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Say I have a 2D integral in rectangular coordinates and want to convert to polar. I transform the dx dy starting with:

$$\begin{array}{c} x = r\cos\theta \\ y = r\sin\theta \\ \end{array} $$ So ... $$\begin{array}{c} dx = \cos\theta\,dr - r \sin\theta\,d\theta \\ dy = \sin\theta\,dr + r \cos\theta\,d\theta \\ \end{array} $$ Then expanding out the multiplication ... $$\begin{align} dx\,dy &= \cos\theta\,\sin\theta\,dr^2 - r^2\,\cos\theta\,\sin\theta\,d\theta^2 + r\,\cos^2\theta\,dr\,d\theta - r\,\sin^2\,\theta\,dr\,d\theta \\ &= r\,dr\,d\theta + \cos\theta\,\sin\theta\,(dr^2 - r^2\,d\theta^2) - 2\,r\,\sin^2\theta\,dr\,d\theta \hspace{3em}\text{... uh oh!} \end{align} $$ So there's $r\,dr\,d\theta$ but with some extra terms. Why doesn't this naive and direct approach work out?

Mark S.
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  • I think there's a bit of abuse of notation going on here. Note that when we say something like ${t=x^2}$ so ${dt=2xdx}$, this is really just shorthand for ${\frac{dt}{dx}=2x}$. In your case - you have differentiated with respect to ? Not to mention treating differentials as if they were numbers always leads to some... dodgy things – Riemann'sPointyNose Jul 26 '20 at 15:31
  • depends on how you understand the $dx,dy$ and a thing symbolized by $dx\wedge dy$ called area element or area proyector – janmarqz Jul 26 '20 at 15:31
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    The reason is because $dx,dy$ etc are NOT real numbers, so you cannot treat them like real numbers. See Polar Coordinate Transformation - Motivation or Why is this derivation of the transformation from Cartesian to polar coordinates not correct？. There are also many other answers on this site which address this issue. – peek-a-boo Jul 26 '20 at 15:34
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    Intuitively, $\mathrm{d}x,\mathrm{d}y$ is the area of the rectangle with width $\mathrm{d}x$ and height $\mathrm{d}y$. Now when you apply the change of coordinates $(x,y)=r(\cos\theta,\sin\theta)$, this infinitesimal rectangle will transform into an "almost" parallelogram whose area is no longer need to be the product of $\mathrm{d}r$ and $\mathrm{d}\theta$. The concept of Jacobian (and more generally, wedge product of differential forms) is introduced to address this issue systematically. I would also recommend you to read the above links. – Sangchul Lee Jul 26 '20 at 15:58
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    In your calculation, on the first line, you have two terms with $dr d\theta$. One of them should be $d\theta dr$. They are not the same...they are actually negatives of each other: $dr d\theta = - d\theta dr$. Also $dr^2 = 0$ and $d\theta^2 = 0$. This has to do with "differential forms" and the "wedge product". – Nick Jul 26 '20 at 16:15

1 Answers1

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Intro

There are two main types of ways to think about things like $\mathrm{d}x\mathrm{d}y$ in multivariable calculus, and we often switch between them depending on the context. (This clarification was inspired in part by Terry Tao's preprint on “differential forms and integration”.) $\mathrm{d}x$ can either act kind of like a number, or act kind of like a vector.

For the “number” interpretation, there are things like limit arguments or infinitesimals in nonstandard analysis to make things rigorous. For the “vector” interpretation, there are things like “differential forms” or “geometric calculus” to make things rigorous. But I'm going to gloss over those details because there are many ways to make things formal, and the exact choices don't affect the intuition here.

Numbers

One way to think about things is that $\mathrm{d}x$ and $\mathrm{d}y$ are in some way like tiny numbers representing the width and length of a tiny rectangle, so that $\mathrm{d}x\mathrm{d}y$ is the area of a tiny rectangle. Then when we write something like $\iint f\left(x,y\right)\,\mathrm{d}x\mathrm{d}y$ or $\iint g\left(r,\theta\right)\,\mathrm{d}r\mathrm{d}\theta$, we just add up the signed volumes (in case $f$ or $g$ is negative) of thin rectangular prisms with cross-sectional area represented by $\mathrm{d}x\mathrm{d}y$ or $\mathrm{d}r\mathrm{d}\theta$.

