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How would I plot in 3D: $\begin{cases} (4 - z^2) = x^2 + y^2, 2 \le z \le 4 \\ x^2 + y^2 = 4, -2 \le z \le 2 \end{cases}$.

I tried http://reference.wolfram.com/mathematica/ref/Piecewise.html

ContourPlot3D[ Piecewise[{(4 - z)^2 == x ^2 + y^2, 2 \le z \le 4}, {x^2 + y^2 = 4, -2 \le z \le 2}], {x, -3, 3}, {y, -3 , 3}, {z, -2, 4}]

I realise there's tex in my code but I don't know how otherwise to symbolise my attempt.

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    $\begingroup$ Your code omits the function Piecewise -- how did you try to use it? Also, your code mixes Mathematica with TeX -- I don't think that's going to work. $\endgroup$
    – Michael E2
    Apr 5, 2014 at 15:23
  • $\begingroup$ Your first equation has only one point of trueness, so just delete it and do ContourPlot3D[4 == x^2 + y^2, {x, -3, 3}, {y, -3, 3}, {z, -2, 2}] $\endgroup$
    – Coolwater
    Apr 5, 2014 at 15:24
  • $\begingroup$ @MichaelE2: Thanks. I emended my question. Piecewise still doesn't function though. $\endgroup$
    – NNOX Apps
    Apr 5, 2014 at 16:11
  • $\begingroup$ @Coolwater: Would you please explain "one true point of trueness"? $\endgroup$
    – NNOX Apps
    Apr 5, 2014 at 16:11
  • $\begingroup$ I thought your first equation would give an empty contour plot, because only one point in R^3 satisfy it. The thing is I was looking at your latex equations which says (4-z^2) rather than (4-z)^2 $\endgroup$
    – Coolwater
    Apr 5, 2014 at 19:40

2 Answers 2

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I would plot each separately and combine them:

Show[
 ContourPlot3D[(4 - z)^2 == (x^2 + y^2), {x, -3, 3}, {y, -3, 3}, {z, 2, 4}],
 ContourPlot3D[x^2 + y^2 == 4, {x, -3, 3}, {y, -3, 3}, {z, -2, 2}],
 PlotRange -> All
 ]

Mathematica graphics

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Piecewise version :

ContourPlot3D[ x^2 + y^2 == 
                Piecewise[{{(4 - z)^2, 2 <= z <= 4}, {4, -2 <= z <= 2}}],
                 {x, -3, 3}, {y, -3, 3}, {z, -2, 4}, PlotRange -> All ]
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