Onion plot of a ball












5












$begingroup$


Consider the following simple example.



RegionPlot3D[x^2 + y^2 + z^2 <= 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 

  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 

  Mesh -> {{0.25, .5, .75}}, PlotStyle -> Opacity[.5], Axes -> False]


enter image description here



I would like to show some of the surfaces described by the mesh function which lie inside the ball. Is it possible to visualize only these surfaces?






plotting regions meshfunction







share|improve this question











$endgroup$

















    5












    $begingroup$


    Consider the following simple example.



    RegionPlot3D[x^2 + y^2 + z^2 <= 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
    
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
    
  Mesh -> {{0.25, .5, .75}}, PlotStyle -> Opacity[.5], Axes -> False]


    enter image description here



    I would like to show some of the surfaces described by the mesh function which lie inside the ball. Is it possible to visualize only these surfaces?






    plotting regions meshfunction







    share|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      Consider the following simple example.



      RegionPlot3D[x^2 + y^2 + z^2 <= 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
      
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
      
  Mesh -> {{0.25, .5, .75}}, PlotStyle -> Opacity[.5], Axes -> False]


      enter image description here



      I would like to show some of the surfaces described by the mesh function which lie inside the ball. Is it possible to visualize only these surfaces?






      plotting regions meshfunction







      share|improve this question











      $endgroup$




      Consider the following simple example.



      RegionPlot3D[x^2 + y^2 + z^2 <= 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
      
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
      
  Mesh -> {{0.25, .5, .75}}, PlotStyle -> Opacity[.5], Axes -> False]


      enter image description here



      I would like to show some of the surfaces described by the mesh function which lie inside the ball. Is it possible to visualize only these surfaces?






      plotting regions meshfunction




      plotting regions meshfunction






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 15 at 10:33









      m_goldberg

      87.6k872198




      87.6k872198










      asked Feb 15 at 9:12









      Ulrich NeumannUlrich Neumann

      9,548617




      9,548617






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Update: Post-process RegionPlot outputs to remove the walls:



          colors = ColorData[97] /@ Range[4];
          
radii = {1, 3/4, 1/2, 1/4};
          
regionplots = RegionPlot3D[x^2 + y^2 + z^2 <= #^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     Mesh -> None, BoundaryStyle -> None, Axes -> False] & /@ radii; 
          

          
Graphics3D[{EdgeForm, FaceForm[{Opacity[.5], #[[2]]}], 
          
  Cases[Normal[#[[1]]][[1]], _GraphicsGroup, All][[1]]}&/@Transpose[{regionplots, colors}],
          
  Boxed -> False]


          enter image description here



          Alternatively, delete the Polygons with constant VertexNormals:



          Show[DeleteCases[Normal@RegionPlot3D[x^2 + y^2 + z^2 <= #^2,
          
     {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     BoundaryStyle -> None, Mesh -> None, BaseStyle -> Opacity[.5], 
          
     PlotStyle -> #2, Axes -> False], 
          
    Polygon[_, VertexNormals -> {{a_, b_, c_} ..}], All] & @@@ 
          
  Transpose[{radii, colors}], Boxed -> False]


          enter image description here



          Original answer:



          You can use a combination of ImplicitRegion and DiscretizeRegion as follows:



          ir[r_] := ImplicitRegion[x^2 + y^2 + z^2 == r^2, {{x, 0, 1}, {y, 0, 1}, {z, 0, 1}}]
          
radii = {1, 3/4, 1/2, 1/4};
          
colors = ColorData[97] /@ Range[4];
          
boundaries = RegionPlot3D[x^2 + y^2 + z^2 <= 1.1,
          
  {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
          
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
          
  Mesh -> {Transpose[{radii, colors}]}, MeshStyle -> Thick, 
          
  PlotStyle -> Opacity[0], Axes -> False, BoundaryStyle -> None];
          
i = 1;
          
surfaces = DiscretizeRegion[ir[#], 
          
  MeshCellStyle -> {{2, All} -> Opacity[0.5, colors[[i++]]]}] & /@ radii;
          
Legended[Show[surfaces, boundaries], 
          
  SwatchLegend[colors, "radius = " <> ToString[#, StandardForm] & /@ radii]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 9:12












          • $begingroup$
            @UlrichNeumann, does the approach in the update work in your problem?
            $endgroup$
            – kglr
            Feb 16 at 10:28










          • $begingroup$
            Yes , thanks for the two solution ideas!
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 10:37



















          8












          $begingroup$

          Like this?



