Is there an algebraic derivation of the area element in polar coordinates?

[SamRoss]

TL;DR

There is a simple geometric derivation of the area element in polar coordinates. Is there an algebraic derivation as well?

There is a simple geometric derivation of the area element $r dr d\theta$ in polar coordinates such as in the following link: http://citadel.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html

Is there an algebraic derivation as well beginning with Cartesian coordinates and using $x=rcos\theta$ and $y=rsin\theta$ to transform the Cartesian area element $dx dy$?

 

[WWGD]

There is a simple geometric derivation of the area element $r dr d\theta$ in polar coordinates such as in the following link: http://citadel.sjfc.edu/faculty/kgreen/vector/Block3/jacob/node4.html

Is there an algebraic derivation as well beginning with Cartesian coordinates and using $x=rcos\theta$ and $y=rsin\theta$ to transform the Cartesian area element $dx dy$?

I am not sure what you mean by algebraic derivation, this is usually called a change of coordinates. In this case, between polars and Cartesian.

 

[SamRoss]

I am not sure what you mean by algebraic derivation, this is usually called a change of coordinates. In this case, between polars and Cartesian.

Yes, that is what I'm looking for.

 

[WWGD]

The setup ( generalized to n variables) is usually of the sort:
$x=f(a,b),\text{ so } dx=f_a(a,b)da$
$y=g(a,b)\text , \text{ so } dy=f_b(a,b)db$
Where $f_a, f_b$ are the partials with respect to $a,b$. In this case we use $x=rcos\theta$, etc.

Is that helpful?

 

[pasmith]

The setup ( generalized to n variables) is usually of the sort:
$x=f(a,b),\text{ so } dx=f_a(a,b)da$
$y=g(a,b)\text , \text{ so } dy=f_b(a,b)db$
Where $f_a, f_b$ are the partials with respect to $a,b$. In this case we use $x=rcos\theta$, etc.

Is that helpful?

<br /> dx = f_a(a,b)\,da + f_b(a,b)\,db etc...

EDIT: Just expanding dx and dy in the new coordinates and multiplying those expressions will not yield the correct answer; instead you must calculate the exterior product dx \wedge dy.

 

[WWGD]

<br /> dx = f_a(a,b)\,da + f_b(a,b)\,db etc...

EDIT: Just expanding dx and dy in the new coordinates and multiplying those expressions will not yield the correct answer; instead you must calculate the exterior product dx \wedge dy.

OP is about change of coordinates, not about finding the total derivative.

 

[pasmith]

OP is about change of coordinates, not about finding the total derivative.

The OP wants an algebraic derivation of the expression for the area element, and the relvant algebra is that of 2-forms. The derivation is then to compute dx \wedge dy in terms of da \wedge db, for which purpose dx must be expressed as dx = f_a\,da + f_b\,db and similarly for dy.

Alternatively the OP could start from dx\,dy = \left \| \frac{\partial \mathbf{r}}{\partial a} \times<br /> \frac{\partial \mathbf{r}}{\partial b}\right\| da\,db.

In any event, I'm not sure how the OP is supposed to get from dx = f_a\,da and dy = f_b\,db to dx\,dy = |f_a g_b - f_b g_a|da\,db.

 

[WWGD]

I referred specifically to the change of coordinates, not to the area element. You expect an answer for a question I did not address. I take things step by step, and start with a change of coordinates. As stated, this is correct. Yes, you did complete the needed steps; I was doing it more gradually.Edit. But, yes, this is correct, the new area is scaled by the norm of the determinant of the change of variables.

 

[WWGD]

So, yes, to clarify, what pasmith says is correct. The change of area element is given by dadb=dxdy|J(x,y)| , where J(x,y) is the Jacobian of the change of variables matrix.

 

[romsofia]

$x=r\cos\theta, y=r\sin\theta \rightarrow dx=\cos \theta dr + -r\sin\theta d\theta, dy=\sin\theta dr + r\cos\theta d\theta$

thus, $dx \wedge dy = \cos \theta dr + -r\sin\theta d\theta \wedge sin\theta dr + r\cos\theta d\theta \rightarrow \cos \theta dr \wedge sin\theta dr + \cos \theta dr \wedge r\cos\theta d\theta + -r\sin\theta d\theta \wedge \sin\theta dr + -r\sin\theta d\theta \wedge r\cos\theta d\theta \rightarrow r \cos^2 \theta dr \wedge d\theta + -r \sin^2 \theta d\theta \wedge dr \rightarrow r \cos^2 \theta dr \wedge d\theta + r \sin^2 \theta dr \wedge d\theta$

Which finally gives us: $(\cos^2 \theta + \sin^2 \theta) r dr\wedge d\theta \rightarrow r dr\wedge d\theta \rightarrow r drd\theta$

You may wonder why some terms went to zero and how i can switch the signs, and that would be due to some properties of exterior derivatives (and wedge products), which i suggest you look into. Hopefully this is what you were looking for.

 

[WWGD]

I believed s/he wanted to go through it gradually , so I first went over the change of variables, but I guess we have to wait and see if they come back and tell us what they wanted.

 

[SamRoss]

Hi everybody. Thanks for all the help and sorry it took me so long to reply back. I had attempted the problem using a change of coordinates and was unable to get it to work. Thank you WWGD for bringing up the Jacobian. I had not thought to use that. Thank you Pasmith for bringing up the exterior product. That is a concept I was unfamiliar with and I will be sure to read up on it. And thank you Romsofia for the full solution. I appreciate everyone's assistance with this.