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14

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]]; ...


14

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 + ...


13

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. ...


13

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]], ...


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 = ...


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, ...


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, ...


7

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 ...


6

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. ...


6

All you're missing is the option for BoundaryStyle RegionPlot[1 <= x^2 + y^2 < 9, {x, -3.5, 3.5}, {y, -3.5, 3.5}, Frame -> False, AxesOrigin -> {0, 0}, Axes -> True, BoundaryStyle -> {Dotted, Thickness[0.005]}] If you want the unit circle filled, you can RegionPlot that separately: Show[RegionPlot[1 <= x^2 + y^2 < 9, {x, ...


6

u = {x, x^3, 0}; v = {0, 0, z2}; l = (u - v) t + v; w = l /. Solve[xs xs + ys ys + (zs - 1)^2 == 1 /. Thread[{xs, ys, zs} -> l], t][[2]]; Manipulate[Show[ ParametricPlot3D[{Cos[u] Sin[v], Sin[u] Sin[v], 1 - Cos[v]}, {v, ArcCos[1 - z1], 0}, {u, 0, 2 Pi}, PlotStyle -> {Opacity[.3], FaceForm[Red, Yellow]}, Mesh -> False, ...


6

Another way to generate all the points is by using Reduce: points = {x, y} /. List@ToRules@ Reduce[x >= 4 y && x <= 4 y + 3 && 0 < x < 63 && 0 < y < 15, {x, y}, Integers] If you give bounds (and thus constrain the possible solutions to a finite set), Reduce will typically list all solutions. Then just plot ...


5

Solution using Show needs to rearrange the order of ranges in ContourPlot3D, e.g. : Show[RegionPlot3D[(s - 3 q*s + q > 0 && p == 0) || (s - 3 q*s + q <= 0 && p == 1), {q, 0, 1}, {s, 0, 1}, {p, 0, 1}, AxesLabel -> Automatic], ContourPlot3D[s - 3 q*s + q == 0, {q, 0, 1}, {s, 0, 1}, {p, 0, ...


5

Here is a workaround: r1 = RegionDifference[Rectangle[{0, 0}, {10, 10}], Rectangle[{4, 4}, {8, 8}]]; r2 = TransformedRegion[r1, RotationTransform[45 \[Degree], {5, 5}]]; mr = DiscretizeRegion[r2] And then: RegionPlot[mr]


5

Clear["Global`*"] expr = y^2 - 2 x^2 y + 6 x^3 - 3*x*y + 2 y - 6; Integrate[ expr Boole[y >= 2*x^2 - 2 && y <= 3*x], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] -(36625/2688) Reduce[y >= 2*x^2 - 2 && y <= 3*x, y] (x == -(1/2) && y == -(3/2)) || (-(1/2) < x < 2 && -2 + 2 x^2 ...


5

Try to define your matrix as a function of $(x,q,r,t)$ variables S[x_, q_, r_, t_] := {{1/r^2, 1, 1, t Sqrt[x]}, {1, 1/t^2, 1, Sqrt[x]/t}, {1, 1, 1, Sqrt[q]}, {t Sqrt[x], Sqrt[x]/t, Sqrt[q], 1}}; Then Manipulate does what you want Manipulate[ RegionPlot[ PositiveDefiniteMatrixQ@S[x, q, r, t], {x, 0, 1}, {q, 0, 1}], {{r, 1/2}, 0, 1}, {{t, 1/2}, 0, ...


5

By looking at the hyperbolic behavior of the branches, you may try a simple RegionFunction[]: 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]], ImageSize -> 550, RegionFunction -> (Abs[#1 #2 ...


4

One easy way will be to use Opacity! Plot[{Sin[x], Cos[x]}, {x, -\[Pi], +\[Pi]}, Filling -> {1 -> {Axis, Directive[Opacity[.7], Blue]}, 2 -> {Axis, Directive[Opacity[.7], Red]}}] Update: Here comes a better solution for your problem. The function is pretty much self explanatory as far as the argument names are concerned. Given a list of ...