          ContourPlot3D[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
 ContourStyle -> Directive[Opacity[.5], ColorData[97][1]],
          
 Axes -> False,
          
 Contours -> {0.25, .5, .75},
          
 Mesh -> None
          
 ]


          enter image description here






          share|improve this answer









          $endgroup$













          • $begingroup$
            Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
            $endgroup$
            – Ulrich Neumann
            Feb 15 at 9:27










          • $begingroup$
            Huh? You have to explain that to me...
            $endgroup$
            – Henrik Schumacher
            Feb 15 at 9:28






          • 1




            $begingroup$
            Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 10:36






          • 2




            $begingroup$
            Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 12:45






          • 1




            $begingroup$
            @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
            $endgroup$
            – Michael E2
            Feb 15 at 13:37











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Update: Post-process RegionPlot outputs to remove the walls:



          colors = ColorData[97] /@ Range[4];
          
radii = {1, 3/4, 1/2, 1/4};
          
regionplots = RegionPlot3D[x^2 + y^2 + z^2 <= #^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     Mesh -> None, BoundaryStyle -> None, Axes -> False] & /@ radii; 
          

          
Graphics3D[{EdgeForm, FaceForm[{Opacity[.5], #[[2]]}], 
          
  Cases[Normal[#[[1]]][[1]], _GraphicsGroup, All][[1]]}&/@Transpose[{regionplots, colors}],
          
  Boxed -> False]


          enter image description here



          Alternatively, delete the Polygons with constant VertexNormals:



          Show[DeleteCases[Normal@RegionPlot3D[x^2 + y^2 + z^2 <= #^2,
          
     {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     BoundaryStyle -> None, Mesh -> None, BaseStyle -> Opacity[.5], 
          
     PlotStyle -> #2, Axes -> False], 
          
    Polygon[_, VertexNormals -> {{a_, b_, c_} ..}], All] & @@@ 
          
  Transpose[{radii, colors}], Boxed -> False]


          enter image description here



          Original answer:



          You can use a combination of ImplicitRegion and DiscretizeRegion as follows:



          ir[r_] := ImplicitRegion[x^2 + y^2 + z^2 == r^2, {{x, 0, 1}, {y, 0, 1}, {z, 0, 1}}]
          
radii = {1, 3/4, 1/2, 1/4};
          
colors = ColorData[97] /@ Range[4];
          
boundaries = RegionPlot3D[x^2 + y^2 + z^2 <= 1.1,
          
  {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
          
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
          
  Mesh -> {Transpose[{radii, colors}]}, MeshStyle -> Thick, 
          
  PlotStyle -> Opacity[0], Axes -> False, BoundaryStyle -> None];
          
i = 1;
          
surfaces = DiscretizeRegion[ir[#], 
          
  MeshCellStyle -> {{2, All} -> Opacity[0.5, colors[[i++]]]}] & /@ radii;
          
Legended[Show[surfaces, boundaries], 
          
  SwatchLegend[colors, "radius = " <> ToString[#, StandardForm] & /@ radii]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 9:12












          • $begingroup$
            @UlrichNeumann, does the approach in the update work in your problem?
            $endgroup$
            – kglr
            Feb 16 at 10:28










          • $begingroup$
            Yes , thanks for the two solution ideas!
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 10:37
















          6












          $begingroup$

          Update: Post-process RegionPlot outputs to remove the walls:



          colors = ColorData[97] /@ Range[4];
          
radii = {1, 3/4, 1/2, 1/4};
          
regionplots = RegionPlot3D[x^2 + y^2 + z^2 <= #^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     Mesh -> None, BoundaryStyle -> None, Axes -> False] & /@ radii; 
          

          
Graphics3D[{EdgeForm, FaceForm[{Opacity[.5], #[[2]]}], 
          
  Cases[Normal[#[[1]]][[1]], _GraphicsGroup, All][[1]]}&/@Transpose[{regionplots, colors}],
          
  Boxed -> False]


          enter image description here



          Alternatively, delete the Polygons with constant VertexNormals:



          Show[DeleteCases[Normal@RegionPlot3D[x^2 + y^2 + z^2 <= #^2,
          
     {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     BoundaryStyle -> None, Mesh -> None, BaseStyle -> Opacity[.5], 
          
     PlotStyle -> #2, Axes -> False], 
          
    Polygon[_, VertexNormals -> {{a_, b_, c_} ..}], All] & @@@ 
          
  Transpose[{radii, colors}], Boxed -> False]


          enter image description here



          Original answer:



