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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

                                      

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

asked 4 hours ago

Brandon MyersBrandon Myers

1305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

                                      

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

asked 4 hours ago

Brandon MyersBrandon Myers

1305

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

                                      

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

asked 4 hours ago

Brandon MyersBrandon Myers

1305

$endgroup$

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];

share|improve this question

asked 4 hours ago

Brandon MyersBrandon Myers

1305

share|improve this question

asked 4 hours ago

Brandon MyersBrandon Myers

1305

asked 4 hours ago

Brandon MyersBrandon Myers

1305

asked 4 hours ago

Brandon MyersBrandon Myers

1305

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

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

$endgroup$

add a comment | 

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]

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

share|improve this answer

edited 3 hours ago

answered 3 hours ago

RomanRoman

6,57111134

$endgroup$

add a comment | 

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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

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

$endgroup$

add a comment | 

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

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

$endgroup$

add a comment | 

$begingroup$

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

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

$endgroup$

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}]

share|improve this answer

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

answered 2 hours ago

Chip HurstChip Hurst

23.9k15995

answered 2 hours ago

Chip HurstChip Hurst

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]

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

share|improve this answer

edited 3 hours ago

answered 3 hours ago

RomanRoman

6,57111134

$endgroup$

add a comment | 

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]

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

share|improve this answer

edited 3 hours ago

answered 3 hours ago

RomanRoman

6,57111134

$endgroup$

add a comment | 

$begingroup$

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]

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

share|improve this answer

edited 3 hours ago

answered 3 hours ago

RomanRoman

6,57111134

$endgroup$

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]

share|improve this answer

edited 3 hours ago

answered 3 hours ago

RomanRoman

6,57111134

answered 3 hours ago

RomanRoman

6,57111134

answered 3 hours ago

RomanRoman

6,57111134