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How can I plot two equations in one 3D graph. I have to plot x^2+y^2=1 and x^2+y^2+z^2=2 but I want tu plot only their intersection :(

Thank you

didllicek
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  • There were many similar questions so most likely this is a duplicate e.g. take a look at this : http://mathematica.stackexchange.com/questions/5968/plotting-implicitly-defined-space-curves. Perhaps another hint : http://mathematica.stackexchange.com/questions/17799/can-mathematica-solve-this-equation/17864#17864 – Artes Apr 20 '13 at 12:20
  • You are plotting x^2 + y^2 = 1. That's the unit circle. The second equation says 1+z^2=2, so z^2=1. Assuming we're talking about real numbers, this means either z=1 or z=-1. So your plot is: a unit circle at z=1 and another unit circle at z=-1. – bill s Apr 20 '13 at 12:42

1 Answers1

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Using the first link provided by Artes in the comments:

ContourPlot3D[{x^2 + y^2 - 1, x^2 + y^2 + z^2 - 2},
 {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
 ContourStyle -> None, Mesh -> None, 
 BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Black}]
Mike Z.
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