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Domain of parametric plane

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data
    A triangle is defined by the 3 points P=(1,0,0), Q=(0,2,0) and R=(0,0,2).

    Set up the double integral over its area.

    3. The attempt at a solution

    The triangle can be described as a plane 2x+y+z+2=0, with xE [0,1], yE [0,2], zE [0,2].

    I parametrized it into r(u,v) =u*PQ+v*PR+P = u[-1,2,0]+v[-1,0,2]+(1,0,0).

    But what's the domain of r(u,v)? I think it is uE [0,1-v], vE[0,1], but even if it is I found it using only gut feeling.
     
  2. jcsd
  3. May 13, 2013 #2

    LCKurtz

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    What you need to do is draw a ##uv## coordinate system. Then mark which ##uv## points are mapped to which points on your plane. For example ##(u,v)=(0,0)## goes to ##(1,0,0)##. Mark also where ##(u,v) = (1,0)## and ##(u,v)=(0,1)## go. This will give you a picture of your ##uv## domain and you will be able to write your double integral limits for ##u## and ##v##.
     
  4. May 13, 2013 #3
    Could you explain the general strategy in further detail? Why did you pick (u,v) = (0,0), (1,0) and (0,1), for instance? Is it because those are the minimal and maximal values of u and v?

    I tried to draw the u,v plane, but I wasn't sure what to do...
     
  5. May 13, 2013 #4

    LCKurtz

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    Just draw two perpendicular axes and label them u and v instead of x and y.

    If you look at your parameterization, r(u,v) =u*PQ+v*PR+P = u[-1,2,0]+v[-1,0,2]+(1,0,0) it is obvious that that those values of u and v give the corners of your triangle. Joining those the points (u,v) = (0,0), (1,0) and (0,1) in the uv plane gives you a picture of the uv domain.

    It is just like as if you had written the xyz equation of the plane as z = 2-2x-y and used x and y as your parameters. Then you would have used the domain triangle in the xy plane for your xy limits.
     
  6. May 13, 2013 #5
    hmm, OK. Thanks
     
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