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Calculating integration limits of a double integral

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mj702 on 24 Nov 2020
Edited: Rohit Pappu on 30 Nov 2020
I am working on a college assignment to calculate the symbolical value and the numerical approximation of the double integral of over the region defined by and to determine the integration limits in y by symbolical computation.
Unfortunately, I haven't been able to figure out how to calculate the integration limits without knowing the final result. Sorry if this is easy, but I am a beginner in Matlab and my knowledge of double integrals isn't very profound.
Could someone please give me a hint how to solve this problem?
Thank You.


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mj702 on 24 Nov 2020
Yes, exactly! The second x should have been an y. Sorry for the mistake.
Thank You
Bruno Luong
Bruno Luong on 27 Nov 2020
x^2/6 + y^2/4 = 1 is a curve (ellipse). It's 1D manifold. I don't know hat "double integral" is doing here. You probably want a curviline integral.
Bruno Luong
Bruno Luong on 27 Nov 2020
The integral is 0 since the function is odd in x
and the domain is symetric in x.
(x,y) in domain then (-x,y) is in.
You can deduce the same conclusion with y variable.

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Answers (1)

Rohit Pappu
Rohit Pappu on 27 Nov 2020
Edited: Rohit Pappu on 30 Nov 2020
For computing the numerical approximation of the above double integral
%% Define the major and minor axis of ellipse
a = sqrt(6);
b = sqrt(4);
%% Define f(x,y)
fun = @(x,y) sin(x.*y);
%% Convert f(x,y) into f(r,theta)
polarfun = @(theta,r) fun(a*r.*cos(theta), b*r.*sin(theta))*a*b.*r;
%% x = a*r.*cos(theta), y = b*r.*cos(theta) , dxdy = a*b*r*dr(dtheta)
%% Assign upper limit for r
rmax =1;
%% Calculate the integral for r = [0,1] and theta = [0,2*pi]
q = integral2(polarfun,0,2*pi,0,rmax)
Additional documentation about integral2 can be found here


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