Why does ContourPlot refuse to draw all 4 quadrants whenever fractional powers are in the implicit function?

Question

I would like to plot the astroid which is given by:

$x^{2/3} + y^{2/3} = 1$

When I use

ContourPlot[x^(2/3) + y^(2/3) == 1, {x, -1, 1}, {y, -1, 1}, Axes -> True]

it only plots the first quadrant. If I try any other fractional powers the same thing happens. However, as soon as I use integer powers, for example if I want to plot a circle, it draws all 4 quadrants just fine. How do I make this work?

Thanks!

A cube root has three possible values. To ensure you get a real one, you could use Surd. For example Surd[x^2, 3]. — wxffles, Apr 13 '16 at 23:58
Try ContourPlot[(x^2)^(1/3) + (y^2)^(1/3) == 1, {x, -1, 1}, {y, -1, 1}, Axes -> True] and report back. Alternatively, the astroid has the parametric equations {Cos[t]^3, Sin[t]^3} that you can use. — J. M.'s missing motivation, Apr 14 '16 at 00:19
Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Michael E2, Apr 14 '16 at 00:21
You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. You may also find this this meta Q&A helpful — Michael E2, Apr 14 '16 at 00:21
Possible duplicate of Finding real roots of negative numbers (for example, $\sqrt[3]{-8}$) — xzczd, Apr 14 '16 at 08:07

Why does ContourPlot refuse to draw all 4 quadrants whenever fractional powers are in the implicit function?

0 Answers0