int integral with three parameters

2023 10 17
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更新时间：2023 10 17
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Hello everyone;
I am struggling to solve this symbolic integral, but it does not solve and just writes int at the beginning of the equation in the Command Window. The equation is: (R_p and R make the f function)
clc
H = 50;
z = H/2;
Z = z/H;
Pe = 4;
syms x y Z_p
R_p = ((x^2 + y^2)^0.5)/H;
R = (sqrt(R_p + (Z - Z_p)^2));
f(x,y,Z_p) = (1/R) * exp (Pe * R/2);
fx = int(f,Z_p,[0 1]) - int(f,Z_p,[-1 0])
fx(x, y) = 
The result is: fx(x, y) = int(exp(2*((Z_p -.....

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Saeid Bina
Saeid Bina 2023-10-17
Thank you so much for your answer. Does this work? because in my case again shows (in command window):
fx(x, y) = int(exp(2*((Z_p - 1/2)^2 + (x^2 + y^2)^(1/2)/50)^(1/2)...
I think the answer shuld be only based on x,y since the equation is integral on Z_p and from 0 to 1 and -1 and 0.
Am I right? I mean why we have int and Z_p in the answer?
Thank you so much once again for your prompt reply.
Saeid Bina
Saeid Bina 2023-10-17
Thank you dear Walter. In my case do I need to double then?
Walter Roberson
Walter Roberson 2023-10-17
double() will not work, as x and y have not been given definite values
Saeid Bina
Saeid Bina 2023-10-17
Actually these parameters are not know and I am gonna find them when the equation is equal to 1.
fx = int(f,Z_p,[0 1]) - int(f,Z_p,[-1 0])==1
[x y ] = solve(fx,[x y]);
But does not work. Sorry for my many questions. I am a beginner in MATLAB.

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Walter Roberson
Walter Roberson 2023-10-17
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Infinite number of solutions.
H = 50;
z = H/2;
Z = z/H;
Pe = 4;
syms x y Z_p
R_p = ((x^2 + y^2)^0.5)/H;
R = (sqrt(R_p + (Z - Z_p)^2));
f(x,y,Z_p) = (1/R) * exp (Pe * R/2);
fx = int(f,Z_p,[0 1]) - int(f,Z_p,[-1 0])
fx(x, y) = 
fimplicit(fx - 1)

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