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Visualizing a complicated Region
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Visualizing a complicated Region
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$begingroup$
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
asked 6 hours ago
Brandon MyersBrandon Myers
1305
$endgroup$
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
asked 6 hours ago
Brandon MyersBrandon Myers
1305
$endgroup$
$begingroup$
Your code works for me, but very slowly. I obtained R1 in several minutes.
$endgroup$
– user64494
4 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
asked 6 hours ago
Brandon MyersBrandon Myers
1305
$endgroup$
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
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
regions visualization
asked 6 hours ago
Brandon MyersBrandon Myers
1305
asked 6 hours ago
Brandon MyersBrandon Myers
1305
asked 6 hours ago
Brandon MyersBrandon Myers
1305
asked 6 hours ago
Brandon MyersBrandon Myers
1305
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
$endgroup$
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 $cos(theta)=x/r$.
answered 6 hours ago
RomanRoman
6,58111134
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered 4 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
answered 4 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
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
$endgroup$
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
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
answered 4 hours ago
Chip HurstChip Hurst
23.9k15995
23.9k15995
add a comment |
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 $cos(theta)=x/r$.
answered 6 hours ago
RomanRoman
6,58111134
$endgroup$
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 $cos(theta)=x/r$.
answered 6 hours ago
RomanRoman
6,58111134
$endgroup$
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 $cos(theta)=x/r$.
answered 6 hours ago
RomanRoman
6,58111134
$endgroup$
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]
The trick is to use $r=sqrt{x^2+y^2}$ and $cos(theta)=x/r$.
answered 6 hours ago
RomanRoman
6,58111134
edited 6 hours ago
answered 6 hours ago
RomanRoman
6,58111134
answered 6 hours ago
RomanRoman
6,58111134
answered 6 hours ago
RomanRoman
6,58111134
6,58111134
add a comment |
add a comment |
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$begingroup$
Your code works for me, but very slowly. I obtained R1 in several minutes.
$endgroup$
– user64494
4 hours ago