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Visualizing a complicated Region

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Visualizing a complicated Region


Visualizing directoriesVisualizing a 2-dimensional PDFVisualizing Type System OperationsRegion from pointsVisualizing Bendixson’s criterionVisualizing region unionsHalfplane region intersection anomalyVisualizing the complex logarithmParametric Region only showing half of regionVisualizing Trained Filters













3












$begingroup$


I was given the problem to find the area bounded by the polar curve



$$r=frac{1}{2}+frac{3}{2}cos(theta)$$



which looks like



                                      
enter image description here



To be clear, the region meant is the lighter of the two here



                                      
enter image description here



I know the integrals needed to actually find the area, and that really isn't the question anymore. Instead, visualization is. I believe that can define the region that I am interested in with the code block (a and b were defined for generality)



a=1/2;

b=3/2;

v={(a c+b c Cos[t])Cos[t],(a c+b c Cos[t])Sin[t]};

R1=ParametricRegion[v,{{c,0,1},{t,0,(2[Pi])/3}}];

R2=ParametricRegion[v,{{c,0,1},{t,2[Pi])/3,(4[Pi])/3}}];

R3=ParametricRegion[v,{{c,0,1},{t,(4[Pi])/3,2[Pi]}}];



R=RegionDifference[RegionUnion[R1,R3],R3];


I can look at these individually, but when I use Region to visualize the whole thing, it doesn't return even a picture, just some internal expression. How can I visualize just the region of interest?






regions visualization







share|improve this question









$endgroup$












  • $begingroup$
    Your code works for me, but very slowly. I obtained R1 in several minutes.
    $endgroup$
    – user64494
    4 hours ago
















3












$begingroup$


I was given the problem to find the area bounded by the polar curve



$$r=frac{1}{2}+frac{3}{2}cos(theta)$$



which looks like



                                      
enter image description here



To be clear, the region meant is the lighter of the two here



                                      
enter image description here



I know the integrals needed to actually find the area, and that really isn't the question anymore. Instead, visualization is. I believe that can define the region that I am interested in with the code block (a and b were defined for generality)



a=1/2;

b=3/2;

v={(a c+b c Cos[t])Cos[t],(a c+b c Cos[t])Sin[t]};

R1=ParametricRegion[v,{{c,0,1},{t,0,(2[Pi])/3}}];

R2=ParametricRegion[v,{{c,0,1},{t,2[Pi])/3,(4[Pi])/3}}];

R3=ParametricRegion[v,{{c,0,1},{t,(4[Pi])/3,2[Pi]}}];



R=RegionDifference[RegionUnion[R1,R3],R3];


I can look at these individually, but when I use Region to visualize the whole thing, it doesn't return even a picture, just some internal expression. How can I visualize just the region of interest?






regions visualization







share|improve this question









$endgroup$












  • $begingroup$
    Your code works for me, but very slowly. I obtained R1 in several minutes.
    $endgroup$
    – user64494
    4 hours ago














3












3








3





$begingroup$


I was given the problem to find the area bounded by the polar curve



$$r=frac{1}{2}+frac{3}{2}cos(theta)$$



which looks like



                                      
enter image description here



To be clear, the region meant is the lighter of the two here



                                      
enter image description here



I know the integrals needed to actually find the area, and that really isn't the question anymore. Instead, visualization is. I believe that can define the region that I am interested in with the code block (a and b were defined for generality)



a=1/2;

b=3/2;

v={(a c+b c Cos[t])Cos[t],(a c+b c Cos[t])Sin[t]};

R1=ParametricRegion[v,{{c,0,1},{t,0,(2[Pi])/3}}];

R2=ParametricRegion[v,{{c,0,1},{t,2[Pi])/3,(4[Pi])/3}}];

R3=ParametricRegion[v,{{c,0,1},{t,(4[Pi])/3,2[Pi]}}];



R=RegionDifference[RegionUnion[R1,R3],R3];


I can look at these individually, but when I use Region to visualize the whole thing, it doesn't return even a picture, just some internal expression. How can I visualize just the region of interest?






regions visualization







share|improve this question









$endgroup$




I was given the problem to find the area bounded by the polar curve



$$r=frac{1}{2}+frac{3}{2}cos(theta)$$



which looks like



                                      
enter image description here



To be clear, the region meant is the lighter of the two here



                                      
enter image description here



I know the integrals needed to actually find the area, and that really isn't the question anymore. Instead, visualization is. I believe that can define the region that I am interested in with the code block (a and b were defined for generality)



a=1/2;

b=3/2;

v={(a c+b c Cos[t])Cos[t],(a c+b c Cos[t])Sin[t]};

R1=ParametricRegion[v,{{c,0,1},{t,0,(2[Pi])/3}}];

R2=ParametricRegion[v,{{c,0,1},{t,2[Pi])/3,(4[Pi])/3}}];

R3=ParametricRegion[v,{{c,0,1},{t,(4[Pi])/3,2[Pi]}}];



R=RegionDifference[RegionUnion[R1,R3],R3];


I can look at these individually, but when I use Region to visualize the whole thing, it doesn't return even a picture, just some internal expression. How can I visualize just the region of interest?






regions visualization




regions visualization






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 6 hours ago









Brandon MyersBrandon Myers

1305




1305












  • $begingroup$
    Your code works for me, but very slowly. I obtained R1 in several minutes.
    $endgroup$
    – user64494
    4 hours ago


















