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Surface Integral

  1. Jan 29, 2010 #1
    1. The problem statement, all variables and given/known data I am to use a substitution of variables u = x, v = x + 2y to evaluate the surface integral

    int(0,1/2)int(0,1-y) exp(x/(x+y))dxdy

    where int(a,b) means integral sign with lower limit a and upper limit b.



    2. Relevant equations



    3. The attempt at a solution I used the substitution and calculated the Jacobian to be 1/2. However, in the new variables, the integral is now

    int(0,1/2)int(0,1+u-v)exp(u/v)dudv

    Both variables appearing in one of the integrals doesn't seem right to me - what has gone wrong with the calculation?
     
  2. jcsd
  3. Jan 30, 2010 #2
    Are you sure you have all factors of 2 correct? E.g. in the denominator of the exponential function, and the integration limit for the inner integral?

    It is not a problem that the outer integration variable is part of the inner integral. The opposite however, doesn't make sense.

    Torquil
     
  4. Jan 30, 2010 #3

    HallsofIvy

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    I get different things for both integrand and limits of integration. Putting x= u into v= x+ 2y gives v= u+ 2y so that y= (v- u)/2. Then x+ y= (v- u)/2+ 2u/2= (u+ v)/2. x/(x+y)= 2u/(u+v), not u/v.

    In the x,y plane, the region to be integrated is the quadrilateral with sides x= 0, y= 0, y= 1/2 and x+ y= 1.

    With x= 0, u= 0, v= 2y or just u= 0. With y= 0, u= x, v= x or v= u. With y= 1/2, u= x, v= x+1 or v= u+1. Adding u= x and v= x+ 2y gives u+v= 2x+ 2y= 2(x+ y) so with x+y= 1, u+v= 2.

    In the xy, plane, the region is a quadrilateral bounded by u= 0, v= u, v=u+1, and u+ v= 2. Now, whether you integrate with respect to u or v first, you are going to have to break that into two pairs of integrals. If you integrate with respect to v first, you will have to have one pair of integrals with u going from 0 to 1/2 and then another with u going from 1/2 to 1. If you integrate with respect to u first, you will have to have a pair of integrals with v going from 0 to 1 and then from 1 to 3/2.
     
  5. Jan 30, 2010 #4
    Sorry, I made a typo, the integral was supposed to be int(0,1/2)int(0,1-2y)exp(x/(x+2y)), hence the substitution.
    Thanks for the help though!
     
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