Integrating a multivariable function one variable at a time

Pere Garau Burguera

2016 7 13

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I have a function which depends on x and y, f(x,y) (returns a scalar) and I need to integrate it so that for a particular point (x0,y0), I get a function p(theta), I basically convert it to polar, change the origin and integrate over r, so the remaining function only depends on theta:

 %fun = @(x,y) Some function
 fun = @(x,y) fun(x+x0,y+y0);
 polarfun = @(r,theta) fun(r.*cos(theta),r.*sin(theta)).*r;
 p = @(theta) integral(@(r)polarfun(r,theta),0,Inf);

After I get this function, I need to integrate it again (w.r.t theta). This doesn't work (a and b are the limits of the integral)

value = integral(p,a,b);

I need to set ArrayValued = true for it to work

value = integral(p,a,b,'ArrayValued',true);

I don't understand why I need to specify it's an ArrayValued function when it's not meant to be, also the integral takes much more time, and I need it to be calculated as fast as possible. Can you explain me why it is ArrayValued, and whether there is a way to make it without the ArrayValued set to true?

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Torsten 2016-7-14

Is there anything that prevents you from using "integral2" ?

Take a look at the example

"Evaluate double integral in polar coordinates"

under

http://de.mathworks.com/help/matlab/ref/integral2.html

Best wishes

Torsten.

Pere Garau Burguera 2016-7-14

Hi Torsten

Well I put what I was doing before I figured out what I had to do, but I don't need to integrate it right after. After I get the p function, I need to perform some operations on it, creating a new function (which adds one variable), integrate the new function with respect to theta and then find a minimum with respect to the other variable.

Best

Pere

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