Under this interpretation, $\mathrm{d}x=\cos\theta\mathrm{d}r-r\sin\theta\mathrm{d}\theta$ doesn't make too much sense. For example, if $\theta=\pi/2$, then we would have $\mathrm{d}x=-r\mathrm{d}\theta$, so that $\mathrm{d}\theta$ and $\mathrm{d}x$ couldn't both represent positive lengths. But we can still understand the relationship between the areas $\mathrm{d}x\mathrm{d}y$ and $\mathrm{d}r\mathrm{d}\theta$ with arguments like the geometric one in this answer by Mike Spivey.

Vectors

The other way to think about things is that $\mathrm{d}x$ and $\mathrm{d}y$ are in some way like tiny vectors whose direction we care about, and this leads to a slightly different discussion. To emphasize this vector idea, I will use some nonstandard notation. Let's write $\overrightarrow{\mathrm{d}x}=\left\langle \Delta x,0,0\right\rangle$ for some positive $\Delta x$, and $\overrightarrow{\mathrm{d}y}=\left\langle 0,\Delta y,0\right\rangle$ for some positive $\Delta y$. So $\overrightarrow{\mathrm{d}x}$ points to the right in the $xy$-plane and $\overrightarrow{\mathrm{d}y}$ points “up” in the $xy$-plane. Then the area of the little rectangle they make is $\left\Vert \overrightarrow{\mathrm{d}x}\times\overrightarrow{\mathrm{d}y}\right\Vert =\Delta x\Delta y$.

However, now that we have vectors, we could choose to care about the orientation. When we think about a usual integral like $\int_{\left[a,b\right]}f\left(x\right)\,\mathrm{d}x$ when $f$ is negative, we decide to count that area in a negative way. Similarly, we could count the “signed area” of the oriented rectangle with sides $\overrightarrow{\mathrm{d}x},\overrightarrow{\mathrm{d}y}$ as the positive value $\Delta x\Delta y$ (which is just the $z$-component of $\overrightarrow{\mathrm{d}x}\times\overrightarrow{\mathrm{d}y}$). But the signed area of the oriented rectangle with sides $\overrightarrow{\mathrm{d}y},\overrightarrow{\mathrm{d}x}$ would be the negative value $-\Delta x\Delta y$ (which is just the $z$-component of $\overrightarrow{\mathrm{d}y}\times\overrightarrow{\mathrm{d}x}$).

Then when we write something like $\iint f\left(x,y\right)\,\mathrm{d}x\mathrm{d}y$ or $\iint g\left(r,\theta\right)\,\mathrm{d}r\mathrm{d}\theta$, we could (if we chose) care about the order of things, with something like an implied $\overrightarrow{\mathrm{d}x}\times\overrightarrow{\mathrm{d}y}$ or $\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}$ in our heads, if not always our notation.

When using things like vectors, then something like $\overrightarrow{\mathrm{d}x}=\cos\theta\overrightarrow{\mathrm{d}r}-r\sin\theta\overrightarrow{\mathrm{d}\theta}$ makes a lot of sense. $\overrightarrow{\mathrm{d}r}$ points away from the origin, and $\overrightarrow{\mathrm{d}\theta}$ points perpendicularly to $\overrightarrow{\mathrm{d}r}$ in the counterclockwise way (so that $\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}$ points in the same positive-$z$ direction as $\overrightarrow{\mathrm{d}x}\times\overrightarrow{\mathrm{d}y}$).

Finally, we can correct your calculation:

$$\overrightarrow{\mathrm{d}x}\times\overrightarrow{\mathrm{d}y}=\left(\cos\theta\overrightarrow{\mathrm{d}r}-r\sin\theta\overrightarrow{\mathrm{d}\theta}\right)\times\left(\sin\theta\overrightarrow{\mathrm{d}r}+r\cos\theta\overrightarrow{\mathrm{d}\theta}\right)$$ $$=\cos\theta\sin\theta\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}r}-r^{2}\sin\theta\cos\theta\overrightarrow{\mathrm{d}\theta}\times\overrightarrow{\mathrm{d}\theta}+r\cos^{2}\theta\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}-r\sin^{2}\theta\overrightarrow{\mathrm{d}\theta}\times\overrightarrow{\mathrm{d}r}$$ $$=\overrightarrow{0}-\overrightarrow{0}+r\cos^{2}\theta\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}-r\sin^{2}\theta\left(-\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}\right)=\boxed{r\,\overrightarrow{\mathrm{d}r}\times\overrightarrow{\mathrm{d}\theta}}$$

Mark S.
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