3
$\begingroup$

I'm trying to show that $dx\,dy=r\,dr\,d\theta$ using differentials.

$x=r\cos(\theta)$ and $y=r\sin(\theta)$ thus $dx=\cos(\theta)dr-r\sin(\theta)d\theta$ and $dy=\sin(\theta)dr+r\cos(\theta)d\theta$

$\begin{align}dx\,dy&=(\cos(\theta)dr-r\sin(\theta)d\theta)(\sin(\theta)dr+r\cos(\theta)d\theta)\\& =\cos(\theta)\sin(\theta) dr^2+r\cos^2(\theta)drd\theta-r\sin^2(\theta)d\theta dr-r^2\cos(\theta)\sin(\theta)d\theta^2\\&=\cos(\theta)\sin(\theta) dr^2-r^2\cos(\theta)\sin(\theta)d\theta^2+rdrd\theta(1-\sin(\theta)^2-\sin(\theta)^2)\end{align}$

If my calculations are correct, $\cos(\theta)\sin(\theta) dr^2-r^2\cos(\theta)\sin(\theta)d\theta^2-2 \sin(\theta)^2rdrd\theta=0$ but how am I suppose to show that?

$\endgroup$
1

2 Answers 2

8
$\begingroup$

$drdr$ and $d\theta d\theta$ are both zero, so forget about them. Also, $drd\theta=-d\theta dr,$ and you did not account for this at one point. If you make these changes, you'll have $$dxdy=r(\sin^2\theta+\cos^2\theta)drd\theta=rdrd\theta.$$

I'm abusing notation a bit, as $drd\theta$ should actually be written as $dr\wedge d\theta$, where $\wedge$ denotes the wedge product of the differential forms $dr$ and $d\theta$ (see https://en.wikipedia.org/wiki/Differential_form), which is alternating by construction, so the wedge of the same form twice gives $0$. See https://en.wikipedia.org/wiki/Exterior_algebra#Formal_definitions_and_algebraic_properties for more details.

$\endgroup$
3
  • 1
    $\begingroup$ I suggest mentioning that $drd\theta$ is actually $dr \wedge d\theta$, a wedge product of differential forms: en.wikipedia.org/wiki/Differential_form $\endgroup$
    – lisyarus
    Commented Jun 20, 2019 at 12:43
  • 1
    $\begingroup$ @lisyarus Seconded; I was about to post that link.One could argue the example in this question is a good gentle introduction to this topic. $\endgroup$
    – J.G.
    Commented Jun 20, 2019 at 12:44
  • $\begingroup$ Agreed. I was on the fence about it, but both of your comments pushed me over; I've added some references in. $\endgroup$
    – cmk
    Commented Jun 20, 2019 at 12:46
2
$\begingroup$

You are missing the important point on the difference between the partition elements in Cartesian and Polar systems. While $dxdy$ is the area of a rectangle $rdrd\theta $ is the area of the curved section between circles of radii $r$ and $r+dr$ and the central angle of $d\theta$

The so called Jacobian gives you the multiplier of your transformation. The Jacobian is the determinant of a matrix whose terms are partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .