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Change of variables

  1. Nov 14, 2009 #1
    Let R be the elliptical regin in the first quadrant by x^2/3^2+y^2/2^2=1. x=3u, y=2v, evaluate the area of R.

    How do I setup the double integral? By this I mean what do I put in for f(g(u,v),h(u,v))jacobian dudv?

    I know the bounds of u and v and the Jacobian but I not sure what to do for g and h. Would that be just pluging 3u and 2v into x^2 and y^2, respectively? But what happens to the 1 in the eqaution?

    So by solving the (u,v) cordinates, one comes up with (0,0); (1,0); and (0,1).
    The Jacobian is 6.
     
  2. jcsd
  3. Nov 14, 2009 #2

    LCKurtz

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    When you substitute your u's and v's for the x's and y's of the equation, for your region A in the xy plane you get a new region B in the uv plane. The point of the substitution is that instead of integrating

    [tex]Area = \int \int_A 1\ dydx[/tex]

    over the region A in the xy plane, which is "ugly", you can instead calculate

    [tex]Area = \int \int_B 1 J\ dudv[/tex]

    using u-v limits in the uv plane. If your new region B is nice, you may easily work out the new integral, or even avoid working it at all.
     
  4. Nov 14, 2009 #3
    So the integral is just 0<=u<=1 and 0<=v<=-u+1 dvdu with no equation in the the integral?
     
  5. Nov 14, 2009 #4

    LCKurtz

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    No. You need to draw a graph of your uv equation in the uv plane and use it to figure out correct limits.
     
  6. Nov 14, 2009 #5
    The drawing is a triangle and those are the bounds. I am not concerned about them. I am trying to figure out what f(g(u,v),h(u,v)) are inside the integrals. Is it 1? Is it u+v not considering the 1 or is it u+v-1?
     
    Last edited: Nov 14, 2009
  7. Nov 14, 2009 #6

    LCKurtz

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    Are you trying to tell me that when you substitute x=3u, y=2v into the equation:

    [tex]\frac {x^2}{3^2}+\frac {y^2}{2^2}=1[/tex]

    that the graph you get in the uv plane is a triangle???

    Your integrand, once you figure out the proper area, will be 1 times the Jacobian.
     
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