calculating Double integral over a region

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Ibraheem
Ibraheem am 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

Akzeptierte Antwort

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

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Weitere Antworten (1)

Carlos Guerrero García
Carlos Guerrero García am 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

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