Double Integration Problem Simpson1/3

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

1 Comment

Start anew with coding using the composite formula given here:
http://mathfaculty.fullerton.edu/mathews/n2003/SimpsonsRule2DMod.html

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Answers (0)

Asked:

on 29 Jan 2019

Commented:

on 29 Jan 2019

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