Double integral - change of variables

  • Thread starter Benny
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  • #1
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Hi, I'm having trouble evaluating the following integral.

[tex]
\int\limits_{}^{} {\int\limits_R^{} {\cos \left( {\frac{{y - x}}{{y + x}}} \right)} } dA
[/tex]

Where R is the trapezoidal region with vertices (1,0), (2,0), (0,2) and (0,1).

I a drew a diagram and found that R is the region bounded by the lines y = - x + 2, y = -x + 1, x = 0 and y = 0.

Firstly, I would like to ask when attempting questions requiring a change of variables, is the priority to make the region of integration simpler or the integrand simpler? I ask this because I am often not sure how to approach these sorts of questions so if I have the correct starting point it should make things easier.

Anyway I think an obvious substitution u = x + y which converts to the lines u = 1 and u = 2. But how about the 'v' substitution? I have accounted for the lines y = -x + 1 and y = -x + 2 with the u substitution but there are also the sides of R given by the equations y = 0 and x = 0. I cannot figure out a suitable substitution. Can someone please help me with that?
 

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  • #2
Tom Mattson
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Benny said:
Firstly, I would like to ask when attempting questions requiring a change of variables, is the priority to make the region of integration simpler or the integrand simpler? I ask this because I am often not sure how to approach these sorts of questions so if I have the correct starting point it should make things easier.

The two questions are intertwined. You of course want the integrand to be easy to integrate, but once you perform that first integration your choice of region influences your integrand. I don't know of any general procedure for determining the region that will best facilitate integration. Experience seems to be the best guide. Mind you, I am trained not as a mathematician, but as a physicist. But as my old calculus teacher always said, "Differentiation is mechanics, integration is art."

Anyway I think an obvious substitution u = x + y which converts to the lines u = 1 and u = 2.

Good start!

But how about the 'v' substitution? I have accounted for the lines y = -x + 1 and y = -x + 2 with the u substitution but there are also the sides of R given by the equations y = 0 and x = 0. I cannot figure out a suitable substitution. Can someone please help me with that?

In this case: Let the integrand be your guide. Inside that cosine function you've got a rational function. You already used the denominator as one substitution. Why not let the numerator give you the other one?

Let v=y-x. It should come out pretty easy that way.

Let us know if you get stuck again.
 
  • #3
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Thanks for your explanations Tom. It's been a few days since I posted this question and during that time I've had a look at some other questions from the same section of the book. I've noticed that the integrands and the regions are set up so that the form of the integrand pretty much gives away the required substitutions. I asked about a procedure to do these sorts of problems because I think that in many cases it doesn't always turn out so well. Anyway thanks again for your help.
 

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