          You can use a combination of ImplicitRegion and DiscretizeRegion as follows:



          ir[r_] := ImplicitRegion[x^2 + y^2 + z^2 == r^2, {{x, 0, 1}, {y, 0, 1}, {z, 0, 1}}]
          
radii = {1, 3/4, 1/2, 1/4};
          
colors = ColorData[97] /@ Range[4];
          
boundaries = RegionPlot3D[x^2 + y^2 + z^2 <= 1.1,
          
  {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
          
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
          
  Mesh -> {Transpose[{radii, colors}]}, MeshStyle -> Thick, 
          
  PlotStyle -> Opacity[0], Axes -> False, BoundaryStyle -> None];
          
i = 1;
          
surfaces = DiscretizeRegion[ir[#], 
          
  MeshCellStyle -> {{2, All} -> Opacity[0.5, colors[[i++]]]}] & /@ radii;
          
Legended[Show[surfaces, boundaries], 
          
  SwatchLegend[colors, "radius = " <> ToString[#, StandardForm] & /@ radii]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 9:12












          • $begingroup$
            @UlrichNeumann, does the approach in the update work in your problem?
            $endgroup$
            – kglr
            Feb 16 at 10:28










          • $begingroup$
            Yes , thanks for the two solution ideas!
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 10:37














          6












          6








          6





          $begingroup$

          Update: Post-process RegionPlot outputs to remove the walls:



          colors = ColorData[97] /@ Range[4];
          
radii = {1, 3/4, 1/2, 1/4};
          
regionplots = RegionPlot3D[x^2 + y^2 + z^2 <= #^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     Mesh -> None, BoundaryStyle -> None, Axes -> False] & /@ radii; 
          

          
Graphics3D[{EdgeForm, FaceForm[{Opacity[.5], #[[2]]}], 
          
  Cases[Normal[#[[1]]][[1]], _GraphicsGroup, All][[1]]}&/@Transpose[{regionplots, colors}],
          
  Boxed -> False]


          enter image description here



          Alternatively, delete the Polygons with constant VertexNormals:



          Show[DeleteCases[Normal@RegionPlot3D[x^2 + y^2 + z^2 <= #^2,
          
     {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     BoundaryStyle -> None, Mesh -> None, BaseStyle -> Opacity[.5], 
          
     PlotStyle -> #2, Axes -> False], 
          
    Polygon[_, VertexNormals -> {{a_, b_, c_} ..}], All] & @@@ 
          
  Transpose[{radii, colors}], Boxed -> False]


          enter image description here



          Original answer:



          You can use a combination of ImplicitRegion and DiscretizeRegion as follows:



          ir[r_] := ImplicitRegion[x^2 + y^2 + z^2 == r^2, {{x, 0, 1}, {y, 0, 1}, {z, 0, 1}}]
          
radii = {1, 3/4, 1/2, 1/4};
          
colors = ColorData[97] /@ Range[4];
          
boundaries = RegionPlot3D[x^2 + y^2 + z^2 <= 1.1,
          
  {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
          
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
          
  Mesh -> {Transpose[{radii, colors}]}, MeshStyle -> Thick, 
          
  PlotStyle -> Opacity[0], Axes -> False, BoundaryStyle -> None];
          
i = 1;
          
surfaces = DiscretizeRegion[ir[#], 
          
  MeshCellStyle -> {{2, All} -> Opacity[0.5, colors[[i++]]]}] & /@ radii;
          
Legended[Show[surfaces, boundaries], 
          
  SwatchLegend[colors, "radius = " <> ToString[#, StandardForm] & /@ radii]]


          enter image description here






          share|improve this answer











          $endgroup$



          Update: Post-process RegionPlot outputs to remove the walls:



          colors = ColorData[97] /@ Range[4];
          
radii = {1, 3/4, 1/2, 1/4};
          
regionplots = RegionPlot3D[x^2 + y^2 + z^2 <= #^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     Mesh -> None, BoundaryStyle -> None, Axes -> False] & /@ radii; 
          

          
Graphics3D[{EdgeForm, FaceForm[{Opacity[.5], #[[2]]}], 
          
  Cases[Normal[#[[1]]][[1]], _GraphicsGroup, All][[1]]}&/@Transpose[{regionplots, colors}],
          
  Boxed -> False]


          enter image description here



          Alternatively, delete the Polygons with constant VertexNormals:



