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












1














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
























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










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
















1














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
























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










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














1












1








1







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















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 at 9:05









Robert Z

93.2k1061132




93.2k1061132










asked Nov 27 at 13:13









MathLover

45710




45710












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










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


















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










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
















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




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












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




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










3 Answers
3






active

oldest

votes


















1














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





















  • Nice explanation (+1).
    – Robert Z
    Nov 28 at 9:03



















0














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





















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












  • 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 $left(frac12,frac12right)$.
    – José Carlos Santos
    Nov 27 at 13:28



















0














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





















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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

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    active

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    1














    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





















    • Nice explanation (+1).
      – Robert Z
      Nov 28 at 9:03
















    1














    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





















    • Nice explanation (+1).
      – Robert Z
      Nov 28 at 9:03














    1












    1








    1






    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












    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 at 15:33









    Christian Blatter

    172k7112325




    172k7112325












    • Nice explanation (+1).
      – Robert Z
      Nov 28 at 9:03


















    • Nice explanation (+1).
      – Robert Z
      Nov 28 at 9:03
















    Nice explanation (+1).
    – Robert Z
    Nov 28 at 9:03




    Nice explanation (+1).
    – Robert Z
    Nov 28 at 9:03











    0














    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





















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












    • 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 $left(frac12,frac12right)$.
      – José Carlos Santos
      Nov 27 at 13:28
















    0














    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





















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












    • 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 $left(frac12,frac12right)$.
      – José Carlos Santos
      Nov 27 at 13:28














    0












    0








    0






    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












    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 at 13:22









    José Carlos Santos

    150k22120221




    150k22120221












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












    • 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 $left(frac12,frac12right)$.
      – José Carlos Santos
      Nov 27 at 13:28


















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












    • 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 $left(frac12,frac12right)$.
      – José Carlos Santos
      Nov 27 at 13:28
















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






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














    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 $left(frac12,frac12right)$.
    – José Carlos Santos
    Nov 27 at 13:28




    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 $left(frac12,frac12right)$.
    – José Carlos Santos
    Nov 27 at 13:28











    0














    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


























      0














      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
























        0












        0








        0






        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












        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 at 10:11









        gandalf61

        7,693623




        7,693623






























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