About double integral which includes non-linear equations in it
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Mehmet
el 23 de Nov. de 2013
Comentada: Mehmet
el 28 de Nov. de 2013
Hi I am a mechanical engineer and I am dealing with double integrals which includes nonlinear eauation but Matlab can t solve the problem exactly. I have been using int(int) but it doesn t work at a message appears like "Explicit integral could not be found". How can I solve this problem;
syms x y alpha beta V
K=1/(le/2-x*(sin(alpha*pi/180))*sin(beta*pi/180)-y*cos(alpha*pi/180))^2;
f=int ((K)*x,x,0,lm/2)
M=int(f,y,0,lm/2)
I also attached the real question. In this question I would like to have an equation related to V,alpha and beta.
E0 lm and Le are constants.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/153202/image.jpeg)
2 comentarios
Walter Roberson
el 23 de Nov. de 2013
It is possible to find a closed form for f, but you need to go piecewise as some of the combinations of conditions can generate infinities. Some of the failing conditions are based mostly on the lm and le values, but others are based more on the angles. For example at -Pi to +Pi radians there get to be exceptions.
I do not have the formula up on my screen now as I made the mistake of trying to visualize at too high of a density and choked my system to a crawl by using up tons of memory :(
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Roger Stafford
el 23 de Nov. de 2013
Editada: Roger Stafford
el 23 de Nov. de 2013
The symbolic toolbox is quite capable of solving your double integral. For simplicity of notation I substituted a = Le/2, b = -sin(alpha)*cos(beta), c = -cos(alpha), and d = lm/2. Here is the expression it gives after simplification:
int(int(x/(a+b*x+c*y)^2,x,0,d),y,0,d) =
(a+c*d)*log((a+b*d+c*d)/(a+c*d))-a*log((a+b*d)/a))/c/b^2
(In fact it is quite possible to solve this merely by looking it up in integral tables together with a certain amount of algebraic manipultion.)
There is one pitfall you need to be careful about. In the entire square range of 0<=x<=d and 0<=y<=d the quantity a+b*x+c*y must not cross zero. Otherwise the double integral becomes divergent and the above formula is invalid.
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Roger Stafford
el 28 de Nov. de 2013
If you used the formula I gave you, the only way you could get a complex result would be to have a negative argument in one or both of the two logarithms, and that in turn would mean that the denominator of your integrand must have equaled zero somewhere within your square region, 0<=x<=d and 0<=y<=d. If you recall I warned you about that. It would mean that you have a divergent integral, and there is no remedy for that. In that sense your professor would be right. But surely you can easily check for that directly by considering the values of a, b, and c for x and y over the range from 0 to d. By the way, remember when using matlab's sin and cos, these assume that their arguments are in radians. You need to take that into account when evaluating b and c from alpha and beta.
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