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I have the following surface to plot $$(1-\frac{z^2}{4})^{16}=x^2+y^2$$

I decied to express $x$ as $x=x(y,z)$. And met the difficulty when combining 2 plots. Here is what I've done.enter image description here

As u see I get only half a picture. Could somebody tell me what I am doing wrong and how to combine this 2 plots?

Samsonov D
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  • Please post code not, screenshots of code. That said, add PlotRange-> {-1,1} as an option to Show. Does that fix it? – march Mar 25 '16 at 19:48
  • Alternatively, you can use Plot3D[{-1, 1} yourSqrtFunction, {x, -1, 1}, {y, -1, 1}]. – MarcoB Mar 25 '16 at 19:51
  • @march Thx, it helped. But I got some ripples on the edge between 2 plots. How to fix this? – Samsonov D Mar 25 '16 at 19:53
  • @SamsonovD Increase the number of points used in the plot, e.g. adding PlotPoints -> 100 to Plot3D. Play with the number of PlotPoints until you get a smooth edge. – MarcoB Mar 25 '16 at 19:55
  • @MarcoB thx, it also helped. – Samsonov D Mar 25 '16 at 19:55
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    ContourPlot3D[(1 - z^2/4)^16 - x^2 - y^2 == 0, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}] – Artes Mar 25 '16 at 19:57
  • @Artes that's the best option, thank you very much – Samsonov D Mar 25 '16 at 19:58
  • @Artes That would be the best way indeed. Would you post it as an answer for future reference? – MarcoB Mar 25 '16 at 20:02
  • @MarcoB I don't think it should be posted since I'm sure this is a duplicate (I guess there are at least a few such questions on this site). However if you like you may post this answer. – Artes Mar 25 '16 at 20:07
  • @MarcoB sorry for the duplicate, but still, all your comments were very helpful – Samsonov D Mar 25 '16 at 20:08
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    @SamsonovD You shouldn't worry. There's nothing very wrong with duplicate questions: marking as duplicate is a way of weaving connections among questions on the site, so that people can get to the answer they seek from multiple entry points. I'm glad we could help! – MarcoB Mar 25 '16 at 20:09

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