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Math Help - unsolvable in a normal way integral mmn 15 5

  1. #1
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    unsolvable in a normal way integral mmn 15 5

    how to use the clue that f(x,y)=-f(x,y)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Express

    \iint_Rf(x,y)\;dxdy=\iint_{R_1}f(x,y)\;dxdy+\iint_  {R_2}f(x,y)\;dxdy

    where

    R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    Express

    \iint_Rf(x,y)\;dxdy=\iint_{R_1}f(x,y)\;dxdy+\iint_  {R_2}f(x,y)\;dxdy

    where

    R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}
    how to write R1 and R2
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by transgalactic View Post
    how to write R1 and R2

    I don't understand.
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  5. #5
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    i dont know how to continue with what you hve said
    dont know how to write R1 and R2 integrals
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Using the substitution x=v,\;y=u the corresponding Jacobian is \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1 so,

    \iint_{R_1}f(x,y)\;dxdy=\iint_{R_2}f(v,u)|-1|\;dudv=

    -\iint_{R_2}f(u,v)\;dudv=-\iint_{R_2}f(x,y)\;dxdy
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  7. #7
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    i cant understand this expression \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1

    i know for example that:
    \frac{\partial f(x,y)}{\partial x}
    means the partial derivetive of "f" by x

    but here the is no clear structure of what is the function
    and by what variable we derive it by
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  8. #8
    Behold, the power of SARDINES!
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    Quote Originally Posted by transgalactic View Post
    i cant understand this expression \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1

    i know for example that:
    \frac{\partial f(x,y)}{\partial x}
    means the partial derivetive by x

    but here the is no clear structure of what is the function
    and by what variable we derive it by
    This is the Jacobian matrix

    Jacobian matrix and determinant - Wikipedia, the free encyclopedia
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  9. #9
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    i saw the link and i know what is jacobian
    but this is not a determinant
    its a fracture of some undefined stuff
    and in the link i didnt see it
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  10. #10
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    how this determinant of what functions??
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by transgalactic View Post
    how this determinant of what functions??

    You should study the double integrals corresponding theory. It is almost useless to select parts of a theory with the only aim of solving a specific problem.
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  12. #12
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    what this \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1
    means
    what function what determinants
    ?
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  13. #13
    MHF Contributor FernandoRevilla's Avatar
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    \det \dfrac{{\partial (x,y)}}{{\partial (u,v)}}=\det \begin{bmatrix}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{bmatrix}=\det \begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}=...=-1
    Last edited by FernandoRevilla; May 26th 2011 at 12:35 AM.
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  14. #14
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    Quote Originally Posted by FernandoRevilla View Post
    Using the substitution x=v,\;y=u the corresponding Jacobian is \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1 so,

    \iint_{R_1}f(x,y)\;dxdy=\iint_{R_2}f(v,u)|-1|\;dudv=

    -\iint_{R_2}f(u,v)\;dudv=-\iint_{R_2}f(x,y)\;dxdy
    on what basis you change the area from R1 to R2
    in the integral
    ??
    and in the end why you go back to x y coordinates withoud multiplying by the yacobian
    ?

    and you didnt use the riven R
    its between y^2=x and x^2=y
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  15. #15
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by transgalactic View Post
    on what basis you change the area from R1 to R2 in the integral??

    The domain R is symmetric with respect to y=x .


    and in the end why you go back to x y coordinates withoud multiplying by the yacobian

    Try to find the Jacobian of u=x,v=y .


    and you didnt use the riven R its between y^2=x and x^2=y

    If you mean the given R , yes I did : R=R_1\cup R_2 .
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