3

If I understand your question correctly - as judged from the text typed - the answer should read (writing - 6x instaed of just - 6 as some others have done, put the function in brackets before multiplying with Boole, also use +- Infinity which is better than some arbitrary (?) finite value like +-2) Integrate[(y^2 - 2*x^2*y + 6*x^3 - 3*x*y + 2*y - 6*x)* ...


3

Amplifying on the answer by Chenmingi: Boole will automatically restrict the integral to the appropriate region so you can integrate from -Infinity to Infinity for each of the variables. int1 = Integrate[ (y^2 - 2 x^2 y + 6 x^3 - 3*x*y + 2 y - 6) * Boole[y >= 2*x^2 - 2 && y <= 3*x], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] ...


3

The reason the B appears incorrectly is not because of the shape of G, but rather because the horizontal length of G is less than the vertical length of B, resulting in B being decapitated. Here's how the letters look at present when one is laid on top of the other at 90º: Notice that in the actual GEB logo, they use a squarish font, which solves this ...


3

Here is a way to do what you want using DiscretizeGraphics as suggested by @user21. No need to call NDSolveFEM` r = RegionDifference[Ball[{0, 0, 0}], Ball[{0, 0, 1}]] rp = RegionPlot3D[r, PlotPoints -> 50] Now we discretize the Graphics object with the following clever replacements that somehow works: DiscretizeGraphics[Normal[rp /. {(PlotRange ...


3

Maybe something like this? I don't know if this is what you meant? sol=Reduce[x1 >= 0 &&-4*x1 <= 16 &&4*x1 >= 16 ||x1 <= 0 &&4*x1 <= 16 &&-4*x1 >= 16, {x1}]; sol[[1, 2]] (*-4*) sol[[2, 2]] (*4*)


2

Another, using smart and fast functions like Array and Tuples, thus a bit more recommended way : RegionPlot[ x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}, Epilog -> { Red, PointSize[0.005], Point[ Join @@ Tuples /@ Array[ {Range[4 #, 4 # + 3], {#}} &, {16}, 0]]}, AspectRatio -> 15/63 ]


2

Alternatively you can generate just the points you want and then plot them : data = DeleteCases[Flatten[Outer[Boole[4 #2 <= #1 <= 4 #2 + 3] {#1, #2} &, Range[0, 63], Range[0, 15]], 1], {0, 0}]; Show[RegionPlot[x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}], ListPlot[data]]


2

More generalized version: newRegionPlot3D[ expr_, {u_, umin_, umax_}, {v_, vmin_, vmax_}, {w_, wmin_, wmax_}, tr_, opts___] := Module[{x, y, z, newExpr, xyz}, newExpr = TransformedField[tr -> "Cartesian", expr, {u, v, w} -> {x, y, z}]; xyz = CoordinateTransform[tr -> "Cartesian", {u, v, w}]; {xmax, ymax, zmax} = NMaxValue[{#, ...


2

A simple approach would be to plot a third function which defines the overlap region. overlap[f_List] := Piecewise[{{Sign[First@f] Min[Abs@f], Equal @@ Sign[f]}}, I] Plot[{Sin[x], Cos[x], overlap[{Sin[x], Cos[x]}]}, {x, -π, π}, Filling -> {1 -> {0, Blue}, 2 -> {0, Red}, 3 -> {0, Thread[Red + Blue, RGBColor]}}]


2

I can get the integral by using the first quadrant (for positivity), placing an assumption on n, and recasting without Boole (I don't know why that was needed). in = 4 Integrate[(x^(2 n) + y^(2 n)), {x, 0, 1}, {y, 0, (1 - x^(2 n))^(1/(2*n))}, Assumptions -> n >= 1] (* Out[112]= (2^(2 - 1/n) Sqrt[\[Pi]] Gamma[1 + 1/(2 n)])/((1 + n) Gamma[(1 + ...


2

I don't want to compete with Algohi's nice answer, but - as to my experience - Reduce can be almost always replaced with Simplify or FullSimplify: res = Simplify[x1 >= 0 && -4*x1 <= 16 && 4*x1 >= 16 || x1 <= 0 && 4*x1 <= 16 && -4*x1 >= 16] Cases[res, _?NumberQ, -1] {-4, 4}



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