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As the output of the following code shows, by default Mathematica does not necessarily place the axes label for 3D graphics at the positive ends of the axes, for axes emanating from the origin.

In the graphical output, the y-axis label is at the negative end, contrary to the way a mathematician would ordinarily label such a figure.

Is there some option to have the labels put at the positive ends?

 ParametricPlot3D[{2 Cos[θ], 2 Sin[θ], 0}, {θ, 0, π/2}, 
    PlotStyle -> Thick, PlotRange -> {{-3, 3}, {-3, 3}, {-2, 2}},
    Boxed -> False, AxesOrigin -> {0, 0, 0}, 
    AxesLabel -> (Style[#, 16] &) /@ {"x", "y", "z"}]

y-axis label is at negative end of axis

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  • $\begingroup$ I don't think so, but it shouldn't be difficult to do with Text. $\endgroup$
    – C. E. ♦
    Commented Nov 30, 2018 at 22:12
  • $\begingroup$ I think this is a duplicate of mathematica.stackexchange.com/q/105715/9490 $\endgroup$
    – Jason B.
    Commented Jan 1, 2019 at 16:12

2 Answers 2

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ParametricPlot3D[{2 Cos[\[Theta]], 2 Sin[\[Theta]], 0}, {\[Theta], 
  0, \[Pi]/2}, PlotStyle -> Thick, 
 PlotRange -> {{-3, 3}, {-3, 3}, {-1, 2}}, Boxed -> False, 
 AxesOrigin -> {0, 0, 0}, 
 AxesLabel -> (Style[#, 16] &) /@ {"x", "y", "z"}, ViewAngle -> Pi/8]

fig1

Second model

ParametricPlot3D[{2 Cos[\[Theta]], 2 Sin[\[Theta]], 0}, {\[Theta], 
  0, \[Pi]/2}, PlotStyle -> Thick, 
 PlotRange -> {{-3, 3}, {-3, 3}, {-1, 2}}, Boxed -> False, 
 AxesOrigin -> {0, 0, 0}, 
 AxesLabel -> {Style[x, 16], Style[y, 16], Style[z, 16]}]

fig2

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  • $\begingroup$ Yes, I know about ViewAngle; and in fact one can just manually drag the 3D graphic to reorient the axes like that and so have axes "properly" labeled at their positive ends. But I was hoping for another option that forced the labels to the positive ends, no matter what the orientation. $\endgroup$
    – murray
    Commented Nov 30, 2018 at 22:59
  • $\begingroup$ And the method of using ViewAngle, unfortunately, does not work if I give the parameter a variable terminal value and put the resulting ParametricPlot expression inside a Manipulate where that terminal value is the control variable. $\endgroup$
    – murray
    Commented Nov 30, 2018 at 23:06
  • $\begingroup$ Correction to my comments: your input does not produce with Mathematica 11.3 the output you show! The axes retain their position as if the ViewAngle option were not there. $\endgroup$
    – murray
    Commented Nov 30, 2018 at 23:10
  • $\begingroup$ @murray I just answered your question. If you want to ask another question ask it. $\endgroup$
    – Alex Trounev
    Commented Nov 30, 2018 at 23:11
  • $\begingroup$ My ` $Version Out[]= "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)"` $\endgroup$
    – Alex Trounev
    Commented Nov 30, 2018 at 23:14
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While I still don't understand why, by default under the macOS front end, the original code oriented the axes with the positive x-axis pointing backwards, the following with use of ViewPoint gives the result I was looking for.

 ParametricPlot3D[{2 Cos[\[Theta]], 2 Sin[\[Theta]], 0}, {\[Theta], 0, \[Pi]/2}, PlotStyle -> Thick, 
   PlotRange -> {{-3, 3}, {-3, 3}, {-1, 2}}, Boxed -> False, 
   AxesOrigin -> {0, 0, 0}, AxesLabel -> (Style[#, 16] &) /@ {"x", "y", "z"}, 
   ViewPoint -> {Pi, Pi/2, 2}]

Orientation with positive x-axis pointing forward

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  • 2
    $\begingroup$ this is not a solution. if you rotate the shape, the position of the labels changes. $\endgroup$
    – Alex Trounev
    Commented Dec 2, 2018 at 15:45

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