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Finding limits of double integral

  1. Dec 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to solve the integral. First I did coordinate transformation by finding jacobian = (1/4)(x2 + y2).

    The problem is, I do not know the limits of integration after transformation...I tried using a graphical approach: by considering 2 cases: y>x and y<x and adding both integrals.

    For y < x it is straightforward, the limits of u: [0,∞] and v: [0,∞].

    However, for y > x: limits of u: [0,∞] which is easy as x,y>0 so v>0.

    But the limits for v i obtained: v: [0,-∞].

    When I add both integrals they sum to zero.

    My suspicion is that the limits of v for y > x should be [-∞,0] and not [0,∞]. But this doesn't make sense as considering the following points:

    1. y > x
    2. y, x are forever increasing

    Combining both, it implies " x and y are increasing with y increasing faster than x". This implies v goes from 0 to -∞..

    2. Relevant equations



    3. The attempt at a solution

    ehdqw8.png
     
    Last edited by a moderator: Dec 22, 2012
  2. jcsd
  3. Dec 22, 2012 #2

    mfb

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    As x,y >= 0, v = 2xy >= 0 in both cases.
    The integral is symmetric with respect to y>x and y<x, you can restrict your integral to one of those cases and multiply with 2 afterwards.
    I don't think you need those cases at all.
     
  4. Dec 22, 2012 #3
    I know that's possible as 1/(1+u2) is an even function, but I don't like using this approach as I feel that I won't always be this lucky.

    I'm interested in finding the limits of du and dv.
     
  5. Dec 22, 2012 #4

    haruspex

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    If you write z = x + iy = r e then z2 = u+iv = r2e2iθ. Thus the transformation from the XY plane to the UV plane fans out the first quadrant, doubling all the polar angles, to cover the first two quadrants.
     
  6. Dec 23, 2012 #5
    I'm not sure what you mean...Usually to find out the limits we consider the boundary curves and work out its shape in the R' region... using the geometric approach.

    But in this case i'm not sure what geometric approach should be taken.
     
  7. Dec 23, 2012 #6

    vela

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    What does the u-axis correspond to in the xy-plane?
     
  8. Dec 23, 2012 #7
    u = x2 - y2 = (x+y)(x-y)

    I would love to say it's the distance from origin squared (x2 + y2) but sadly it's not...

    I have no idea.
     
  9. Dec 23, 2012 #8

    vela

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    Try again. Note that the u-axis is v=0, just like the x-axis is y=0.
     
  10. Dec 23, 2012 #9

    haruspex

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    I'm saying that if you think of both the XY plane and the UV plane as being complex planes, the geometry of the mapping becomes clear. (x+iy)2 = u+iv. A radial line from the origin, polar angle theta say, maps to a radial line from the origin at angle 2 theta. (Radial distances become squared.) Thus the 1st quadrant maps to the upper half plane (v≥0).
     
  11. Dec 24, 2012 #10
    First, we have to assume that u and v are orthogonal...let's say it is.

    For v = 0, u = x2 or -y2

    For u = 0, x = y and v = 2x2 or 2y2

    doesn't this make the limits of v[0,∞] and u[0,-∞] or u[0,∞] which are the cases i mentioned earlier....Summing them would give a result of 0.
     
  12. Dec 24, 2012 #11

    haruspex

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    I think your problem is that you're trying to map x and y ranges individually, in some sense, to u and v ranges. That's the only way I can understand how you get ranges like "u[0,-∞] or u[0,∞]". It doesn't work like that. You need to determine the set of points (u,v) that are in the region, then express that as ranges for integration purposes. The region is v >= 0, which you can express as v range [0,∞) and u range (-∞,+∞).
     
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