How to integrate discrete values over a known x, y coordinate image

I know a digital image f (x, y), and the corresponding power spectrum is obtained by Fourier transform, but I need to request the integration of the power spectrum in an annular area, I want to use integral2 It is, but I only have the value of f (x, y) and the coordinate value (x, y), I do n’t know how to use double integration。

ps：the power spectrum like this

and I want to integrate the power spectrum in an annular region like this

and the integral is

is the power spectrum of the image f(x,y)(A random image)，and i calculate it by this

I=fftshift(ft2(image);
s_power = abs(I).^2;

But s_powers is (x, y) coordinate, I could not calculate its double integral, ask for help!SOS!

[haha mark]

I used to multiply the an annular mask by s_power,ps：1 in the annular area and 0 in other areas

temp = mask.*s_power;
r = sum(sum(temp));

and use r as a result，however，maybe it is not ture。

[John D'Errico]

Don't send me direct e-mail as you did, asking me to answer your question. You should have gotten that hint from my profile, the way you did manage to contact me. It specifically asks people not to contract me. If I respond to it, then you learn to pester me for every future question you have about MATLAB or mathematics. That then forces me to add your name to mail filters to delete any future mail from you - now done. So this is now the only time I will respond to you.

I will only say you cannot just use integral/integral2 on an image, or any set of data. A function is required for those tools. Regardless, integral is not the proper tool to integrate a noisy surface, as it would try to chase every bump in the noise. Instead, consider what rectangle rule is, when used on a surface. Don't forget that for a true integral, you must multiply by the area of each element.

[haha mark]

I'm sorry to bother you, but I didn't see your homepage prompt when I emailed you. I'm too anxious to solve this problem. I'm sorry to trouble you. I'm very sorry. I didn't intend to disturb you, but I found your answer very helpful. I sincerely beg your forgiveness. I really need your help, sorry to bring you a bad mood

[haha mark]

I am so sad,I really didn't mean it.I found a question you answered very useful, so I am too eager to ask you, so that I did not click on your personal homepage, I am really sorry, it is indeed my fault, hope you forgive me.

Answer 1

[darova]

Here is the idea

% calculate increments
dx = x(2)-x(1);
dy = y(2)-y(1);
% assume A is your image
% assume X and Y are your 2d matrices of coordinates
R = hypot(X,Y);             % calculate radius
ix = r1 <= R & R <= r2;     % boundaries of integration
A1 = A*0;                   % preallocation
A1(ix) = A(ix);             % region of interest
S = 0;
for i = 1:size(A1,1)
    for j = 1:size(A1,2)
        s = A1(i:i+1,j:j+1);    % 4 neigbour values
        S = S + sum(s(:));      % sum values
    end
end
S = S*dx*dy/4;              % value of integral

[haha mark]

First of all, I sincerely thank you for your reply.But I still have some questions,

first，Should the value range of i be narrowed

for i = 1:size(A1,1)-1
    for j = 1:size(A1,2)-1
        s = A1(i:i+1,j:j+1);    % 4 neigbour values
        S = S + sum(s(:));      % sum values
    end
end

Otherwise it will exceed the matrix size of A1

second，Do you mean to convert continuous integrals into discrete integral sums?But I don’t understand why you add four adjacent values，you finally divide 4

        s = A1(i:i+1,j:j+1);    % 4 neigbour values
        S = S + sum(s(:));      % sum values

Can you help me answer my doubts？

[darova]

You are calculating volume under the the surface

Volume is average height and dx,dy:

I forgot to add this line. It's because you have a complicated shape, you have to sum only nonzero values

s = A1(i:i+1,j:j+1);    % 4 neigbour values
if all(s(:))
    S = S + sum(s(:));      % sum values
end

Little explanation

[haha mark]

Thank you very much. I appreciate your answer！