I would like to plot two 2d functions in a 3d coordinate system. Examples are z = x^2 and z = y^2. Each function has one independent variable, (either x or y), and a global dependent variable (z).

  • 1
    $\begingroup$ You've seen Plot3D[] already? $\endgroup$ Apr 20, 2013 at 18:48
  • $\begingroup$ yes, of course, but i would like to show two independent 2d functions, eg. z=x^2 and z=y^2 in a xyz coordinate system. $\endgroup$
    – pauleck
    Apr 20, 2013 at 18:56
  • $\begingroup$ Then what's wrong with Plot3D[x^2+y^2,{x,-1,1},{y,-1,1}] ? $\endgroup$
    – Jens
    Apr 20, 2013 at 18:58
  • $\begingroup$ Well, Plot3D[{x^2, y^2}, {x, -2, 2}, {y, -2, 2}] works for me... $\endgroup$ Apr 20, 2013 at 19:02
  • $\begingroup$ The Plot3D Argument merge these functions into one surface object, i would like to show them independent in, zy and zx layer. $\endgroup$
    – pauleck
    Apr 20, 2013 at 19:03

1 Answer 1


For the two curves use the command:

g1 = ParametricPlot3D[{{t, 0, t^2}, {0, u, u^2}}, {t, -10,10}, {u, -10, 10},
     BoundaryStyle -> Thick];

in addition, if you want also to have the two $x-y$ and $y-z$ planes in the plot:

planes = ContourPlot3D[{x == 0, y == 0}, {x, -10, 10}, {y, -10, 10}, {z,0, 100}, 
         Mesh -> False, ContourStyle -> {Directive[Blue, Opacity[0.4]], 
         Directive[Green, Opacity[0.4]]}];


enter image description here

  • $\begingroup$ many, many thanks. that's what i am looking for! $\endgroup$
    – pauleck
    Apr 20, 2013 at 20:29
  • $\begingroup$ @pauleck, if this answers your question you should click on the check mark and the up-arrow next to the answer (to up-vote it - as I just did...) If you still want to wait for other answers, it's OK to leave the check mark unchecked until a day or so later. $\endgroup$
    – Jens
    Apr 20, 2013 at 21:03
  • $\begingroup$ @Jens, OP has rep $1$; unless he is given rep (say, by upvoting his question), he won't be able to upvote Spawn's answer. He can certainly accept Spawn's answer, if it suits his needs. $\endgroup$ Apr 21, 2013 at 0:26

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