calculating Double integral over a region

조회 수: 1 (최근 30일)
Ibraheem
Ibraheem 2022년 11월 24일
댓글: Carlos Guerrero García 2022년 11월 24일
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

채택된 답변

Torsten
Torsten 2022년 11월 24일
syms x y
int(int(2,y,-sqrt(1-(x/3)^2),sqrt(1-(x/3)^2)),x,-3,3)
ans = 
  댓글 수: 1
Carlos Guerrero García
Carlos Guerrero García 2022년 11월 24일
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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추가 답변 (1개)

Carlos Guerrero García
Carlos Guerrero García 2022년 11월 24일
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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