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Obtain RegionIntersection solution for rotated 3D regions?

Posted 7 years ago

I run Mathematica 11.1 on my student license (personal laptop - Windows 10 64-bit).

I have 2 inequalities whose intersection region I need to find when one of those is rotated by a particular angle. The below code tries to accomplish this objective, but the evaluation does not finish. Any changes that can make this work?

ieq1 = 1 > (1.25*Sqrt[x^2 + y^2])/
     Sqrt[6.25*x^2 + (x^2 + y^2)*z^2] + (0.007493*Sqrt[x^2 + y^2]*
       Sqrt[1 - (1.5625*x^2)/(6.25*x^2 + (x^2 + y^2)*z^2)])/
           (Sqrt[6.25*x^2 + (x^2 + y^2)*z^2]*
       Sqrt[((x^2 + y^2)*(-1.5625*x^2*
              z^2 + (x^2 + z^2)*(6.25*x^2 + (x^2 + y^2)*z^2)))/(6.25*
             x^2 + (x^2 + y^2)*z^2)^2]) && 4 <= x^2 + y^2 <= 25/4;

ieq2 = 1 < (1.25*Sqrt[x^2 + y^2])/
     Sqrt[6.25*x^2 + (x^2 + y^2)*z^2] - (0.007493*Sqrt[x^2 + y^2]*
       Sqrt[1 - (1.5625*x^2)/(6.25*x^2 + (x^2 + y^2)*z^2)])/
           (Sqrt[6.25*x^2 + (x^2 + y^2)*z^2]*
       Sqrt[((x^2 + y^2)*(-1.5625*x^2*
              z^2 + (x^2 + z^2)*(6.25*x^2 + (x^2 + y^2)*z^2)))/(6.25*
             x^2 + (x^2 + y^2)*z^2)^2]) && 4 <= x^2 + y^2 <= 25/4;

region1 = ImplicitRegion[ieq1, {{x, -3, 3}, {y, -3, 3}, {z, -3, 3}}];

region2 = ImplicitRegion[ieq2, {{x, -3, 3}, {y, -3, 3}, {z, -3, 3}}];  
region2 = TransformedRegion[region2 , RotationTransform[\[Pi]/3, {0, 1, 0}]];

commonRegion = RegionIntersection[region1 , region2];

AbsoluteTiming[DiscretizeRegion[commonRegion]]
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