calculating Double integral over a region

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Ibraheem
Ibraheem 2022-11-24
评论： Carlos Guerrero García 2022-11-24
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

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采纳的回答

Torsten
Torsten 2022-11-24
Ran in:
syms x y
int(int(2,y,-sqrt(1-(x/3)^2),sqrt(1-(x/3)^2)),x,-3,3)
ans = 
  1 个评论
Carlos Guerrero García
Carlos Guerrero García 2022-11-24
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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更多回答（1 个）

Carlos Guerrero García
Ran in:
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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