Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Object x should be composed of thick lines, object su should have no or a thin border.

With BoundaryStyle, I am not able to differentiate between x and su. I haven't had much luck with PlotStyle -> {Thick, } or PlotStyle -> {EdgeForm[Thick], } either.

x1 := {u, 0, 0}
x2 := {u, 1, 1}
x3 := {0, w, w}
x4 := {1, w, w}
x := {x1, x2, x3, x4}
su := {u, (1 - u) w^2 + u w^2, (1 - u) w + u w}
ParametricPlot3D[{x, su}, {u, 0, 1}, {w, 0, 1}]

enter image description here

share|improve this question
Since your x's are just lines, why not just render them as separate primitives (i.e. with Line[]) and then combine them with your surface using Show[]? – J. M. Apr 14 '13 at 13:31
To be precise, did you want something like this? – J. M. Apr 14 '13 at 13:37
@J.M. Yes, exactly like this. However, x1-x4 do not necessarily have to be straight lines, they can also be functions that meet in the corners. – Frank Apr 14 '13 at 13:39
In that case, you'll have to invoke ParametricPlot3D[] twice; once for your surface, and once for your curves, and then combine all that with Show[]. – J. M. Apr 14 '13 at 13:41
up vote 5 down vote accepted

The "lines" or edges of 3D plots are not controlled by PlotStyle (as one might think from 2D experience), but by BoundaryStyle. Since this is applied to all objects in the plot, I suggest plotting them one at a time and combining with Show:

ParametricPlot3D[#, {u, 0, 1}, {w, 0, 1}, BoundaryStyle -> #2, Mesh -> None] & @@@ 
    {{x, Directive[Red, Thick]}, {su, None}} // Show

enter image description here

share|improve this answer
It somehow seems perverse to me that BoundaryStyle affects curves, seeing that they don't really have a boundary... :) – J. M. Apr 14 '13 at 13:57
@J.M. I agree, and this used to trip me up quite often in the past. – R. M. Apr 14 '13 at 14:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.