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1

$begingroup$

Let $Ssubset mathbb R^2$. If $S$ has the area $dxdy$ in $(x,y)$, then it will have the area $$|det(x(u,v),y(u,v))|dudv$$ in $(u,v)$.

We commonly write $$dxdy=|det(x(u,v),y(u,v)|dudv.$$

I'm not really sure how to interpret it. Would it be the area of $S$ in $(u,v)$ ? But in this case,
$$|S|=iint_Sdxdy=iint_S|det(x(u,v),y(u,v))|dudv,$$
so $dS=dxdy=|det(x(u,v),y(u,v)|dudv$ ? But what does it really mean ?

real-analysis integration multivariable-calculus definite-integrals jacobian

share|cite|improve this question

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

$endgroup$

add a comment | 

1

$begingroup$

Let $Ssubset mathbb R^2$. If $S$ has the area $dxdy$ in $(x,y)$, then it will have the area $$|det(x(u,v),y(u,v))|dudv$$ in $(u,v)$.

We commonly write $$dxdy=|det(x(u,v),y(u,v)|dudv.$$

I'm not really sure how to interpret it. Would it be the area of $S$ in $(u,v)$ ? But in this case,
$$|S|=iint_Sdxdy=iint_S|det(x(u,v),y(u,v))|dudv,$$
so $dS=dxdy=|det(x(u,v),y(u,v)|dudv$ ? But what does it really mean ?

real-analysis integration multivariable-calculus definite-integrals jacobian

share|cite|improve this question

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

$endgroup$

add a comment | 

$begingroup$

Let $Ssubset mathbb R^2$. If $S$ has the area $dxdy$ in $(x,y)$, then it will have the area $$|det(x(u,v),y(u,v))|dudv$$ in $(u,v)$.

We commonly write $$dxdy=|det(x(u,v),y(u,v)|dudv.$$

I'm not really sure how to interpret it. Would it be the area of $S$ in $(u,v)$ ? But in this case,
$$|S|=iint_Sdxdy=iint_S|det(x(u,v),y(u,v))|dudv,$$
so $dS=dxdy=|det(x(u,v),y(u,v)|dudv$ ? But what does it really mean ?

real-analysis integration multivariable-calculus definite-integrals jacobian

share|cite|improve this question

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

share|cite|improve this question

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

edited Mar 31 at 7:32

Rodrigo de Azevedo

13.2k41962

asked Mar 29 at 12:03

user657324user657324

59510

asked Mar 29 at 12:03

user657324user657324

59510

1 Answer
1

active

oldest

votes

1

$begingroup$

You have in the $(x,y)$-plane the standard area measure $rm d(x,y)$ and similarly in the "auxiliar" $(u,v)$-plane the standard area measure $rm d(u,v)$. When you are given a (maybe complicated) domain $S$ in the $(x,y)$-plane and want to compute its area then you often use an essentially 1:1 parametrization of $S$ from an auxiliar domain $hat S$ in the $(u,v)$-plane:
$$psi:quad hat Sto S,qquad (u,v)mapstobigl(x(u,v),y(u,v)bigr) .$$
Such a parametrization will in general not be area conserving. In fact an arbitrary "area element" centered at some point $(u,v)inhat S$ will be mapped to a smaller or larger area element centered at the point $bigl(x(u,v),y(u,v)bigr)in S$. The local area scaling factor turns out to be
$$|J_psi(u,v)|=bigl|rm det(dpsi(u,v))bigr| .$$
This is often written as
$$rm d(x,y)=bigl|rm det(dpsi(u,v))bigr|>rm d(u,v)$$
and appears in the integral as
$$rm area(S)=int_Srm d(x,y)=int_hat Sbigl|rm det(dpsi(u,v))bigr|>rm d(u,v) .$$
Note that I have just listed the usual formulas, I have proven nothing.

share|cite|improve this answer

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So, in some sense, it's the local area variation of an area element when we pass from $(u,v)$ to $(x,y)$, right ? (or when we pass from $(u,v)$ to $(x,y)$ ?)
  $endgroup$
  – user657324
  Mar 31 at 9:17
  

  
  
