Integral evaluation in an alphashape

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berk can acikgoz
berk can acikgoz 2020-3-11
编辑： Matt J 2020-3-12
I have an alphashape created by alphaShape function and an integral. Is there a way to evaluate this volume integral in the alpha shape? i.e. I have a function and I want to find the volume integral of this function in the shape defined by
x coordinates:
0
0.0107
0.0160
0.0101
y coordinates:
0
0
0
0.0106
z coordinates:
0
0.0101
0
0
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berk can acikgoz
berk can acikgoz 2020-3-12
It is actually a volume integral. Also it is a tetrahedral. I know how to integrate 3D but i dont want to since there are too many of these tetrahedrals and each time i will have to calculate the integration boundaries etc.

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回答（1 个）

Matt J
Matt J 2020-3-11
编辑：Matt J 2020-3-11
Perhaps as follows. Here, shp refers to your alphaShape object.
fun=@(x,y,z) (x.^2+y.^2+z.^2).*shp.inShape(x,y,z);
range=num2cell( [min(shp.Points);max(shp.Points)] );
result=integral3(fun,range{:});
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berk can acikgoz
berk can acikgoz 2020-3-12
Integral is calculated allright. But it takes 182 seconds to evaluate the integral
Matt J
Matt J 2020-3-12
编辑：Matt J 2020-3-12
If both versions give the same result, then go back to the first method (the fast one) and ignore the warnings.

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