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22

There are already good answers, but I'm going to improve the performance, generalize to any region in any dimensions and make the function more convenient. The main idea is to use DirichletDistribution (the uniform distribution on a simplex, e.g. triangle or tetrahedron). This idea was implemented by PlatoManiac and me in the related question obtaining ...


17

You can always hide away the coordinate transformations inside a function that calls RegionPlot3D. Here's a quick & dirty sphericalRegionPlot3D: sphericalRegionPlot3D[ ineq_, {r_, rmin_: 0, rmax_: 1}, {th_, thmin_: 0, thmax_}, {ph_, phmin_, phmax_}, opts___] := RegionPlot3D[With[{ r = Sqrt[x^2 + y^2 + z^2], th = ArcCos[z/Sqrt[x^2 + y^2 + ...


16

You can use Show to combine graphics of the same type: g1 = Plot3D[x^2 - y^2, {x, -3, 3}, {y, -3, 3}, RegionFunction -> Function[{x, y, z}, 2 < x^2 + y^2 < 9]]; g2 = SphericalPlot3D[ 1 + Sin[5 θ] Sin[5 φ]/5, {θ, 0, π}, {φ, 0, 2 π}, Mesh -> None, RegionFunction -> (#6 > 0.95 &), PlotStyle -> FaceForm[Orange, Yellow]]; ...


16

It would be nice if UniformDistribution worked on arbitrary regions, then we could simply do RandomVariate[UniformDistribution[region]]. Someone at Wolfram should get on that. In the meantime, it seems we have to write our own sampling routines. @m_goldberg's answer is very nice (vote it up!) and uses rejection sampling, which works for arbitrary regions. ...


15

Update Compare two pictures. First is able to make mistake like you made the code. You need to do like this code using Mod[ArcTan[x, y], 2π]. h[r_,θ_] := 2 < r <= 5 && 3/4 π < θ < 3/2 π RegionPlot[ h[Sqrt[x^2 + y^2], Mod[ArcTan[x, y], 2π]], {x, -6, 6}, {y, -6, 6}] So I suggest to use ParametricPlot like this. rg = 6; mg = ...


14

While not positive, I believe the answer is that RegionPlot does not support spherical (or other non-Cartesian) coordinates natively. If correct, I guess the question becomes "What's the easiest way to plot a region defined in terms of spherical coordinates, without resorting to converting the equations by hand?" V9 has commands to ease this process. ...


14

Βαγγέλη, you can use an appropriate RegionFunction: V = 1/2*(x^2 + y^2 + z^2) + (x^2*y^2 + x^2*z^2 + y^2*z^2 - x^2*y^2*z^2); E0 = 7; S0 = ContourPlot3D[V == E0, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}, PlotPoints -> 70, PerformanceGoal -> "Speed", Mesh -> None, ContourStyle -> Directive[Green, Opacity[0.3], Specularity[White, 30]], ...


13

Update: Using MeshFunctions and Mesh in RegionPlot: RegionPlot[Evaluate[Region`RegionProperty[Rationalize /@ blob, {x, y}, "FastDescription"][[1, 2]]], {x, -3, 3}, {y, -3, 3}, Mesh -> 50, MeshFunctions -> {#1 + #2 &, #1 - #2 &}, MeshStyle -> White, PlotStyle -> Directive[{Thick, Blue}]] With settings MeshStyle -> ...


13

Update Silvia proposed a much faster algorithm that I believe produces I uniform distribution. Here is my implementation of it. pointsInMask2[mask_Image, n_Integer, range : {_, _} : {0, 1/2}] := Reverse @ ImageData @ Binarize[mask, range]\[Transpose] // SparseArray[#]["NonzeroPositions"] & // RandomChoice[#, n] + RandomReal[{-1, 0}, {n, 2}] ...


12

In principal you should be able to do r = RegionDifference[Ball[{0, 0, 0}], Ball[{0, 0, 1}]]; rp = RegionPlot3D[r, PlotPoints -> 50]; DiscretizeGraphics[rp] Unfortunately, this does not work and is hopefully improved in a future version. One thing you can do, however, is use the finite element mesh generator for this: Needs["NDSolve`FEM`"] m = ...


