# Area of a implicit curve

조회 수: 64(최근 30일)
Maxim Bogdan 17 Apr 2021 20:09
댓글: Maxim Bogdan 18 Apr 2021 14:34
Let's say that I have a multiply connected curve defined implicitly. For example:
f=@(x,y) sin(x).*sin(y)-0.5;
fimplicit(f,[-10,10,-10,10]);
How can we find the area of from the rectangle ?
So, if we can calculate the area of a filled figure than it's all done. There is a function for that kind of evaluations?
P.S. I wrote myself a program which detects where the changes from a connected component to another is made by fimplicit data points and I made a loop that sums all the areas of the polygons created using the [k,v]=boundary(Points,1) command.
I have the areas precision up to 4 decimal places, but I cannot solve the problems that appear on the corners of the rectangle , because boundary don't have them included in the points obtained from fimplicit. To be more precis, if I have the area that my code gives is not the area of a quarter of a circle, but the area of that sector minus the area of the triangle determined by it.
Also I have to mention that the integral method fails (it has too little precision). I mean that we can use the formula
integral2(@(x,y)heaviside(f(x,y)),-10,10,-10,10)

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### 채택된 답변

Matt J 17 Apr 2021 21:14
편집: Matt J 18 Apr 2021 4:06
Perhaps as follows?
plotRange=[-1,+1, -1,+1]*8;
plotCorners=[-1 1;1 1;1 -1; -1 -1]*8;
f=@(x,y) (sin(x).*sin(y)-0.5);
fp=fimplicit(f,2*plotRange,'MeshDensity',3000);
XY=[fp.XData;fp.YData].'; close
shpRegions=polyshape(XY);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
shpBounds=polyshape(plotCorners);
shpRegions=intersect(shpBounds,shpRegions);
Area=area(shpRegions)
Area = 48.6866
plot(shpRegions,'FaceColor','r')
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Matt J 18 Apr 2021 12:42
Here is a modification to detect the complementary area
plotRange=[-1,+1, -1,+1]*8;
plotCorners=[-1 1;1 1;1 -1; -1 -1]*8;
f=@(x,y) (sin(x).*sin(y)-0.5);
fp=fimplicit(f,2*plotRange,'Mesh',4000);
XY=[fp.XData;fp.YData].'; close
shpRegions=polyshape(XY);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
shpBounds=polyshape(plotCorners);
shpRegions=intersect(shpBounds,shpRegions);
[x,y]=shpRegions.centroid(1:numboundaries(shpRegions));
s=f(x,y); %test the sign of the centroids
if any( s>0 ) %take complementary region
shpRegions=subtract(shpBounds,shpRegions);
end
Area=area(shpRegions)
Area = 207.3129
plot(shpRegions,'FaceColor','r')

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### 추가 답변(1개)

Matt J 18 Apr 2021 13:39
편집: Matt J 18 Apr 2021 14:22
Here is a very similar strategy based on alphaShape instead of polyshape. It is slower, but IMO handles the bounding rectangle as well as the distinction between and more gracefully.
plotRange=[-1,+1, -1,+1]*8;
f=@(x,y) (sin(x).*sin(y)-0.5);
fp=fimplicit(f,plotRange,'MeshDensity',3000);
XY=unique([fp.XData;fp.YData].','rows','stable'); close
XY( any( isnan(XY),2) , : )=[];
[Xb,Yb]=ndgrid(linspace(plotRange(1), plotRange(end),500));
supp=f(Xb,Yb)<=0;
shp=alphaShape([XY;Xb(supp), Yb(supp)],0.1);
Area=area(shp)
Area = 207.3125
plot(shp,'FaceColor','r','EdgeColor','none')
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Maxim Bogdan 18 Apr 2021 14:34
Thanks a lot Matt J! Now the problem of the area and perimeter of a implicitly defined shape is very clear.

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