int integral with three parameters

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 -.....

[Saeid Bina]

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.

[Walter Roberson]

I described reasons in https://www.mathworks.com/matlabcentral/answers/2032224-why-are-there-different-results-of-int-function-regarding-int-and-double-function#answer_1331409

[Saeid Bina]

Thank you dear Walter. In my case do I need to double then?

[Walter Roberson]

double() will not work, as x and y have not been given definite values

[Saeid Bina]

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.

Answer 1

[Walter Roberson]

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)