double integration with loop

thank you very much if anyone can help me.

so, I need to create a loop that calculates integrals at each point. Say I have the following function handle

D = @(x,y) x.*(Z(x,y).^2);

from that function, I need to calculate a double integral and create the following loop

int1 = @(y) integral(@(x)D(x,y),x1,x2,'ArrayValued',true);
int2 = integral(@(y)int1,y1,y2,'ArrayValued',true);
c1 = int2./g % g is a scalar number

int11 = @(y) integral(@(x)D(x,y),x2,x3,'ArrayValued',true);
int22 = integral(@(y)int11,y1,y2,'ArrayValued',true);
c11 = int22./g % g is a scalar number

int111 = @(y) integral(@(x)D(x,y),x3,x4,'ArrayValued',true);
int222 = integral(@(y)int111,y1,y2,'ArrayValued',true);
c111 = int222./g % g is a scalar number

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.
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i'm trying to do like this

x = 1:0.1:5
for i = 1:length(x)
    D = @(x,y) x.*(Z(x,y).^2);
    int1(i) = @(y) integral(@(x)D(x,y),x(i),x(i)+0.5,'ArrayValued',true);
    int2(i) = integral(@(y)int1(i),y1,y2,'ArrayValued',true);
    c = int2(i)./g
end

I need to plot a graph of c versus x

[darova]

Can you show original equations?

[José Anchieta de Jesus Filho]

dF = @(z1,r) 2.*pi.*v.*ro.*(Brt(z1,r).^2).*r; % v , ro are scalar numbers
p1 = @(r) integral(@(z1) dF(z1,r),z0,z0+e,'ArrayValued',true); % e is also a scalar
F = integral(@(r) p1,ri,re,'ArrayValued',true); 
c = F./v

this is my equation

[José Anchieta de Jesus Filho]

that's my role in response to z0. what I need is for it to be calculated with the z0 varying

z0 = 0:0.001:0.03

for each variation of z0, I will have a new p1, a new F and, consequently, a new c. I want to plot a graph of z0 versus c

[Walter Roberson]

p1 = @(r) integral(@(z1) dF(z1,r),z0,z0+e,'ArrayValued',true); % e is also a scalar

p1 is a function handle.

F = integral(@(r) p1,ri,re,'ArrayValued',true); 

@( r) p1 is a handle to an anonymous function that accepts a single parameter and ignores the parameter and returns the function handle p1. You should be just using

F = integral(p1,ri,re,'ArrayValued',true); 

[José Anchieta de Jesus Filho]

thank you very much, I will modify. But, you know how to change the function to calculate the integral for various values of z0

Answer 1

[darova]

Here is a shot

z0 = 0:0.001:0.03;
c = z0*0;
Brt = @(z1,r) z1+r;
e = 0.1;
ri = 0;
re = 1;
ro = 15;
v = 2;
dF = @(z1,r) 2.*pi.*v.*ro.*(Brt(z1,r).^2).*r; % v , ro are scalar numbers
for i = 1:length(z0);
    F = integral2(dF,z0(i),z0(i)+e,ri,re);
    c(i) = F./v;
end
plot(z0,c)

[José Anchieta de Jesus Filho]

I tried, but an error appears. The Brt function (z1, r), it is a little complicated, as it was calculated with numeric integrals. I can't explain it very well, but this example (4-D Integral of Sphere) is similar to the calculation done for her

https://www.mathworks.com/help/matlab/ref/integral3.html

I can't calculate the integral, in this case, p1, without adding 'Array Valued', because of the integrations I did earlier.

I really appreciate your help and, unfortunately, the code didn't work

[darova]

Please attach the whole code

[José Anchieta de Jesus Filho]

I thank you so much for your help. I changed the code a little and managed to make it work. Thank you

for i = 1:length(z0)
    P = @(r) integral(@(z1) dF(z1,r),z0(i),z0(i)+e,'ArrayValued',true);
    F = integral(P,ri,re,'ArrayValued',true);
    c(i) = F./v;
end
plot(z0,c)