12

Here is what I think the issue is: Let's look at what NDSolve parses. Needs["NDSolve`FEM`"] {state} = NDSolve`ProcessEquations[{op == 0, bc}, u, {x, y} \[Element] reg, Method -> {"FiniteElement", "MeshOptions" -> {MaxCellMeasure -> 0.005}}]; femData = state["FiniteElementData"]; femData["PDECoefficientData"]["All"] {{{{0}}, {{{{0}, ...


12

regplt = RegionPlot[\[CapitalOmega], AspectRatio -> Automatic]; ContourPlot[{2 x^2 + 3 y^2 + 2 x y - 2, x^2 + y^2 - .1}, {x, -1.25, 1.25}, {y, -1.25, 1.25}, Contours -> {{0}}, BaseStyle -> Thick, GridLines -> {xg, yg}, Method -> {"GridLinesInFront" -> True}, MeshFunctions -> {#1 &, #2 &}, Mesh -> {xg, yg}, MeshStyle -> ...


11

I don't like to think too much :P Manipulate[ {#, Composition[ # - 1 &, Length, Union, Flatten, MorphologicalComponents, Binarize, Rasterize ]@#} &@ ParametricPlot[{Sin[ n t], Sin[m t]}, {t, 0, 2 Pi}, Axes -> False, PlotStyle -> Thick] , {n, 2, 10, 1}, {m, 1, 9, 1}]


10

The inertia tensor is defined as an integral of the following tensor over the body region vars = {x, y, z}; r2 = IdentityMatrix[3] Tr[#] - # &@Outer[Times, vars, vars]; r2 // MatrixForm It is very simple to do with integration over a region Integrate[r2, vars ∈ region] It can be wrapped in the following function inertiaTensor[reg_, assum_: {}] ...


10

It sounds like you might like to use a stereographic projection. xy[ϕ_, λ_] := 2 Tan[(π - ϕ)/2] {Cos[λ], Sin[λ]}; Here is a cubic with a free parameter (for fun): cubic[{x_, y_}, b_] := y^3 - b x y + 4 x^3; For the sphere, use SphericalPlot3D of course. Although you could resort to ParametricPlot3D for the cubic curve, it's automatic and much simpler ...


10

Perhaps something like this? p = Plot[x^3/9, {x, -3, 3}, PlotStyle -> Thickness[0.01], Filling -> Axis, AspectRatio -> 1, ImageSize -> 500, BaseStyle -> {FontFamily -> "Calibri", 30}, AxesStyle -> Thick]; With[{k = 3}, SphericalPlot3D[1, {theta, 0, Pi}, {phi, 0, 2 Pi}, Mesh -> None, Boxed -> False, Axes -> False, ...


10

Following up on my comment and borrowing a method from Vectorizing an image like "Trace Bitmap" in Inkscape: mask = Binarize @ Import["http://i.stack.imgur.com/yoPNX.png"]; {row, col} = ImageDimensions[mask]; intf = ListInterpolation @ Reverse @ ImageData @ mask; region = DiscretizeGraphics @ RegionPlot[intf[c, r] < 1/2, {r, 1, row}, {c, ...


8

Use the inequality to sow all integer coordinates that are inside the boundary of the region when iterating through all integer pairs of the full range: pts = First@Last@Reap@Do[If[x >= 4 y && x <= 4 y + 3, Sow@{x, y}], {x, 0, 63}, {y, 0, 15}] RegionPlot[x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}, Epilog -> {Red, ...


8

This is a bug, I think, and I filed it as such: The second region should not evaluate to a RegionQ BoundaryMeshRegion. A BoundaryMeshRegion is valid if it contains a closed surface. The subtle point about BoundaryMeshRegion is that this closed surface is a (sparse) representation of the entire region the surface encloses. Why the first one does not work, I ...


8

Here is a work-around that will let you extract as many randoms points from a region as you wish. SeedRandom[0]; region = DiscretizeRegion @ RegionUnion@Table[Disk[RandomReal[4, {2}], RandomReal[1]], {10}] randomFromRegion[region_, trials_] := Module[{bounds, randPts}, bounds = RegionBounds@region; randPts = Transpose @ ...


