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5

Another way will be to simply take advantage of the CapForm primitives and get those rounded lines. I am using a transformation rule on the Graphics object generated by the PolarPlot function to change the default lines into a rounded one. plot=PolarPlot[theta/2 Pi, {theta, 0, 20 Pi},Axes->None,PlotStyle->Black, PlotRange -> All]; ...


5

Here's how I would go about doing drawing the spirals. image = Graphics[{ AbsoluteThickness[10] , CapForm["Round"] , Line[Table[ theta/(2 Pi) {Cos[theta], Sin[theta]} , {theta, 0, 20 Pi, 20 Pi/599} ]] } ,ImageSize -> {500, 500} ] Changing the argument of AbsoluteThickness[20] allows you to get different thicknesses. Changing the ...


4

parm = CoordinateTransform["Polar" -> "Cartesian", {u/2 Pi, u}]; n = {#2, -#1} & @@ (Normalize@D[parm, u]); ParametricPlot[{{r n + parm}, {parm}}, {u, 0, 10 Pi}, {r, -1.5, 1.5}, MaxRecursion -> 5] ParametricPlot[{{u^2 /400 r n + parm}, {parm}}, {u, 0, 10 Pi}, {r, -1.5, 1.5}, MaxRecursion -> 5]


3

You can also use DiscretizeGraphics with some clever options in your ContourPlot3D. Just add the Option Lighting -> "Neutral": gr = ContourPlot3D[G[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Contours -> {-5}, Lighting -> "Neutral"] // Quiet; Then you can compute the area with: Area @ DiscretizeGraphics @ gr 18.8018 Here is the image ...


3

This problem can be solved out of the box in V10 using the ImplicitRegion function. Lets use your function G and define the bounded 3D region. We can use RegionPlot3D to visualize the object. region = ImplicitRegion[G[x, y, z] <= -5, {x, y, z}]; RegionPlot3D[region, PlotRange -> 1.5, PlotPoints -> 60] Now we will use DiscretizeRegion to form a ...


1

You might consider using Labeled instead of PlotLabel: Labeled[Plot[Sin[x], {x, 0, 2 \[Pi]}], "Fancy label!"]


1

The present state of affairs The kind of manual manipulation demonstrated by kglr should theoretically not be necessary. There may have been some problems in the initial version 10 release but in either 10.0.2 or 10.1.0 many Themes already combine correctly. Emulating his penultimate example: Plot[Evaluate@Table[BesselJ[n, x], {n, 5}], {x, 0, 10}, ...



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