In the double integral, for the integrating area element, we have the transformation equality from Cartesian coordinate system $y=y(x)$ to polar coordinate system $r=r(\theta)$: $${\rm d}\sigma={\rm d}x{\rm d}y=r{\rm d}r{\rm d}\theta.\tag{1}$$ But if we let $$x=r\cos\theta,~~~y=r\sin\theta,$$ then $${\rm d}x={\rm d}(r\cos\theta)=-r\sin\theta{\rm d}\theta+\cos\theta{\rm d}r,~~~{\rm d}y={\rm d}(r\sin\theta)=r\cos\theta{\rm d}\theta+\sin\theta{\rm d}r.$$ Therefore \begin{align*} {\rm d}x{\rm d}y&=(-r\sin\theta{\rm d}\theta+\cos\theta{\rm d}r)(r\cos\theta{\rm d}\theta+\sin\theta{\rm d}r)\\ &=-r^2\sin\theta\cos\theta d^2 \theta+r(\cos^2 \theta-\sin^2 \theta){\rm d}r{\rm d}\theta+\sin\theta\cos\theta{\rm d}^2 r.\tag{2} \end{align*} But the result is not correct. What's the reason? I know the right result $(1)$, and I just wonder why $(2)$ is not right.
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It is not true that $d^2r$ and $d^2\theta$ are negligible in front of $dr,d\theta$. – Jun 26 '19 at 10:19
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But it's not true even though containing them... – mengdie1982 Jun 26 '19 at 10:23
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2The proper way to handle area/volume element as algebraic objects are through differential forms and the generator of them ($dx, dy, dr, d\theta, \ldots$) anti-commute. So expression like $d^2\theta$ should really be $d\theta \wedge d\theta = 0$. – achille hui Jun 26 '19 at 10:27
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@achillehui Why? – mengdie1982 Jun 26 '19 at 10:29
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1The anti-commutativity of the differential forms essentially encode the information that the Jacobian is the absolute value of a determinant. My understanding that Jacobian is a determinant (up to a sign) comes from the geometric fact that area/volume is invariant shear linear transform.... – achille hui Jun 26 '19 at 10:38
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This is very similar to this newer question which is not closed. – Mark S. Jul 31 '20 at 16:54
2 Answers
The short answer is that things like $dx,dy, dr, d\theta$ are not real numbers, so they do not follow the same arithmetic/algebra as real numbers. But, from a more fundamental point of view, I'm pretty sure that you're currently thinking of $dx,dy,dr,d\theta$ as "infinitesimal changes in a coordinate", but this alone isn't a precise definition. So, in this case, writing something like \begin{equation} dx\, dy = r \, dr \, d \theta \end{equation} makes as much sense as the equation \begin{equation} \ddot{\smile} \alpha = \ddot{\frown} @ \dot{\frown} \end{equation} simply because the symbols haven't been defined properly. It is only after you define the symbols $dx,dy,dr,d\theta$ that you can make sense of equations involving them. The standard way of doing this is taught in courses on differential geometry. Once you learn the proper definitions, you can make sense of an expression like $d(\text{something})$. Such an object is called a $1$-form, and then, you can show that it satisfies a lot of the familiar rules of calculus.
One additional thing is that we can define a "multiplication" between two one-forms. i.e if $\omega$ and $\eta$ are one-forms then we can define a multiplication between them, denoted by \begin{equation} \omega \wedge \eta \end{equation} called the "wedge product". This product satisfies a lot of familiar properties of regular multiplication of real numbers (such as distributive laws, bilinearity etc), except that it is anti-commutative, so \begin{equation} \omega \wedge \eta = - \eta \wedge \omega \end{equation} (so in particular $\omega \wedge \omega = 0$). So, now if you run through your computation again, and you keep track of the order of multiplication, you'll find that \begin{equation} dx \wedge dy = r \, dr \wedge d \theta \end{equation} (the minus sign in $-\sin^2 \theta$ will become $+$, if you treat the order of the wedge product carefully).
I doubt you will have understood much of what I said about the $1$-forms, and that's ok. The most direct answer to your question is that "$dx,dy, dr, d\theta$ etc. are NOT real numbers, so you shouldn't manipulate them as if they are" (even though many physics books and some math books treat them as such).
My suggestion to you is to think about things geometrically. If you define the polar coordinate transformation $f$ (with the appropriate domain restrictions) by \begin{equation} f(r,\theta) = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix} \end{equation} Then, the factor of $r$ comes up because it is the absolute value of the determinant of the Jacobian matrix: \begin{equation} \left|\det f'(r,\theta) \right| = r \end{equation}
You should convince yourself/read a book on why this absolute value of determinant of derivative contains the information about how areas and volumes "transform/scale" in different coordinate systems. So, at this point, rather than attaching meaning to symbols like $dx,dy, dr, d\theta$, just think of them as placeholders at the end of an integral (like how we always end sentences with "."). The real geometric meaning lies in $|\det f'(r,\theta)|$, and this is what you should aim to understand for now.
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Take a small circle of radius $\delta$ centered at distance $r$ from the origin.
It approximately spans a square $dx\,dy\sim\delta^2$.
And in polar coordinates,
$$dr\sim\delta,\\d\theta\sim\frac{\delta}{r}.$$
Hence
$$dx\,dy\sim r\,dr\,d\theta.$$