1
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If I try to

h = x^2 + 2 y^2 + 3 z^2 - 1;
s = ContourPlot3D[{h == 0}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1},  
  ContourStyle -> Directive[Blue, Opacity[0.5], Specularity[White, 40]],
  AxesLabel -> {x, y, z}, BaseStyle -> 14] 

Then everything works fine, and I get the expected ellipsoid in 3d.enter image description here

Now I would like to do the same but in a higher dimension. I want to solve a polynomial in 5 variables, namely

h = x^2 + 2 y^2 + 3 z^2  + 2 a^2 + b^2- 1;

and see what happens in the lower dimension (3d) in terms of x,y,z. So I try to evaluate the same line as before

s = ContourPlot3D[{h == 0}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1},  
  ContourStyle -> Directive[Blue, Opacity[0.5], Specularity[White, 40]],
  AxesLabel -> {x, y, z}, BaseStyle -> 14]

But it does not work. I get the same Input line as output and no plots.

I have no idea what is wrong. Can you explain why? and how to get this working?

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closed as off-topic by Daniel Lichtblau, LLlAMnYP, bbgodfrey, MarcoB, user9660 Nov 11 '15 at 17:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, LLlAMnYP, bbgodfrey, MarcoB, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ You will need to assign values to a and b. Otherwise ContourPlot is left with a symbolic expression that it cannot plot. $\endgroup$ – MarcoB Nov 11 '15 at 14:58
  • $\begingroup$ Hi Marco, shouldn't a and b posses values from the solutions of the polynomial equation? $\endgroup$ – Rasheed Nov 11 '15 at 16:06
  • $\begingroup$ There are many possible values a and b could take depending on the values of x y z and vice-versa. The complete set of solutions would then be an ellipsoid in x y z, which isn't a surface, but a 3d region and that's not the function of ContourPlot. $\endgroup$ – LLlAMnYP Nov 11 '15 at 16:22
  • $\begingroup$ I'm not sure I understand. As I can imagine, the solution will be a region in the fifth dimensional space (i.e values for x,y,z, b and a). I'm interested in looking at the values of x, y and z that are in the solution set by a plot in 3D. This means that I want to take all the values of x, y and z from this set (and forget about a, b), then put all these values together in a 3D plot. Are you saying that this will still be an ellipsoid like the example above in my question? $\endgroup$ – Rasheed Nov 11 '15 at 16:26
  • $\begingroup$ No, not quite. In the first example you have the two-dimensional boundary of an ellipsoid in three-dimensional space. With 5 (or even with 4) real valued variables with positive coefficients you will get the entire ellipsoid. $\endgroup$ – LLlAMnYP Nov 11 '15 at 16:34

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