# Is it possible to use PolarPlot together with RegionPlot?

I'm studying a bit about dual integrals by watching this YouTube video.

I was able to plot two polar functions together:

r1=Sqrt[18];R2=6Cos[θ];
g1=PolarPlot[r1,{θ,0,2π}];
g2=PolarPlot[r2,{θ,0,2π}];
Show[{g1,g2}]


I defined the analysis regions using the RegionPlot function:

RegionPlot[x^2+y^2<18,{x,0,10},{y,0,10},PlotStyle->Yellow]
RegionPlot[(x-3)^2+y^2<9,{x,0,10},{y,0,10},PlotStyle->Blue]
RegionPlot[x^2+y^2<18&&(x-3)^2+y^2<9,{x,0,10},{y,0,10}]


Using the ImplicitRegion function, you get the intersection between these two graphs:

ℛ=ImplicitRegion[x^2+y^2<18&&(x-3)^2+y^2<9,{x,y}];
RegionDimension[ℛ]


$2$

The question is as follows:

Is it possible to use the PolarPlot function within the RegionPlot function?

I tried something like this, but it was not possible ...

RegionPlot[PolarPlot[r1,{θ,0,2π}]&&PolarPlot[r2,{θ,0,2π},{x,{Y,0,10}]


I only got by the definition of the circle equation ...

It is not clear why you want to use PolarPlot within RegionPlot

reg1 = ImplicitRegion[x^2 + y^2 <= 18, {x, y}];
reg2 = ImplicitRegion[(x - 3)^2 + y^2 < 9, {x, y}];
reg3 = RegionIntersection[reg1, reg2];

RegionPlot[{reg1, reg2, reg3}, AspectRatio -> 2 Sqrt[18]/(Sqrt[18] + 6),
PlotLegends -> "Expressions"]


However, the colors are changed where there is overlap. This can be corrected by using non-overlapping regions.

reg4 = RegionDifference[reg1, reg2];
reg5 = RegionDifference[reg2, reg3];

RegionPlot[{reg4, reg5, reg3}, AspectRatio -> 2 Sqrt[18]/(Sqrt[18] + 6),
PlotLegends -> "Expressions"]


• The PolarPlot function seemed interesting because if I did not know the circle equation and having only the radius it is easier to insert the inputs. – LCarvalho Jun 21 '17 at 18:03