# calculating Double integral over a region

4 views (last 30 days)
Ibraheem on 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

Torsten on 24 Nov 2022
syms x y
int(int(2,y,-sqrt(1-(x/3)^2),sqrt(1-(x/3)^2)),x,-3,3)
ans =
Carlos Guerrero García on 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!!!

Carlos Guerrero García on 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)
aaa = 18.8496

### Categories

Find more on Calculus in Help Center and File Exchange

### Community Treasure Hunt

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

Start Hunting!