- #1
Benny
- 584
- 0
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?
[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?