          Show[DeleteCases[Normal@RegionPlot3D[x^2 + y^2 + z^2 <= #^2,
          
     {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
     BoundaryStyle -> None, Mesh -> None, BaseStyle -> Opacity[.5], 
          
     PlotStyle -> #2, Axes -> False], 
          
    Polygon[_, VertexNormals -> {{a_, b_, c_} ..}], All] & @@@ 
          
  Transpose[{radii, colors}], Boxed -> False]


          enter image description here



          Original answer:



          You can use a combination of ImplicitRegion and DiscretizeRegion as follows:



          ir[r_] := ImplicitRegion[x^2 + y^2 + z^2 == r^2, {{x, 0, 1}, {y, 0, 1}, {z, 0, 1}}]
          
radii = {1, 3/4, 1/2, 1/4};
          
colors = ColorData[97] /@ Range[4];
          
boundaries = RegionPlot3D[x^2 + y^2 + z^2 <= 1.1,
          
  {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, 
          
  MeshFunctions -> (Sqrt[#1^2 + #2^2 + #3^2] &), 
          
  Mesh -> {Transpose[{radii, colors}]}, MeshStyle -> Thick, 
          
  PlotStyle -> Opacity[0], Axes -> False, BoundaryStyle -> None];
          
i = 1;
          
surfaces = DiscretizeRegion[ir[#], 
          
  MeshCellStyle -> {{2, All} -> Opacity[0.5, colors[[i++]]]}] & /@ radii;
          
Legended[Show[surfaces, boundaries], 
          
  SwatchLegend[colors, "radius = " <> ToString[#, StandardForm] & /@ radii]]


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Feb 16 at 12:57

























          answered Feb 15 at 22:41









          kglrkglr

          189k10206424




          189k10206424












          • $begingroup$
            Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 9:12












          • $begingroup$
            @UlrichNeumann, does the approach in the update work in your problem?
            $endgroup$
            – kglr
            Feb 16 at 10:28










          • $begingroup$
            Yes , thanks for the two solution ideas!
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 10:37


















          • $begingroup$
            Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 9:12












          • $begingroup$
            @UlrichNeumann, does the approach in the update work in your problem?
            $endgroup$
            – kglr
            Feb 16 at 10:28










          • $begingroup$
            Yes , thanks for the two solution ideas!
            $endgroup$
            – Ulrich Neumann
            Feb 16 at 10:37
















          $begingroup$
          Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
          $endgroup$
          – Ulrich Neumann
          Feb 16 at 9:12






          $begingroup$
          Thank you very much for your effort. In my concrete problem I only can build on (don't know why ContourPlot and ImplicitRegion don't evaluate) the result of RegionPlot3D to find the shells.
          $endgroup$
          – Ulrich Neumann
          Feb 16 at 9:12














          $begingroup$
          @UlrichNeumann, does the approach in the update work in your problem?
          $endgroup$
          – kglr
          Feb 16 at 10:28




          $begingroup$
          @UlrichNeumann, does the approach in the update work in your problem?
          $endgroup$
          – kglr
          Feb 16 at 10:28












          $begingroup$
          Yes , thanks for the two solution ideas!
          $endgroup$
          – Ulrich Neumann
          Feb 16 at 10:37




          $begingroup$
          Yes , thanks for the two solution ideas!
          $endgroup$
          – Ulrich Neumann
          Feb 16 at 10:37











          8












          $begingroup$

          Like this?



          ContourPlot3D[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
 ContourStyle -> Directive[Opacity[.5], ColorData[97][1]],
          
 Axes -> False,
          
 Contours -> {0.25, .5, .75},
          
 Mesh -> None
          
 ]


          enter image description here






          share|improve this answer









          $endgroup$













          • $begingroup$
            Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
            $endgroup$
            – Ulrich Neumann
            Feb 15 at 9:27










          • $begingroup$
            Huh? You have to explain that to me...
            $endgroup$
            – Henrik Schumacher
            Feb 15 at 9:28






          • 1




            $begingroup$
            Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 10:36






          • 2




            $begingroup$
            Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 12:45






          • 1




            $begingroup$
            @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
            $endgroup$
            – Michael E2
            Feb 15 at 13:37
















          8












          $begingroup$

          Like this?



