How to determine that a given map from $mathbb{R}^2 to mathbb{R}^2$ preserves area or not?












1












$begingroup$


I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$



I tried to draw the graph but as function is not linear, my approach could not work.



Can any one suggest me some approach so that I can work this out?






analysis multivariable-calculus







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  • $begingroup$
    Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 27 '18 at 13:21










  • $begingroup$
    No Sir .I had just done only real analysis in that reimann steljes intergral only
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:22
















1












$begingroup$


I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$



I tried to draw the graph but as function is not linear, my approach could not work.



Can any one suggest me some approach so that I can work this out?






analysis multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 27 '18 at 13:21










  • $begingroup$
    No Sir .I had just done only real analysis in that reimann steljes intergral only
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:22














1












1








1





$begingroup$


I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$



I tried to draw the graph but as function is not linear, my approach could not work.



Can any one suggest me some approach so that I can work this out?






analysis multivariable-calculus







share|cite|improve this question











$endgroup$




I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$



I tried to draw the graph but as function is not linear, my approach could not work.



Can any one suggest me some approach so that I can work this out?






analysis multivariable-calculus




analysis multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 9:05









Robert Z

95.8k1065136




95.8k1065136










asked Nov 27 '18 at 13:13









MathLoverMathLover

49710




49710












  • $begingroup$
    Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 27 '18 at 13:21










  • $begingroup$
    No Sir .I had just done only real analysis in that reimann steljes intergral only
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:22


















  • $begingroup$
    Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 27 '18 at 13:21










  • $begingroup$
    No Sir .I had just done only real analysis in that reimann steljes intergral only
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:22
















$begingroup$
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 13:21




$begingroup$
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
$endgroup$
– астон вілла олоф мэллбэрг
Nov 27 '18 at 13:21












$begingroup$
No Sir .I had just done only real analysis in that reimann steljes intergral only
$endgroup$
– MathLover
Nov 27 '18 at 13:22




$begingroup$
No Sir .I had just done only real analysis in that reimann steljes intergral only
$endgroup$
– MathLover
Nov 27 '18 at 13:22










3 Answers
3






active

oldest

votes


















1












$begingroup$

A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice explanation (+1).
    $endgroup$
    – Robert Z
    Nov 28 '18 at 9:03



















0












$begingroup$

If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why Sir? if some function preserve area then that means its derivative also preserve same?
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:24












  • $begingroup$
    Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
    $endgroup$
    – José Carlos Santos
    Nov 27 '18 at 13:28



















0












$begingroup$

Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
    $$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
    possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
    $$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
    In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice explanation (+1).
      $endgroup$
      – Robert Z
      Nov 28 '18 at 9:03
















    1












    $begingroup$

    A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
    $$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
    possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
    $$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
    In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice explanation (+1).
      $endgroup$
      – Robert Z
      Nov 28 '18 at 9:03














    1












    1








    1





    $begingroup$

    A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
    $$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
    possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
    $$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
    In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.






    share|cite|improve this answer









    $endgroup$



    A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
    $$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
    possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
    $$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
    In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 15:33









    Christian BlatterChristian Blatter

    173k7113326




    173k7113326












    • $begingroup$
      Nice explanation (+1).
      $endgroup$
      – Robert Z
      Nov 28 '18 at 9:03


















    • $begingroup$
      Nice explanation (+1).
      $endgroup$
      – Robert Z
      Nov 28 '18 at 9:03
















    $begingroup$
    Nice explanation (+1).
    $endgroup$
    – Robert Z
    Nov 28 '18 at 9:03




    $begingroup$
    Nice explanation (+1).
    $endgroup$
    – Robert Z
    Nov 28 '18 at 9:03











    0












    $begingroup$

    If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why Sir? if some function preserve area then that means its derivative also preserve same?
      $endgroup$
      – MathLover
      Nov 27 '18 at 13:24












    • $begingroup$
      Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
      $endgroup$
      – José Carlos Santos
      Nov 27 '18 at 13:28
















    0












    $begingroup$

    If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why Sir? if some function preserve area then that means its derivative also preserve same?
      $endgroup$
      – MathLover
      Nov 27 '18 at 13:24












    • $begingroup$
      Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
      $endgroup$
      – José Carlos Santos
      Nov 27 '18 at 13:28














    0












    0








    0





    $begingroup$

    If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.






    share|cite|improve this answer









    $endgroup$



    If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 13:22









    José Carlos SantosJosé Carlos Santos

    157k22126227




    157k22126227












    • $begingroup$
      Why Sir? if some function preserve area then that means its derivative also preserve same?
      $endgroup$
      – MathLover
      Nov 27 '18 at 13:24












    • $begingroup$
      Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
      $endgroup$
      – José Carlos Santos
      Nov 27 '18 at 13:28


















    • $begingroup$
      Why Sir? if some function preserve area then that means its derivative also preserve same?
      $endgroup$
      – MathLover
      Nov 27 '18 at 13:24












    • $begingroup$
      Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
      $endgroup$
      – José Carlos Santos
      Nov 27 '18 at 13:28
















    $begingroup$
    Why Sir? if some function preserve area then that means its derivative also preserve same?
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:24






    $begingroup$
    Why Sir? if some function preserve area then that means its derivative also preserve same?
    $endgroup$
    – MathLover
    Nov 27 '18 at 13:24














    $begingroup$
    Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
    $endgroup$
    – José Carlos Santos
    Nov 27 '18 at 13:28




    $begingroup$
    Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
    $endgroup$
    – José Carlos Santos
    Nov 27 '18 at 13:28











    0












    $begingroup$

    Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.






        share|cite|improve this answer









        $endgroup$



        Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 10:11









        gandalf61gandalf61

        8,329625




        8,329625






























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