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solving integral with three parameters

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Saeid Bina
Saeid Bina 2023 年 10 月 16 日
閉鎖済み: Dyuman Joshi 2023 年 10 月 17 日
Hello everyone. I have question about parametric integral. I need to find the x and y when teta is especific values like 0.5, 1,2...
clc;clear;
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^2 + (Z - Z_p)^2);
f = (1/R) * exp (Pe * R/2);
teta = exp(Pe/2*R_p*cos(phi)*(int(f,Z_p,[0 1]-int(f,Z_p,[-1 0])
  • phi is polar angle from 0 to pi. for this step lets imagine 0.
  • The results should be pair of x and y that if we plot them will be oval it circular shapes.
  2 件のコメント
Rik
Rik 2023 年 10 月 16 日
What have you tried so far?
Saeid Bina
Saeid Bina 2023 年 10 月 16 日
Thank you so much for your reply. I firstly tried without the first part but does not work again.
clc;clear;
syms x y Z_p
H = 50;
z = H/2;
Z = z/H;
Pe = 4;
R_p = ((x^2 + y^2)^0.5)/50;
R = sqrt(R_p^2 + (Z - Z_p)^2);
f = (1/R) * exp (Pe * R/2)
solve(int(f,Z_p,[0 1])==2)
I deleted the first part.

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