The Jacobian of $(x,y)mapsto (x+y^2,y+x^2)$ under the substitution $u=x+y^2$ and $v=y+x^2$.












-1












$begingroup$


I am given the map $(x,y)mapsto (x+y^2,y+x^2)$. I am unable to find the Jacobian by making the substitution $u=x+y^2$ and $v=y+x^2$. Any hints would be appreciated.



(I am trying to find whether the map is area preserving? I know "the map $f:mathbb{R}^ntomathbb{R}^n$ is area and orientation preserving iff the determinant of the Jacobian is $pm1$".)










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$endgroup$








  • 1




    $begingroup$
    What is the definition of Jacobian?
    $endgroup$
    – zoidberg
    Dec 9 '18 at 2:39
















-1












$begingroup$


I am given the map $(x,y)mapsto (x+y^2,y+x^2)$. I am unable to find the Jacobian by making the substitution $u=x+y^2$ and $v=y+x^2$. Any hints would be appreciated.



(I am trying to find whether the map is area preserving? I know "the map $f:mathbb{R}^ntomathbb{R}^n$ is area and orientation preserving iff the determinant of the Jacobian is $pm1$".)










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the definition of Jacobian?
    $endgroup$
    – zoidberg
    Dec 9 '18 at 2:39














-1












-1








-1





$begingroup$


I am given the map $(x,y)mapsto (x+y^2,y+x^2)$. I am unable to find the Jacobian by making the substitution $u=x+y^2$ and $v=y+x^2$. Any hints would be appreciated.



(I am trying to find whether the map is area preserving? I know "the map $f:mathbb{R}^ntomathbb{R}^n$ is area and orientation preserving iff the determinant of the Jacobian is $pm1$".)










share|cite|improve this question











$endgroup$




I am given the map $(x,y)mapsto (x+y^2,y+x^2)$. I am unable to find the Jacobian by making the substitution $u=x+y^2$ and $v=y+x^2$. Any hints would be appreciated.



(I am trying to find whether the map is area preserving? I know "the map $f:mathbb{R}^ntomathbb{R}^n$ is area and orientation preserving iff the determinant of the Jacobian is $pm1$".)







multivariable-calculus transformation area






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edited Dec 9 '18 at 2:32









Shaun

9,570113684




9,570113684










asked Dec 9 '18 at 1:56









Yadati KiranYadati Kiran

2,1061621




2,1061621








  • 1




    $begingroup$
    What is the definition of Jacobian?
    $endgroup$
    – zoidberg
    Dec 9 '18 at 2:39














  • 1




    $begingroup$
    What is the definition of Jacobian?
    $endgroup$
    – zoidberg
    Dec 9 '18 at 2:39








1




1




$begingroup$
What is the definition of Jacobian?
$endgroup$
– zoidberg
Dec 9 '18 at 2:39




$begingroup$
What is the definition of Jacobian?
$endgroup$
– zoidberg
Dec 9 '18 at 2:39










1 Answer
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$begingroup$

The Jacobian of a multivariable function $f:mathbb{R}^2 rightarrow mathbb{R}^2$ such that $f(x,y)=(u(x,y),v(x,y))$ is:



$$textbf{J}(f(x,y)) = detbegin{pmatrix}
frac{partial u}{partial x} & frac{partial u}{partial y} \
frac{partial v}{partial x} & frac{partial v}{partial y} \
end{pmatrix}$$



where $det$ denotes the determinant of the matrix.



Calculating the partial derivatives of $u(x,y)=x+y^2$ and $v(x,y)=y+x^2$ we have that the Jacobian of your given function is:



$$detbegin{pmatrix}
1 & 2y \
2x & 1 \
end{pmatrix}=1-4xy$$






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    $begingroup$

    The Jacobian of a multivariable function $f:mathbb{R}^2 rightarrow mathbb{R}^2$ such that $f(x,y)=(u(x,y),v(x,y))$ is:



    $$textbf{J}(f(x,y)) = detbegin{pmatrix}
    frac{partial u}{partial x} & frac{partial u}{partial y} \
    frac{partial v}{partial x} & frac{partial v}{partial y} \
    end{pmatrix}$$



    where $det$ denotes the determinant of the matrix.



    Calculating the partial derivatives of $u(x,y)=x+y^2$ and $v(x,y)=y+x^2$ we have that the Jacobian of your given function is:



    $$detbegin{pmatrix}
    1 & 2y \
    2x & 1 \
    end{pmatrix}=1-4xy$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The Jacobian of a multivariable function $f:mathbb{R}^2 rightarrow mathbb{R}^2$ such that $f(x,y)=(u(x,y),v(x,y))$ is:



      $$textbf{J}(f(x,y)) = detbegin{pmatrix}
      frac{partial u}{partial x} & frac{partial u}{partial y} \
      frac{partial v}{partial x} & frac{partial v}{partial y} \
      end{pmatrix}$$



      where $det$ denotes the determinant of the matrix.



      Calculating the partial derivatives of $u(x,y)=x+y^2$ and $v(x,y)=y+x^2$ we have that the Jacobian of your given function is:



      $$detbegin{pmatrix}
      1 & 2y \
      2x & 1 \
      end{pmatrix}=1-4xy$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The Jacobian of a multivariable function $f:mathbb{R}^2 rightarrow mathbb{R}^2$ such that $f(x,y)=(u(x,y),v(x,y))$ is:



        $$textbf{J}(f(x,y)) = detbegin{pmatrix}
        frac{partial u}{partial x} & frac{partial u}{partial y} \
        frac{partial v}{partial x} & frac{partial v}{partial y} \
        end{pmatrix}$$



        where $det$ denotes the determinant of the matrix.



        Calculating the partial derivatives of $u(x,y)=x+y^2$ and $v(x,y)=y+x^2$ we have that the Jacobian of your given function is:



        $$detbegin{pmatrix}
        1 & 2y \
        2x & 1 \
        end{pmatrix}=1-4xy$$






        share|cite|improve this answer











        $endgroup$



        The Jacobian of a multivariable function $f:mathbb{R}^2 rightarrow mathbb{R}^2$ such that $f(x,y)=(u(x,y),v(x,y))$ is:



        $$textbf{J}(f(x,y)) = detbegin{pmatrix}
        frac{partial u}{partial x} & frac{partial u}{partial y} \
        frac{partial v}{partial x} & frac{partial v}{partial y} \
        end{pmatrix}$$



        where $det$ denotes the determinant of the matrix.



        Calculating the partial derivatives of $u(x,y)=x+y^2$ and $v(x,y)=y+x^2$ we have that the Jacobian of your given function is:



        $$detbegin{pmatrix}
        1 & 2y \
        2x & 1 \
        end{pmatrix}=1-4xy$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 3:45

























        answered Dec 9 '18 at 3:20









        MAXMAX

        19218




        19218






























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