          ContourPlot3D[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
 ContourStyle -> Directive[Opacity[.5], ColorData[97][1]],
          
 Axes -> False,
          
 Contours -> {0.25, .5, .75},
          
 Mesh -> None
          
 ]


          enter image description here






          share|improve this answer









          $endgroup$













          • $begingroup$
            Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
            $endgroup$
            – Ulrich Neumann
            Feb 15 at 9:27










          • $begingroup$
            Huh? You have to explain that to me...
            $endgroup$
            – Henrik Schumacher
            Feb 15 at 9:28






          • 1




            $begingroup$
            Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 10:36






          • 2




            $begingroup$
            Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 12:45






          • 1




            $begingroup$
            @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
            $endgroup$
            – Michael E2
            Feb 15 at 13:37














          8












          8








          8





          $begingroup$

          Like this?



          ContourPlot3D[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
 ContourStyle -> Directive[Opacity[.5], ColorData[97][1]],
          
 Axes -> False,
          
 Contours -> {0.25, .5, .75},
          
 Mesh -> None
          
 ]


          enter image description here






          share|improve this answer









          $endgroup$



          Like this?



          ContourPlot3D[x^2 + y^2 + z^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
          
 ContourStyle -> Directive[Opacity[.5], ColorData[97][1]],
          
 Axes -> False,
          
 Contours -> {0.25, .5, .75},
          
 Mesh -> None
          
 ]


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Feb 15 at 9:23









          Henrik SchumacherHenrik Schumacher

          56.8k577157




          56.8k577157












          • $begingroup$
            Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
            $endgroup$
            – Ulrich Neumann
            Feb 15 at 9:27










          • $begingroup$
            Huh? You have to explain that to me...
            $endgroup$
            – Henrik Schumacher
            Feb 15 at 9:28






          • 1




            $begingroup$
            Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 10:36






          • 2




            $begingroup$
            Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 12:45






          • 1




            $begingroup$
            @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
            $endgroup$
            – Michael E2
            Feb 15 at 13:37


















          • $begingroup$
            Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
            $endgroup$
            – Ulrich Neumann
            Feb 15 at 9:27










          • $begingroup$
            Huh? You have to explain that to me...
            $endgroup$
            – Henrik Schumacher
            Feb 15 at 9:28






          • 1




            $begingroup$
            Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 10:36






          • 2




            $begingroup$
            Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
            $endgroup$
            – Sjoerd Smit
            Feb 15 at 12:45






          • 1




            $begingroup$
            @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
            $endgroup$
            – Michael E2
            Feb 15 at 13:37
















          $begingroup$
          Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
          $endgroup$
          – Ulrich Neumann
          Feb 15 at 9:27




          $begingroup$
          Yes like this, thanks. But in my underlying problem I need to use the RegionPlot3D, because ContourPlot3D cannot evaluate.
          $endgroup$
          – Ulrich Neumann
          Feb 15 at 9:27












          $begingroup$
          Huh? You have to explain that to me...
          $endgroup$
          – Henrik Schumacher
          Feb 15 at 9:28




          $begingroup$
          Huh? You have to explain that to me...
          $endgroup$
          – Henrik Schumacher
          Feb 15 at 9:28




          1




          1




          $begingroup$
          Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
          $endgroup$
          – Sjoerd Smit
          Feb 15 at 10:36




          $begingroup$
          Have you tried tweaking the PlotPoints and MaxRecursion options of your ContourPlot3D? Often when MMA takes a long time to produce a plot, it's simply obsessing over detail you're not really interested in and it pays to reduce the values for these options.
          $endgroup$
          – Sjoerd Smit
          Feb 15 at 10:36




          2




          2




          $begingroup$
          Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
          $endgroup$
          – Sjoerd Smit
          Feb 15 at 12:45




          $begingroup$
          Trying MaxRecursion -> 0 and PlotPoints -> 25 is always a good start. The result will look very blocky, but the number of function evaluations should be low.
          $endgroup$
          – Sjoerd Smit
          Feb 15 at 12:45




          1




          1




          $begingroup$
          @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
          $endgroup$
          – Michael E2
          Feb 15 at 13:37




          $begingroup$
          @UlrichNeumann The default for MaxRecursion is 2, and time complexity, where significant refinement is called for, can grow exponentially. E.g. Table[First@ AbsoluteTiming[ ContourPlot3D[ x^2 + Sin[6 y] + z^2 == 1, {x, -1.1, 1.1}, {y, -1.1, 1.1}, {z, -1.1, 1.1}, MaxRecursion -> r]], {r, 0, 4}] --> {0.096103, 1.23869, 4.76033, 20.5314, 118.243}
          $endgroup$
          – Michael E2
          Feb 15 at 13:37


















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