Flemish Posted January 4, 2023 Share Posted January 4, 2023 (edited) I have recently been studying double integrals. One of the topics were double variable substitutions. I find it hard to find the right double variable substitutions. For example: I substituted y = 1/2(u+v) and x = u+v, in hopes of getting rid of (2y-x)^3/2 and making the integral easier. However, this led me down a very long process and ended up being too complicated for it to work. Any advice? Edited January 4, 2023 by Flemish Link to comment Share on other sites More sharing options...
Genady Posted January 4, 2023 Share Posted January 4, 2023 This was not a good substitution anyway, because it simply makes y=0.5x. You want two linearly independent variables. They better relate linearly with x, y. Then you can easily replace limits of integration. You have two linear terms in x, y. I guess they are a hint. Link to comment Share on other sites More sharing options...
Flemish Posted January 4, 2023 Author Share Posted January 4, 2023 (edited) 26 minutes ago, Genady said: This was not a good substitution anyway, because it simply makes y=0.5x. You want two linearly independent variables. They better relate linearly with x, y. Then you can easily replace limits of integration. You have two linear terms in x, y. I guess they are a hint. So in this case what would make a good substitution and why? Also what do you mean my linearly independent variable? Edited January 4, 2023 by Flemish Link to comment Share on other sites More sharing options...
Genady Posted January 4, 2023 Share Posted January 4, 2023 41 minutes ago, Flemish said: what do you mean my linearly independent variable? If you don't know, doesn't matter, I'll say it differently. y=0.5x makes a line, not a parallelogram; you need variables that run independently over the parallelogram. 44 minutes ago, Flemish said: in this case what would make a good substitution and why? A good substitution would separate integration variables rather than "get rid of". I don't know what "very long process" you're talking about. If you "get rid" of 2y-x by making it equal 0, as your substitution did, the whole function becomes simply =0. If you separate variables, you can treat the integrals one by one. Link to comment Share on other sites More sharing options...
Lorentz Jr Posted January 4, 2023 Share Posted January 4, 2023 (edited) 2 hours ago, Flemish said: I find it hard to find the right double variable substitutions. Do them one at a time. Single variable substitutions. (That is, introduce one new variable at a time, but it can depend on both x and y.) If you don't like the complicated expression inside the fractional power, start with that one. Edited January 4, 2023 by Lorentz Jr Link to comment Share on other sites More sharing options...
Flemish Posted January 4, 2023 Author Share Posted January 4, 2023 9 hours ago, Lorentz Jr said: Do them one at a time. Single variable substitutions. (That is, introduce one new variable at a time, but it can depend on both x and y.) If you don't like the complicated expression inside the fractional power, start with that one. Alright, I'll try that next time! 10 hours ago, Genady said: If you don't know, doesn't matter, I'll say it differently. y=0.5x makes a line, not a parallelogram; you need variables that run independently over the parallelogram. A good substitution would separate integration variables rather than "get rid of". I don't know what "very long process" you're talking about. If you "get rid" of 2y-x by making it equal 0, as your substitution did, the whole function becomes simply =0. If you separate variables, you can treat the integrals one by one. Ah, that makes a whole lot of sense. The reason why it was such a long process was because I was trying to find the transformation and Jacobian. I had to find all the lines that would make the parallelogram and then find the shape that it transform into. Also had to find the Jacobian. In the end of the day, two of the lines ended up equal something like 4=0 which made no sense meaning I was unable to find the new shape. Thank you, have you been very useful! Link to comment Share on other sites More sharing options...
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