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

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• SamRoss
The derivation is 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\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 - ff

#### SamRoss

Gold Member
TL;DR Summary
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##?

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.

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.

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.

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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.

$$dx = f_a(a,b)\,da + f_b(a,b)\,db$$ 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 $dx \wedge dy$.

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• Cryo
$$dx = f_a(a,b)\,da + f_b(a,b)\,db$$ 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.

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 $$dx\,dy = \left \| \frac{\partial \mathbf{r}}{\partial a} \times \frac{\partial \mathbf{r}}{\partial b}\right\| da\,db.$$

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 $dx\,dy = |f_a g_b - f_b g_a|da\,db$.

• fresh_42
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.

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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.

##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.

• fresh_42
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.

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.

• WWGD