Integral "int" function not evaluating
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Ali Almakhmari
am 3 Mär. 2022
Kommentiert: Walter Roberson
am 6 Mär. 2022
Why is the "int" function not evaluating my integral? The "r" variable still has the "int" in it.
clear
clc
theta_s = 0:0.5:pi/2;
syms theta_v phi
for i = 1:length(theta_s)
integ2(i) = ((((1/(2*pi))*((pi-phi)*cos(phi) + sin(phi))*tan(theta_s(i))*tan(theta_v)-(1/pi)*(tan(theta_s(i))+tan(theta_v)+sqrt(tan(theta_v)^2 + tan(theta_s(i))^2 - 2*tan(theta_s(i))*tan(theta_v)*cos(phi)))))*cos(theta_v)*sin(theta_v));
r(i)= int((integ2(i)), theta_v, [0 pi/2]);
end
1 Kommentar
Walter Roberson
am 3 Mär. 2022
What reason do you have to lead you to expect that there is a closed form integral?
Akzeptierte Antwort
David Hill
am 3 Mär. 2022
theta_s = 0:0.5:pi/2;
syms theta_v
phi=pi/6;%choose a phi or loop for various values of phi
for i = 1:length(theta_s)
integ2(i) = ((((1/(2*pi))*((pi-phi)*cos(phi) + sin(phi))*tan(theta_s(i))*tan(theta_v)-(1/pi)*(tan(theta_s(i))+tan(theta_v)+sqrt(tan(theta_v)^2 + tan(theta_s(i))^2 - 2*tan(theta_s(i))*tan(theta_v)*cos(phi)))))*cos(theta_v)*sin(theta_v));
r(i)= vpaintegral((integ2(i)), theta_v, [0 pi/2]);
end
6 Kommentare
David Hill
am 4 Mär. 2022
Works fine.
theta_s = 0:0.5:pi/2;
syms theta_v phi
for i = 1:length(theta_s)
integ2(i) = ((((1/(2*pi))*((pi-phi)*cos(phi) + sin(phi))*tan(theta_s(i))*tan(theta_v)-(1/pi)*(tan(theta_s(i))+tan(theta_v)+sqrt(tan(theta_v)^2 + tan(theta_s(i))^2 - 2*tan(theta_s(i))*tan(theta_v)*cos(phi)))))*cos(theta_v)*sin(theta_v));
r(i)= vpaintegral(vpaintegral((integ2(i)), theta_v, [0 pi/2]),phi,[0 pi]);
end
Walter Roberson
am 6 Mär. 2022
By the way, integrating first with respect to phi is faster.
At one point I saw a closed form solution for integration with respect to phi, but I was not able to reproduce that later.
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!