  • $begingroup$
    Your code works for me, but very slowly. I obtained R1 in several minutes.
    $endgroup$
    – user64494
    4 hours ago
















$begingroup$
Your code works for me, but very slowly. I obtained R1 in several minutes.
$endgroup$
– user64494
4 hours ago




$begingroup$
Your code works for me, but very slowly. I obtained R1 in several minutes.
$endgroup$
– user64494
4 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

We can construct such a region with CrossingPolygon and in this instance Polygon will work too.



c = 1;

pts = Table[v, {t, Subdivide[0., 2π, 200]}];

pts[[-1]] = pts[[1]];



Graphics[{EdgeForm[ColorData[97][1]], Opacity[.3], ColorData[97][2], Polygon[pts]}]






CrossingCount[Polygon[pts], {0.5, 0}]





2






share|improve this answer









$endgroup$





















    2












    $begingroup$

    I think your plot is not quite correct, as the region should only extend to $x=2$ on the horizontal axis.



    Here's how to plot it using RegionPlot:



    RegionPlot[-1/2 + 3/2 x/Sqrt[x^2 + y^2] < Sqrt[x^2 + y^2] < 1/2 + 3/2 x/Sqrt[x^2 + y^2],
    
  {x, -1/2, 2}, {y, -5/4, 5/4}, PlotPoints -> 100]


    enter image description here



    The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.






    share|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      We can construct such a region with CrossingPolygon and in this instance Polygon will work too.



      c = 1;
      
pts = Table[v, {t, Subdivide[0., 2π, 200]}];
      
pts[[-1]] = pts[[1]];
      

      
Graphics[{EdgeForm[ColorData[97][1]], Opacity[.3], ColorData[97][2], Polygon[pts]}]
      





      CrossingCount[Polygon[pts], {0.5, 0}]
      




      2






      share|improve this answer









      $endgroup$


















        4












        $begingroup$

        We can construct such a region with CrossingPolygon and in this instance Polygon will work too.



        c = 1;
        
pts = Table[v, {t, Subdivide[0., 2π, 200]}];
        
pts[[-1]] = pts[[1]];
        

        
Graphics[{EdgeForm[ColorData[97][1]], Opacity[.3], ColorData[97][2], Polygon[pts]}]
        





        CrossingCount[Polygon[pts], {0.5, 0}]
        




        2






        share|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          We can construct such a region with CrossingPolygon and in this instance Polygon will work too.



          c = 1;
          
pts = Table[v, {t, Subdivide[0., 2π, 200]}];
          
pts[[-1]] = pts[[1]];
          

          
Graphics[{EdgeForm[ColorData[97][1]], Opacity[.3], ColorData[97][2], Polygon[pts]}]
          





          CrossingCount[Polygon[pts], {0.5, 0}]
          




          2






          share|improve this answer









          $endgroup$



          We can construct such a region with CrossingPolygon and in this instance Polygon will work too.



          c = 1;
          
pts = Table[v, {t, Subdivide[0., 2π, 200]}];
          
pts[[-1]] = pts[[1]];
          

          
Graphics[{EdgeForm[ColorData[97][1]], Opacity[.3], ColorData[97][2], Polygon[pts]}]
          





          CrossingCount[Polygon[pts], {0.5, 0}]
          




          2







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          Chip HurstChip Hurst

          23.9k15995




          23.9k15995























              2












              $begingroup$

              I think your plot is not quite correct, as the region should only extend to $x=2$ on the horizontal axis.



              Here's how to plot it using RegionPlot:



              RegionPlot[-1/2 + 3/2 x/Sqrt[x^2 + y^2] < Sqrt[x^2 + y^2] < 1/2 + 3/2 x/Sqrt[x^2 + y^2],
              
  {x, -1/2, 2}, {y, -5/4, 5/4}, PlotPoints -> 100]


              enter image description here



              The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.






              share|improve this answer











              $endgroup$


















                2












                $begingroup$

                I think your plot is not quite correct, as the region should only extend to $x=2$ on the horizontal axis.



                Here's how to plot it using RegionPlot:



                RegionPlot[-1/2 + 3/2 x/Sqrt[x^2 + y^2] < Sqrt[x^2 + y^2] < 1/2 + 3/2 x/Sqrt[x^2 + y^2],
                
  {x, -1/2, 2}, {y, -5/4, 5/4}, PlotPoints -> 100]


                enter image description here



                The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.






                share|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I think your plot is not quite correct, as the region should only extend to $x=2$ on the horizontal axis.



                  Here's how to plot it using RegionPlot:



                  RegionPlot[-1/2 + 3/2 x/Sqrt[x^2 + y^2] < Sqrt[x^2 + y^2] < 1/2 + 3/2 x/Sqrt[x^2 + y^2],
                  
  {x, -1/2, 2}, {y, -5/4, 5/4}, PlotPoints -> 100]


                  enter image description here



                  The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.






                  share|improve this answer











                  $endgroup$



                  I think your plot is not quite correct, as the region should only extend to $x=2$ on the horizontal axis.



                  Here's how to plot it using RegionPlot:



                  RegionPlot[-1/2 + 3/2 x/Sqrt[x^2 + y^2] < Sqrt[x^2 + y^2] < 1/2 + 3/2 x/Sqrt[x^2 + y^2],
                  
  {x, -1/2, 2}, {y, -5/4, 5/4}, PlotPoints -> 100]


                  enter image description here



                  The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 6 hours ago

























                  answered 6 hours ago









                  RomanRoman

                  6,58111134




                  6,58111134






























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