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Math Help - Jacobian help

  1. #1
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    Jacobian help

    My double integral is (x+y)
    where R is the region on the xy plane bounded by the lines
    x+y=3, x-y=3, and the hyperbola x^2-y^2=3
    (u=x^2 - y^2....v=x+y)

    I calculated the jacobian d(u,v)/d(x,y) determinant to be 2(x+y)=2v
    and the reciprical = 1/2v

    I dont understand what the new integral is and its limits?
    my book gives like one example then goes to the next section
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  2. #2
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    so x=v/y implies v goes from -3 to 3 right
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  3. #3
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    You have to find the functions x=g(u,v) and y=h(u,v) by
    solving the equations for u and v and then plug those back into the function you're integrating, which is f(x,y)=x+y

    u=x^2-y^2=(x+y)(x-y)=v(x-y)=vx-vy

    u=vx-vy
    v=x+y

    This gives x=\frac{u+v^2}{2v} and y=\frac{v^2-u}{2v}

    So now the integral will take the form

    \int_S\int f(\frac{u+v^2}{2v},\frac{v^2-u}{2v})\mid \frac{\partial (x,y)}{\partial (u,v)}\mid dudv
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  4. #4
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    I calculated the jacobian d(u,v)/d(x,y) determinant to be 2(x+y)=2v
    and the reciprical = 1/2v
    I don't see what you did here. The defintion of the Jacobian is:

    \mid \frac{\partial (x,y)}{\partial (u,v)}\mid =\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}
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  5. #5
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    partial derivatives of u wrt x,y
    partial derivaitves of v wrt x,y
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  6. #6
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    Quote Originally Posted by begin View Post
    partial derivatives of u wrt x,y
    partial derivaitves of v wrt x,y
    I think you're confused by the notation. The defintion in terms of a determinant is this:

    \frac{\partial (x,y)}{\partial (u,v)}=\left( \begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right)
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  7. #7
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    yes and the reciprical =d(u,v)/d(x,y) which is easier to compute
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  8. #8
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    I calculated it to \frac{1}{2v}, which is what you obtained, but I don't know what your method was. I'm not familiar with it...
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  9. #9
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    yea thats what i got..thanks for the help so far...i got the limits for v=-3,3 and u=0-sqrt(3)
    is that right?
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  10. #10
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    Quote Originally Posted by adkinsjr View Post
    I calculated it to \frac{1}{2v}, which is what you obtained, but I don't know what your method was. I'm not familiar with it...
    i think if you do d(u,v)/(x,y)=1/d(x,y)/(u,v)
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  11. #11
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    Find your bounds before transformation and then convert those using the selected transformation. You can follow this example here: Jacobians
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