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Jacobian transformation

  1. Apr 15, 2014 #1
    jacobian.png 1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    Umm what just happened?
    I understand as far as u=x+y and v = y/x and when he does the 2d curl. What I don't get is the step thereafter when he flips it. How does he know to flip it? Further, when he flips it wouldn't that make the dvdu inside the integral cancel and hence leave him with dxdy?
     
  2. jcsd
  3. Apr 15, 2014 #2
    The Jacobian has the property
    [itex]J=\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{ \partial (x,y)}}[/itex]
    so the change of variables can be written as:
    [itex]\int_{S}f(x,y)dxdy=\int_{S'}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{ \partial(u,v)}\right|dudv=\int_{S'}f(x(u,v),y(u,v))\frac{1}{\left|\frac{\partial(u,v)}{ \partial (x,y)}\right|}dudv[/itex]


    This is why you can first write [itex]u=u(x,y); v=v(x,y)[/itex] and then find the inverse of the corresponding Jacobian. Of course, you can easily verify that the result comes out the same if you use their inverses directly, that is, write [itex]x=x(u,v);y=y(u,v)[/itex] and find the corresponding Jacobian, it's just that in this particular example the first method works faster. As you can see, the [itex]dudv[/itex] part remains unaffected either way.
     
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