8

ClearAll[mpbmdcg,ljr]; mpbmdcg[k_]:= Composition[MeshPrimitives[#,k]&, BoundaryMesh, DiscretizeGraphics, Graphics]; ljr = Composition[Line, Join[#, {#[[1]]}]&, Replace[#,Line[{a_,b_}]:>a, {0, Infinity}]&]; poly = First@mpbmdcg[2]@(Rectangle/@(-1+Normal[october2014[Values]])); fastdesc= FullSimplify[ Reduce[ ...


7

One option would be to restrict the function from funky regions with a Condition liks this f[x_, y_] /; Abs[x - y] > 5 := (Sin[x] - Sin[y])/(x - y); Plot3D[f[x, y], {x, -10, 10}, {y, -10, 10}] Out: One can easily see that Plot3D will also sample points in the region which is "forbidden", the points are just not drawn due to the RegionFunction. ...


7

This appears to be equivalent to your previous question. ybeltukov's beautiful answer can be adapted. His answer provides volume above x-y plane. volume[p_, abc_] := π Times @@ abc (2/3 + # - #^3/3) & @@ Normalize[abc p] an[a_] := {Cos[Pi/2 - a], Sin[Pi/2 - a], 0} vol[a_, abc_] := 4 Pi Times @@ abc/3 - volume[an[a], abc] a in the above is the angle ...


7

SeedRandom[0]; region = RegionUnion @ Table[Disk[RandomReal[4, {2}], RandomReal[1]], {10}]; DiscretizeRegion @ region points = RandomReal[{-1, 5}, {10000, 2}]; circles = List @@ (region /. Disk -> Circle); Graphics[{AbsolutePointSize[2], Transpose[{RegionMember[region, points] /. {False -> White, True -> Gray}, Point /@ points}], ...


7

Here is a clean way to do it. The idea is to specify all circular boundary regions and then to explicitly set those that are region holes, such that those are not meshed. Needs["NDSolve`FEM`"] \[CapitalOmega] = ImplicitRegion[(9/10)^2 <= x^2 + y^2 <= 1^2, {x, y}]; ToElementMesh[\[CapitalOmega]]["Wireframe"] ToElementMesh[\[CapitalOmega], ...


7

This is the same idea as unlikely's, which has same following drawback. The discretization of the region makes the boundary piecewise linear and thereby reduces the accuracy. The points found on the grid are points on the linear approximations of the boundary. This causes the boundary to contract further (on the concave side), thereby further reducing the ...


7

Consider A = {{1, 2}, {3, 4}}; B = {{5, 6}, {7, 8}}; Then ParametricPlot[ { {Cos[θ], Sin[θ]}.A.{Cos[θ], Sin[θ]}, {Cos[θ], Sin[θ]}.B.{Cos[θ], Sin[θ]} }, {θ, 0, 2 π}] Or more generally, A = {{1, 2}, {3, 4}}; B = {{5, 6}, {x, 8}}; ParametricPlot[ Table[{{Cos[θ], Sin[θ]}.A.{Cos[θ], Sin[θ]}, {Cos[θ], Sin[θ]}.B.{Cos[θ], Sin[θ]}}, ...


7

You might create a NearestFunction to help pick the particular boundary you want. You can use it to mark the boundary elements of an ElementMesh (FEM). plot = ParametricPlot[ bezierfunc[ξ, η], {ξ, 0, 1}, {η, 0, 1}]; edges = Map[ First@Cases[ Normal@ParametricPlot[#, {t, 10^-5, 1 - 10^-5}, PlotPoints -> 100], Line[p_] :> p, ...


7

I suppose MeshRegion and Graphics3D have similar methods of the visualization. However, Graphics3D is more convenient and GraphicsComplex can increase the performance: mesh = ReadFSMesh@"fsmesh.bin"; brain = GraphicsComplex[MeshCoordinates@#, MeshCells[#, 2]] &@mesh; Graphics3D[{EdgeForm[], Gray, brain}, Lighting -> "Neutral", Boxed -> False] ...


7

You could triangulate the polygon, pick out the triangles inside the polygon, and compute their area. area[poly_Polygon] := With[{mpoly = TriangulateMesh[DiscretizeGraphics[poly], "MaxCellMeasure" -> Area@Graphics`Mesh`BoundingAxisAlignedRectangle[poly]]}, With[{pts = MeshCoordinates[mpoly], tri = MeshCells[mpoly, 2]}, ...



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