegionPlot3D[f[x, y, z] < 0, {x, -3, 3}, {y, -3, 3}, {z, 0, 1}, ViewPoint -> {0, 0, ∞}]

enter image description here

In this way you can see which $(x,y)$ values allow $f(x,y,z)<0$ for any $z\in[-1,1]$.

Using @ChipHurst's suggestion seems a bit more tricky and gives a poor quality plot:

Region[ImplicitRegion[Exists[z, 0<z<1 && f[x,y,z]<0], {x, y}],
  PlotRange -> {{-3,3},{-3,3}}]

enter image description here


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