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Putting a 2D region plot under a 3D plot



The Next CEO of Stack OverflowPutting ticks with values on a parametric plotPlot above desired regionIntersecting RegionPlotsRegion plot with predicatePlot region from given coordinatesPutting arrows in 3d plotFilling the region under a lineHow Plot this RegionPlot a function under constraintregion plot with region function










2












$begingroup$


I am trying to plot this 3D function over a hexagonal region:



a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]


This works fine, but I would also like to draw the region under the 3D graph, something like:



bz = Graphics[RegularPolygon[R, 6]];


or



bz = RegionPlot[RegularPolygon[R, 6]];


However, using



Show[ep, bz]


doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.






plotting regions







share|improve this question











$endgroup$











  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
    $endgroup$
    – Alrubaie
    4 hours ago















2












$begingroup$


I am trying to plot this 3D function over a hexagonal region:



a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]


This works fine, but I would also like to draw the region under the 3D graph, something like:



bz = Graphics[RegularPolygon[R, 6]];


or



bz = RegionPlot[RegularPolygon[R, 6]];


However, using



Show[ep, bz]


doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.






plotting regions







share|improve this question











$endgroup$











  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
    $endgroup$
    – Alrubaie
    4 hours ago













2












2








2





$begingroup$


I am trying to plot this 3D function over a hexagonal region:



a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]


This works fine, but I would also like to draw the region under the 3D graph, something like:



bz = Graphics[RegularPolygon[R, 6]];


or



bz = RegionPlot[RegularPolygon[R, 6]];


However, using



Show[ep, bz]


doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.






plotting regions







share|improve this question











$endgroup$




I am trying to plot this 3D function over a hexagonal region:



a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]


This works fine, but I would also like to draw the region under the 3D graph, something like:



bz = Graphics[RegularPolygon[R, 6]];


or



bz = RegionPlot[RegularPolygon[R, 6]];


However, using



Show[ep, bz]


doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.






plotting regions




plotting regions






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 12 mins ago









J. M. is slightly pensive♦

98.7k10310467




98.7k10310467










asked 5 hours ago









AshAsh

185




185











  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
    $endgroup$
    – Alrubaie
    4 hours ago
















  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
    $endgroup$
    – Alrubaie
    4 hours ago















$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago




$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago












$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago




$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    4 hours ago










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    4 hours ago











  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    3 hours ago










  • $begingroup$
    @mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
    $endgroup$
    – MarcoB
    3 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    4 hours ago










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    4 hours ago











  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    3 hours ago










  • $begingroup$
    @mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
    $endgroup$
    – MarcoB
    3 hours ago















4












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    4 hours ago










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    4 hours ago











  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    3 hours ago










  • $begingroup$
    @mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
    $endgroup$
    – MarcoB
    3 hours ago













4












4








4





$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$



Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]


Mathematica graphics







share|improve this answer












share|improve this answer



share|improve this answer










answered 4 hours ago









MarcoBMarcoB

38k556114




38k556114











  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    4 hours ago










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    4 hours ago











  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    3 hours ago










  • $begingroup$
    @mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
    $endgroup$
    – MarcoB
    3 hours ago
















  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    4 hours ago










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    4 hours ago











  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    3 hours ago










  • $begingroup$
    @mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
    $endgroup$
    – MarcoB
    3 hours ago















$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago




$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago












$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago





$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago













$begingroup$
That's nice! What AspectRatio did you use?
$endgroup$
– mjw
3 hours ago




$begingroup$
That's nice! What AspectRatio did you use?
$endgroup$
– mjw
3 hours ago












$begingroup$
@mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago




$begingroup$
@mjw Thank you! The aspect ratio is inherited from ep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago

















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