# Visualizing a complicated Region

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?

• Your code works for me, but very slowly. I obtained R1 in several minutes. – user64494 Apr 29 at 19:02

## 2 Answers

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


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 $$\cos(\theta)=x/r$$.