# $dxdy=rdrd\theta$

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?

$$drdr$$ 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.

• I suggest mentioning that $drd\theta$ is actually $dr \wedge d\theta$, a wedge product of differential forms: en.wikipedia.org/wiki/Differential_form Jun 20, 2019 at 12:43
• @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.
– J.G.
Jun 20, 2019 at 12:44
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 $$\theta$$