What's an intuition behind $\lambda=1$ for $dx\;dy = \lambda r\;dr\;d\theta$?

Question

I've read a bunch of articles here on converting between rectangular and polar coordinates in integrals. I get the intuition about how the natural infinitesimal area segment in rectangular coordinates is an infinitesimal square (ergo $dx\;dy$) but is an infinitesimal chunk of an annulus in polar coordinates (ergo $r\;d\theta\cdot dr$). What I don't get is why we can just swap the two out without some kind of scaling factor. That is, why is it that if we write $dx\;dy = \lambda{}r\;dr\;d\theta$, we have $\lambda=1$?

E.g., when I imagine making $\theta$ infinitesimal first, the infinitesimally thin sector in polar coordinates has area $\frac{1}{2}r^{2}\theta$ whereas the rectangle with infinitesimal height has area $xy \approx r^2\theta$ since the part of the sector falling outside the rectangle vanishes and what we're left with is practically a triangle within the rectangle. In other words, the rectangular coordinate area is twice that of the polar coordinate area. If we look at the infinitesimal band of the annulus at the end (i.e., looking at the "crust" of the "pie slice"), won't that cause the part of the sector falling outside the rectangle to play a dominant role again? Why would that just happen to have the same area as a rectangle with the same infinitesimal width?

I've read that this is "explained" by using the Jacobian in converting between rectangular and polar coordinates since the determinant of the Jacobian is $r$ and not $\lambda r$ for some $\lambda \neq 1$. But this doesn't really inform my intuition. I get that the Jacobian of a function $f$ at $p$ is $f$'s best linear approximation at $p$, but I don't get why the thing to do here is to take its determinant and inject that into an integral when you're switching between coordinate systems. And that seems so much more general than what I'm looking for here. Is it actually necessary to understand all square Jacobians to get why the infinitesimal areas of rectangular and polar coordinates just happen to be equal?

So$\ldots$ why is $\lambda=1$?

Answer 1

If $0 \leq r_{0} < r_{0} + dr$ and $0 < d\theta \leq 2\pi$, the area of the "polar element" defined by $$ r_{0} \leq r \leq r_{0} + dr,\qquad \theta_{0} \leq \theta \leq \theta_{0} + d\theta $$ is exactly the difference of the areas of two "pie-shaped" sectors, namely $$ \text{Area} = \tfrac{1}{2}(r_{0} + dr)^{2})\, d\theta - \tfrac{1}{2}(r_{0})^{2}\, d\theta = r_{0}\, dr\, d\theta + \tfrac{1}{2}\, dr^{2}\, d\theta. $$ (N.B. $dr$ and $d\theta$ are positive real numbers.)

If $dr \ll r_{0}$, then the "third-order" term $\frac{1}{2}\, dr^{2}\, d\theta$ is negligible compared to the "second-order" term, so $$ \text{Area} \approx r_{0}\, dr\, d\theta. $$

Geometrically, if $r_{0} > 0$ then "taking $d\theta$ small" does not give a quasi-triangular region, but a quasi-rectangular one.

(The "polar element" is triangular if $r_{0} = 0$, but in that case $r_{0}\, dr\, d\theta = 0$, so there's no inconsistency with multiplicative factors.)

Answer 2

The area element $r\,dr\,d\theta$ is really $(r\,d\theta)\,dr$.

One writes $r\,dr\,d\theta$ for a reason, and that reason facilitates evaluating integrals, but it doesn't explain why it's $r\,dr\,d\theta$. The way to see why it is that is to write it as $(r\,d\theta)\,dr$.

When you write $\displaystyle\int_{\theta_1}^{\theta_2} \left( \int_{r_1}^{r_2} (\text{some function of $r$ and $\theta$})\cdot r\,dr\right)\,d\theta$ with the inside integral with respect to $r$, then it needs to be written as $r\,dr\,d\theta$.

But now go back to what radians are: the radius times the radian measure of the angle is the length of the arc of the circle. When the radius is $r$ and there is an infinitely small change $d\theta$ in the angle $\theta$, then the infinitely small arc length is $r\,d\theta$. The infinitely small change $dr$ in $r$ is in a direction perpendicular to the arc. So you have two infinitely short lines at right angles to each other: one whose length is $r\,d\theta$ and one whose length is $dr$. Multiplying those gives the area of the infinitely small rectangle.

