Visualize manifold specified by equalities












4












$begingroup$


Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:



enter image description here



How can I plot the curve defined by the intersection in 3D?






graphics3d







share|improve this question









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    4












    $begingroup$


    Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:



    enter image description here



    How can I plot the curve defined by the intersection in 3D?






    graphics3d







    share|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:



      enter image description here



      How can I plot the curve defined by the intersection in 3D?






      graphics3d







      share|improve this question









      $endgroup$




      Suppose I have two nonlinear equalities $x^3 = y^2, y = z^3$. How can I visualize the manifold in $mathbb{R}^3$ that is generated by simultaneously satisfying the two equalities? I think ContourPlot3D is the one to use but I couldn't get it to work show the set of points in $mathbb{R}^3$ that satisfy the two equalities. The best I can do is make it show the intersection of the surfaces:



      enter image description here



      How can I plot the curve defined by the intersection in 3D?






      graphics3d




      graphics3d






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 31 mins ago









      ITAITA

      1575




      1575






















          2 Answers
          2






          active

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          5












          $begingroup$

          You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:



          ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
          
 Mesh -> None, ContourStyle -> Opacity[.3], 
          
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]


          enter image description here



          Also



          SliceContourPlot3D[y - z^3,  x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
          
 Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None, 
          
 ContourStyle -> Directive[Red, Thick]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
            $endgroup$
            – ITA
            12 mins ago



















          3












          $begingroup$

          r = 1;
          
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
          
Region[R]





          share|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:



            ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Mesh -> None, ContourStyle -> Opacity[.3], 
            
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]


            enter image description here



            Also



            SliceContourPlot3D[y - z^3,  x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None, 
            
 ContourStyle -> Directive[Red, Thick]]


            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
              $endgroup$
              – ITA
              12 mins ago
















            5












            $begingroup$

            You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:



            ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Mesh -> None, ContourStyle -> Opacity[.3], 
            
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]


            enter image description here



            Also



            SliceContourPlot3D[y - z^3,  x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None, 
            
 ContourStyle -> Directive[Red, Thick]]


            enter image description here






            share|improve this answer











            $endgroup$













            • $begingroup$
              That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
              $endgroup$
              – ITA
              12 mins ago














            5












            5








            5





            $begingroup$

            You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:



            ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Mesh -> None, ContourStyle -> Opacity[.3], 
            
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]


            enter image description here



            Also



            SliceContourPlot3D[y - z^3,  x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None, 
            
 ContourStyle -> Directive[Red, Thick]]


            enter image description here






            share|improve this answer











            $endgroup$



            You can use the option BoundaryStyle to mark the intersection of the two contour surfaces as follows:



            ContourPlot3D[{x^3 == y^2, y == z^3}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Mesh -> None, ContourStyle -> Opacity[.3], 
            
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Directive[Thick, Red]}]


            enter image description here



            Also



            SliceContourPlot3D[y - z^3,  x^3 == y^2, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
            
 Contours -> {{0}}, BoundaryStyle -> None, ContourShading -> None, 
            
 ContourStyle -> Directive[Red, Thick]]


            enter image description here







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 10 mins ago

























            answered 22 mins ago









            kglrkglr

            186k10202421




            186k10202421












            • $begingroup$
              That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
              $endgroup$
              – ITA
              12 mins ago


















            • $begingroup$
              That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
              $endgroup$
              – ITA
              12 mins ago
















            $begingroup$
            That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
            $endgroup$
            – ITA
            12 mins ago




            $begingroup$
            That's exactly what I needed. I follow most of it: Since a list was passed as first argument the '{1 -> None, 2-> None ... }' but how Mathematica knew to handle {1,2} -> is just magic!
            $endgroup$
            – ITA
            12 mins ago











            3












            $begingroup$

            r = 1;
            
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
            
Region[R]





            share|improve this answer









            $endgroup$


















              3












              $begingroup$

              r = 1;
              
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
              
Region[R]





              share|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                r = 1;
                
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
                
Region[R]





                share|improve this answer









                $endgroup$



                r = 1;
                
R = ImplicitRegion[{x^3 == y^2, y == z^3}, {{x, -r, r}, {y, -r, r}, {z, -r, r}}];
                
Region[R]






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 21 mins ago









                Henrik SchumacherHenrik Schumacher

                55.1k475154




                55.1k475154






























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