Double integral using matlab multiplication

2 views (last 30 days)
Hello, I want to calculate the double integral of an expression (not given here) using matrix multiplication. Using integral2 doesnot work for my case because of singularity issues I will have in that problem. So, first I'm trying out some basic double integral using matrix multiplication but those are not giving me the accurate answer I get calculating analytically.
Integration of sin(th).cos(phi) d(th)d(phi) in the limit of th=0 to pi and phi=0 to 2pi
th=0:pi/180:pi; %theta
phi=0:pi/180:2*pi; %phi
dth2=th(6)-th(5); dphi=phi(6)-phi(5);
int= (dphi.*ones(1,length(phi)))* (cos(phi)'*sin(th))*(dth.*ones(1,length(th)))'; %integration by matrix multiplication
I get answer of 0.0349 instead of 0. Increasing step size to a very massive number helps but that's unreasonable.
Can anyone help, where I'm doing wrong

Accepted Answer

Torsten
Torsten on 12 Apr 2022
th=pi/(2*180):pi/180:pi-pi/(2*180); %theta
phi=pi/(2*180):pi/180:2*pi-pi/(2*180); %phi
dth=th(6)-th(5); dphi=phi(6)-phi(5);
intvalue= (dphi.*ones(1,length(phi)))* (cos(phi)'*sin(th))*(dth.*ones(1,length(th)))'
  2 Comments
Saqeeb Adnan
Saqeeb Adnan on 12 Apr 2022
Thank you so much. So, the singularity at pi/2 was the reason.
Torsten
Torsten on 12 Apr 2022
Edited: Torsten on 13 Apr 2022
No. In order to approximate a two-dimensional integral, you always have to take the value of the function at the center of a cell, not at the boundary, and multiply it with the area of that cell.

Sign in to comment.

More Answers (0)

Categories

Find more on Mathematics in Help Center and File Exchange

Products


Release

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!