Double Integration Problem Simpson1/3

Kevin E Vazquez

2019 1 29

0 个回答

8 次查看（30 天）

0 个投票

Hi, So im having a problem with simpson 1/3 for double integration. I dont know if the problem is in the for loop that i created or where. The answer should be pi. Im really stuck on this problem and have been for weeks. If anybody could help I could appreciate it. Here I included the program that im having trouble with.

% For Simpson's 1/3 for double integrals
f=@(x,y) 1; R=1;
a=@(x)-sqrt(R^2-x^2);b=@(x)sqrt(R^2-x^2); % Inner integral bounds
c=-R;d=R; % Outer integral limits
N=600;M=600;
dx=(d-c)/N; 
SUM=0;
v = [];
for i = 1:N %loop for Outer
    if i == 0 || i == N
        k=1;
    elseif mod(i,2) ~= 0 
        k=4;
    else
        k=2;
    end
 x = c + i*dx;
 v = [v,x];
end
x = sum(v);
dy=(b(x)-a(x))/M;
    for j = 1:M %loop for Inner
       if j == 0 || j == M
        l = 1;
       elseif mod(j,2) ~= 0
        l = 4;
       else
        l = 2;
       end
       y = -sqrt(R^2-x^2) + j*dy; 
       v = [v,y];
    end
y = sum(y);
SUM = SUM + (k*l) * f(x,y); d = (dx/dy)*9; f = (f(a)+f(b)); r = d*SUM;
DS13 = d*SUM + f