there, I encountered this problem while coding:

I generate a table in the r-z plane of cylindrical coordinates:

tbl = Flatten[Table[{r, z}, {r, 0, 1, .1}, {z, -.8, .8, .1}], 1];

and the boundary curve of the region is defined parametrically in the bispherical coordinates:

r[η_, ξ_] := (c Sin[η])/(Cosh[ξ] - Cos[η]);
z[η_, ξ_] := (c Sinh[ξ])/(Cosh[ξ] - Cos[η]);

with the constant:

c = ρ Sin[η0];(*ρ Sin[η0]*)
ρ = (2/(3 (π - η0) Cos[η0] + 3 Sin[η0] - 

Thus, given the fixed η0, such as η0=π/3, plot the curve:

ParametricPlot[{r[η0, ξ], z[η0, ξ]}, {ξ, -30, 30}, PlotRange -> All]

I want to select the points that lie inside of the region bounded by the curve and the y-axis, but I don't know how to add this condition to "Select". Please help!

  • $\begingroup$ Your curve isn't closed. I assume you want to bound the region with $x>0$ ? $\endgroup$ Oct 4, 2017 at 20:12
  • $\begingroup$ yes, sorry, I didn't describe it correctly, I want the points inside of the region bounded by the curve and the y-axis $\endgroup$
    – Ying Zhang
    Oct 4, 2017 at 20:16
  • $\begingroup$ Look up RegionFunction and define your region by statements such as $0<x<1.2$ and $-X(x) < y < + X(x)$, where $X(x)$ is derived from you implicit equations. Once you have a region function you can use RegionMember applied to each point. $\endgroup$ Oct 4, 2017 at 20:33
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    $\begingroup$ Alternativly to Vitaliy's answer, you could also solve the equation to a $y(x)$ function as mentioned by David with xiSols = Normal[Solve[x == r[\[Eta]0, xi], xi, Reals]] ySol = z[\[Eta]0, xi] /. xiSols // Simplify $\endgroup$ Oct 4, 2017 at 20:37
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2 Answers 2


As suggested i present an alternative solution.

But keep in mind, this depends on the ability to solve your implicit equations and therefore, does not work everytime. But when it works it yields a fast Boolean condition.

We doing the following. Consider your set of equations, depending on the parameter $\xi$.


We solve the first one for the parameter and plug this solution into the other equation. We hope to not only solve this, but to get at least 2 equations which bound your region in $y(x)$.

$$\xi=r^{-1}(x)\\ y(x)=z(r^{-1}(x))$$

We can get the $r^{-1}$ with


which has two solutions. So we've already won. Plug this into the second equation:


The sign signals us that ySolL < ySolR and therefore ySolL < y < ySolR

We can easily check this:



So you could now either:

  • create a region and check with Element
  • Use the fast Boolean condition directly ySolL<=y<=ySolR && x > 0


    With the plot

    Show[RegionPlot[ySolL <= y <= ySolR, {x, 0, 1.2}, {y, -1, 1}], ListPlot[tblIn]]


  • $\begingroup$ +1. The lack of an upper limit on $x$ becomes a problem if there are points to the right of the region though. $\endgroup$
    – C. E.
    Oct 4, 2017 at 21:34

You should parametrize your region. c looks to me like sort of a radius and it will help to cut off half a plane.


r[η_,ξ_, c_]:=(c Sin[η])/(Cosh[ξ]-Cos[η]);
z[η_,ξ_, c_]:=(c Sinh[ξ])/(Cosh[ξ]-Cos[η]);

ρ=(2/(3 (π-η0) Cos[η0]+3 Sin[η0]-Sin[η0]^3))^(1/3);
c0=ρ Sin[η0];

rig = Region[ParametricRegion[{{r[η0, ξ, c], z[η0, ξ, c]}, 
    0 < c < c0}, {{ξ, -30, 30}, c}], Epilog -> Point[tbl]]

enter image description here



enter image description here

I am not sure why some points look they slightly sticking out of the region, - might be drawing imprecision or an issue. You should check this as I am out of time, I hope it's a start.

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    $\begingroup$ beat me by just 1 minute. $\endgroup$ Oct 4, 2017 at 20:35
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    $\begingroup$ @JulienKluge it is always a race here ;-) $\endgroup$ Oct 4, 2017 at 20:37
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    $\begingroup$ Many appreciations to both of you! $\endgroup$
    – Ying Zhang
    Oct 4, 2017 at 20:38
  • $\begingroup$ @Julien Kluge can you please also post your answer if you're not using the same method? $\endgroup$
    – Ying Zhang
    Oct 4, 2017 at 20:38
  • 1
    $\begingroup$ @JulienKluge yes if you are doing something else feel free to post. $\endgroup$ Oct 4, 2017 at 20:39

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