Integral evaluation in an alphashape

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

[darova]

• I want to find the volume integral of this function in the shape defined by

are those coordinated kind of boundaries? What do they mean?

[berk can acikgoz]

Exactly! I am sorry I forgot to mention. These are the coordinates of corners of a tetrahedral

[darova]

I plotted the sphere and points. Looks as following:

You want to calculate the volume numerically?

I found some links for area calculation: wolfram

[berk can acikgoz]

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.

[darova]

What about triangulation?

Answer 1

[Matt J]

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{:});

[berk can acikgoz]

I tried this way. However, integration was unseccesful due to the result failing the global error test.

[Matt J]

What is the "global error test" and why should it be trusted?

[Matt J]

Try

result=integral3(fun,range{:},'Method','iterated');

[berk can acikgoz]

Well now it works but unacceptably slow. Maybe there is a faster way :/

[Matt J]

Is the calculated integral value very different in each case? And what does "unacceptably slow" mean? How fast is it meant to be?

[berk can acikgoz]

Integral is calculated allright. But it takes 182 seconds to evaluate the integral

[Matt J]

If both versions give the same result, then go back to the first method (the fast one) and ignore the warnings.