Answer 3

You might have been told that integrands transform under a Jacobian determinant, but it sounds like you haven't connected that Jacobian with any kind of geometric understanding. Here's what's going on:

Let's say you have some double integral in Cartesian like so:

$$\int \int f(x,y) \, dx \, dy$$

You can embed this into a 3d space to get the same result:

$$\int \int f(x,y) \hat z \cdot [ (\hat x \times \hat y) \, dx \, dy]$$

Of course, $\hat x \times\hat y$ is $\hat z$, which will cancel the $\hat z$, and you'll get the same answer. Why do I group the $(\hat x \times \hat y)$ with the $dx \, dy$? To capture the idea that, in integrating over a surface (here, the $xy$-plane), we must keep account of how that surface is oriented, and how large an area is traced out as you sweep through the coordinates.

Indeed, the $\hat x$ and $\hat y$ here do not derive from taking infinitesimal areas, but by looking at the direction and length traced out by coordinate curves. Here, we use $\hat x$ because $\partial \vec r/\partial x = \hat x$. We use $\hat y$ because $\partial \vec r/\partial y = \hat y$. That is to say, we could write the integral equivalently as this:

$$\int \int f(x,y) \hat z \cdot \left[ \frac{\partial \vec r}{\partial x} \times \frac{\partial \vec r}{\partial y} \right] \, dx \, dy$$

And similarly, you can write the integral in polar coordinates like this:

$$\int \int \tilde f(r, \theta) \hat z \cdot \left[\frac{\partial \vec{\tilde r}}{\partial r} \times \frac{\partial \vec{\tilde r}}{\partial \theta} \right] \, dr \, d\theta$$

(I use tildes here because $\vec r(x,y)$ is not the same as $\vec{\tilde r}(r, \theta)$, and the same is true for $f, \tilde f$. That said, most people would understand implicitly if I weren't this pedantic.)

Of course, you should be able to see that the cross product gives us $\hat r \times (r\hat \theta) = r \hat z$, and we get the well known result of $r \, dr \, d\theta$.

What I want to impress upon you is that this has nothing to do with thinking of $dr, d\theta$ as infinitesimal lengths. We're just using tangent vectors to construct some notion of area spanned by vectors--area that gets larger as vectors used to span it gets larger.

Answer 4

A reason why someone might want you to understand something about Jacobians in general in order to explain $r\;dr\,d\theta$ to you is that the general fact about the Jacobian isn't very much harder to understand intuitively (they hope) than to understand any other explanation of $r\;dr\,d\theta$, and once you do understand it you have the answer for $r\;dr\,d\theta$ and a whole lot of other change-of-variables questions in addition to that.

Unfortunately sometimes the explanation of a Jacobian is "intuitively obvious" to the person who wrote it and not so much to the person reading it. I encourage you to give it more thought, and maybe you will get over that hurdle.

Meanwhile, if you just want to consider the rectangular-to-polar change of variables by looking at infinitesimal "pieces" of the plane, the piece you look at not only have to be compared in size, but we also want them to be in the same place and distributed in the same way around that place. We do not take $N$ units of area from somewhere around $(0,1)$ and claim that you can substitute $N$ units of area from somewhere around $(1,0)$, because the integral over the first $N$ units might be much greater than (or less than) the other. We also do not cut a long, thin rectangle into two triangles along a diagonal and claim the integral over one triangle is half the integral over the rectangle, because the triangle's integral depends mostly on the values of the integrand in the end of the rectangle where the triangle's "fat end" is, whereas the rectangle is equally affected by the integrand's values at both ends and this may result in a different value of the integral.

In your comparison of a sector and a rectangle, the regions rarely overlap much, and even when they do, the integral over the sector gives less relative weight to the function values nearer the origin than the rectangle does.

If you must think in terms of infinitesimal "pieces" of the plane, at least make both pieces be contained within a small neighborhood of a single point so that the function values over the two regions are comparable. When you do that, the obvious region to look at in polar coordinates is a small chunk taken out of a sector or an annulus (in fact, it is the intersection of a sector and an annulus). Visit any point in the plane other than the origin itself, restrict yourself to a small enough neighborhood around that point (the radius of the neighborhood should be a tiny fraction of the distance to the origin), and the polar-coordinates regions within that neighborhood look roughly rectangular, although not usually parallel to the axes. One side of such a "rectangle" has length $dr$; the adjacent "side" has length $r\,d\theta$. The area is therefore $r\;dr\,d\theta$ (in more mathematical terms, the area is asymptotically $r \,\Delta r\,\Delta\theta$ as both $\Delta r$ and $\Delta\theta$ tend to zero).