  
  

add a comment | 

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1

$begingroup$

You have in the $(x,y)$-plane the standard area measure $rm d(x,y)$ and similarly in the "auxiliar" $(u,v)$-plane the standard area measure $rm d(u,v)$. When you are given a (maybe complicated) domain $S$ in the $(x,y)$-plane and want to compute its area then you often use an essentially 1:1 parametrization of $S$ from an auxiliar domain $hat S$ in the $(u,v)$-plane:
$$psi:quad hat Sto S,qquad (u,v)mapstobigl(x(u,v),y(u,v)bigr) .$$
Such a parametrization will in general not be area conserving. In fact an arbitrary "area element" centered at some point $(u,v)inhat S$ will be mapped to a smaller or larger area element centered at the point $bigl(x(u,v),y(u,v)bigr)in S$. The local area scaling factor turns out to be
$$|J_psi(u,v)|=bigl|rm det(dpsi(u,v))bigr| .$$
This is often written as
$$rm d(x,y)=bigl|rm det(dpsi(u,v))bigr|>rm d(u,v)$$
and appears in the integral as
$$rm area(S)=int_Srm d(x,y)=int_hat Sbigl|rm det(dpsi(u,v))bigr|>rm d(u,v) .$$
Note that I have just listed the usual formulas, I have proven nothing.

share|cite|improve this answer

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So, in some sense, it's the local area variation of an area element when we pass from $(u,v)$ to $(x,y)$, right ? (or when we pass from $(u,v)$ to $(x,y)$ ?)
  $endgroup$
  – user657324
  Mar 31 at 9:17
  

  
  
  
  

add a comment | 

1

$begingroup$

You have in the $(x,y)$-plane the standard area measure $rm d(x,y)$ and similarly in the "auxiliar" $(u,v)$-plane the standard area measure $rm d(u,v)$. When you are given a (maybe complicated) domain $S$ in the $(x,y)$-plane and want to compute its area then you often use an essentially 1:1 parametrization of $S$ from an auxiliar domain $hat S$ in the $(u,v)$-plane:
$$psi:quad hat Sto S,qquad (u,v)mapstobigl(x(u,v),y(u,v)bigr) .$$
Such a parametrization will in general not be area conserving. In fact an arbitrary "area element" centered at some point $(u,v)inhat S$ will be mapped to a smaller or larger area element centered at the point $bigl(x(u,v),y(u,v)bigr)in S$. The local area scaling factor turns out to be
$$|J_psi(u,v)|=bigl|rm det(dpsi(u,v))bigr| .$$
This is often written as
$$rm d(x,y)=bigl|rm det(dpsi(u,v))bigr|>rm d(u,v)$$
and appears in the integral as
$$rm area(S)=int_Srm d(x,y)=int_hat Sbigl|rm det(dpsi(u,v))bigr|>rm d(u,v) .$$
Note that I have just listed the usual formulas, I have proven nothing.

share|cite|improve this answer

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So, in some sense, it's the local area variation of an area element when we pass from $(u,v)$ to $(x,y)$, right ? (or when we pass from $(u,v)$ to $(x,y)$ ?)
  $endgroup$
  – user657324
  Mar 31 at 9:17
  

  
  
  
  

add a comment | 

$begingroup$

You have in the $(x,y)$-plane the standard area measure $rm d(x,y)$ and similarly in the "auxiliar" $(u,v)$-plane the standard area measure $rm d(u,v)$. When you are given a (maybe complicated) domain $S$ in the $(x,y)$-plane and want to compute its area then you often use an essentially 1:1 parametrization of $S$ from an auxiliar domain $hat S$ in the $(u,v)$-plane:
$$psi:quad hat Sto S,qquad (u,v)mapstobigl(x(u,v),y(u,v)bigr) .$$
Such a parametrization will in general not be area conserving. In fact an arbitrary "area element" centered at some point $(u,v)inhat S$ will be mapped to a smaller or larger area element centered at the point $bigl(x(u,v),y(u,v)bigr)in S$. The local area scaling factor turns out to be
$$|J_psi(u,v)|=bigl|rm det(dpsi(u,v))bigr| .$$
This is often written as
$$rm d(x,y)=bigl|rm det(dpsi(u,v))bigr|>rm d(u,v)$$
and appears in the integral as
$$rm area(S)=int_Srm d(x,y)=int_hat Sbigl|rm det(dpsi(u,v))bigr|>rm d(u,v) .$$
Note that I have just listed the usual formulas, I have proven nothing.

share|cite|improve this answer

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

$endgroup$

share|cite|improve this answer

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328

answered Mar 31 at 8:57

Christian BlatterChristian Blatter

176k8115328