calculating Double integral over a region
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
Ibraheem
il 24 Nov 2022
Commentato: Carlos Guerrero García
il 24 Nov 2022
I am trying to plot this double integral but i keep getting an error, can someone help me out thanks.
ymax = @(x) sqrt((9-x.^2)/9);
ymin =@(x) -1.*sqrt((9-x.^2)/9);
fun = @(x,y) aa;
aaa =integral2(fun,-3,3,ymin,ymax);
aa = 2
0 Commenti
Risposta accettata
Torsten
il 24 Nov 2022
syms x y
int(int(2,y,-sqrt(1-(x/3)^2),sqrt(1-(x/3)^2)),x,-3,3)
1 Commento
Carlos Guerrero García
il 24 Nov 2022
I think that Torsten answer is better than mine. I was trying to answer the question with minor changes in the original code, but Torsten code is easier and more elegant than mine. +1 to Torsten!!!
Più risposte (1)
Carlos Guerrero García
il 24 Nov 2022
When you define "fun", the variable "aa" is undefined yet. Also, because the variables "x" and "y" doesn't appear in the "fun" definition, the compiler provides another error. I suggest (avoiding the usage of the unnecesary declaration of the "aa" value) the following code, resulting the expected numerical value of 6*pi that is two times the area of an ellipse of semiaxes 3 and 1, as expected:
syms x y;
ymax = @(x) sqrt((9-x.^2)/9);
ymin =@(x) -1.*sqrt((9-x.^2)/9);
fun = @(x,y) 2+0*x+0*y; % Avoiding the "aa" declaration and incluing "x" and/or "y" in the function "fun"
aaa =integral2(fun,-3,3,ymin,ymax)
0 Commenti
Vedere anche
Categorie
Scopri di più su Calculus in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!