calculating Double integral over a region
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Ibraheem
el 24 de Nov. de 2022
Comentada: Carlos Guerrero García
el 24 de Nov. de 2022
I am trying to plot this double integral but i keep getting an error, can someone help me out thanks.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1205188/image.jpeg)
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
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Torsten
el 24 de Nov. de 2022
syms x y
int(int(2,y,-sqrt(1-(x/3)^2),sqrt(1-(x/3)^2)),x,-3,3)
1 comentario
Carlos Guerrero García
el 24 de Nov. de 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!!!
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Carlos Guerrero García
el 24 de Nov. de 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)
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