Proving $dx.dy = r.drd\theta$ using x and y

Question

I can prove that $dx dy = r.dr d\theta$ by drawing a circle and calculating the area of a small square in the polar coordinates, but when I try proving it using the equations below, I fail to prove it. What is my mistake?

$x = r\cos(\theta) => dx = \cos(\theta).dr - \sin(\theta)r.d\theta$

$y = r\sin(\theta)=>dy=\sin(\theta).dr + \cos(\theta)r.d\theta$

$=> dxdy = r(\cos^2(\theta)-\sin^2(\theta))drd\theta=r\cos(2\theta).drd\theta$

---

I actually saw a similar question in this site: how to get $dx\; dy=r\;dr\;d\theta$ My problem was that I didn't understand why $drd\theta = -d\theta dr$

Answer 1

You can think about $dxdy$ as a small patch of area that is the parallelogram defined by the vectors $dx$ and $dy$. The key is that the area has a sign; so if we switch the order of $dx$ and $dy$, we get $dydx = -dxdy$. (One way to think about it is that the multiplication operator between $dx$ and $dy$ behaves like a cross product; it is anti-commutative.) Thus $drdr = -drdr$ so $drdr = 0$, similarly $d\theta d\theta = 0$.

Thus $$dxdy = (\cos(\theta)dr-r\sin(\theta)d\theta)(\sin(\theta)dr+r\cos(\theta)d\theta)\\=\cos(\theta)\sin(\theta)drdr-r\sin^2(\theta)d\theta dr+r\cos^2(\theta)drd\theta -r^2\sin(\theta)\cos(\theta)d\theta d\theta\\=0+r\sin^2(\theta)drd\theta +r\cos^2(\theta)drd\theta+0\\=rdrd\theta.$$ (The key is changing the $d\theta dr$ to $-drd\theta$ on the third line.)

Answer 2

Specializing the general formula for a change of coordinates to this coordinate change gives $$dx \,dy = \left\vert\det\pmatrix{\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}}\right\vert dr \,d\theta .$$ Computing the determinant gives the expected result.

Alternatively, if you know a little exterior algebra, you probably think of $dx \,dy$ as $dx \wedge dy$, and so $$dx \wedge dy = d(r \cos \theta) \wedge d(r \sin \theta) = (\cos \theta \,dr - r \sin \theta\,d\theta) \wedge (\sin \theta \,dr + r \cos \theta \,d\theta).$$ In particular, the antisymmetry of $\wedge$ gives $d\theta \wedge dr = -dr \wedge d\theta$.

Answer 3

If you want to do things algebraically then you'll have to resort to the exterior product:

$$\Bbb d x \wedge \Bbb d y = (\cos \theta \ \Bbb d r - r \sin \theta \ \Bbb d \theta) \wedge (\sin \theta \ \Bbb d r + r \cos \theta \ \Bbb d \theta) = \\ \cos \theta \sin \theta \ \Bbb d r \wedge \Bbb d r + r \cos^2 \theta \ \Bbb d r \wedge \Bbb d \theta - r \sin^2 \theta \ \Bbb d \theta \wedge \Bbb d r - r^2 \sin \theta \cos \theta \ \Bbb d \theta \wedge \Bbb d \theta$$

and, since $\Bbb d r \wedge \Bbb d r = \Bbb d \theta \wedge \Bbb d \theta = 0$ and $\Bbb d r \wedge \Bbb d \theta = - \Bbb d \theta \wedge \Bbb d r$ because the exterior product is graded-commutative (in particular anti-commutative for $1$-forms), we get

$$\Bbb d x \wedge \Bbb d y = r \ \Bbb d r \wedge \Bbb d \theta .$$