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On the answers of the question Is $\frac{dy}{dx}$ not a ratio? it was told that $\frac{dy}{dx}$ cannot be seen as a quotient, even though it looks like a fraction. My question is: does $dxdy$ in the double integral represent a multiplication of differentials? The problem than can be generalized for a multiple integral.

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For «$dxdy$» to be a product, first $dx$ and $dy$ have to be things which can be multiplied —numbers, or number-values functios. And they are not: they are just notational devices to indicate with respect to which variables integration is being done, exactly in the same way that $dx/dy$ is not a quotient. (And yes, when you move along you will learn about differential forms and what not, but that doesn't chanhe absolutely anything) –  Mariano Suárez-Alvarez Mar 7 at 17:10
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@Emin, since you included the nonstandard analysis tag I thought you were looking for an answer in this framework. If you wish an answer in a traditional framework, you should specify it. The problem then would be to explain the meaning of your term "differential", which only has a kind of a tautological meaning in the traditional framework. Basically what is going on is that one is paritioning the domain of integration into tiny squares, whose area is indeed the product of the sides. This basic idea can be formalized in various ways, but you have to be clearer about what you are looking for. –  user72694 Mar 9 at 17:53
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Something is usually defined as multiplication when it distributes with addition, not just because it is an extension of the common numbering system. You could; for example, multiply two polynomials and get another polynomial, but a polynomial isn't a number. As to the original question; it would be better phrased as "can we assume multiplication properties for dx dy" rather than "is dx dy multiplication". –  DanielV Mar 15 at 20:16
    
@MarianoSuárez-Alvarez : I disagree. See my answer below. –  Michael Hardy Jul 29 at 17:07
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I have seldom seen anyone miss a point more thoroughly that Mariano Suárez-Alvarez in comments further below. I wrote that intuitive ideas can be made rigorous but the intuitions exist independently of the ways of making them rigorous. He seems to have construed my comment as "Intuitive ideas can be made rigorous so let's do that and then explain the rigorous ideas to students." That the opposite of my point. My point was they exist independently of ways of making them rigorous and so can be presented in the classroom to students who can't understand rigorous arguments. –  Michael Hardy Jul 29 at 19:40

9 Answers 9

In a double integral, you are actually integrating a differential two-form:

$$\int_R \mathrm{f}(x,y) \ \mathrm{d}x \wedge \mathrm{d}y$$

Here, $\mathrm{d}x$ and $\mathrm{d}y$ are the basis differential one-forms and $\mathrm{d}x \wedge \mathrm{d}y$ is their exterior product.

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I'm well aware of differential forms. Nevertheless, what you've written is more of a convenient abbreviation. If $e_x, e_y$ are basis vectors, then what you're really integrating is $\int_R f(x,y) (\mathrm dx \wedge \mathrm dy)(e_x \wedge e_y \, dx \, dy)$. The question pertains to the latter $dx, dy$, not to the basis one-forms $\mathrm dx, \mathrm dy$. –  Muphrid Mar 8 at 19:37
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@Muphrid Your point of view disagrees with the Wikipedia article, and I have never seen the notation that you are using in the setting of integral calculus. Can you please supply a reliable third-party reference for you claims? –  Fly by Night Mar 8 at 21:42
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I have no doubt that the methods are mathematically equivalent, but I still think it is worth emphasizing that the tangent $k$-vector is involved in integration. The apparent relationship between basis forms and differentials cannot exist otherwise. Moreover, sometimes you may wish to integrate something other than a form. Maybe you want to integrate a $k$-vector. Knowing that integration inherently involves the tangent $k$-vector of the manifold, you can properly identify the factors of the metric that should be there. Differentials will be in this integral, even though there are no forms. –  Muphrid Mar 8 at 23:01
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@Muphrid Please could you continue your conversation with Hurkyl elsewhere. The comments section is to ask the person making the post for more information and/or to suggest improvements. It is not to host a forum. Thank you. –  Fly by Night Mar 10 at 21:41
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@Hurkyl Please could you continue your conversation with Muphrid elsewhere. The comments section is to ask the person making the post for more information and/or to suggest improvements. It is not to host a forum. Thank you. –  Fly by Night Mar 10 at 21:41

The OP listed non-standard analysis as one of the tags. It is therefore not inappropriate to point out that, just as one can think of the derivative as a true ratio $\frac{dy}{dx}$ modulo an infinitesimal error (eliminated by applying the shadow), so also one can think of a single-variable integral as an infinite sum of infinitesimal terms of type $dx$ (again up to applying shadow). Double integrals can naturally be viewed as double (infinite) sums, where $dx\,dy$ is most decidedly an ordinary product. And of course this generalizes to multiple integrals as the OP suggested. If one is in Euclidean space, talking about differential forms is an unnecessary obfuscation.

Edit 1: For finite Riemann sums approxiating the double integral, it is obvious that the term $\Delta x \Delta y$ is a product; it seems nobody in his right mind would deny this. The difference is that one cannot deduce the value of the integral from a finite Riemann sum. On the other hand, with an infinite Riemann sum when $\Delta x$ is replaced by $dx$, etc., the value of the integral is deduced from the value of the Riemann sum by taking the shadow (see above). That's the advantage of having the richer syntax of the hyperreal approach.

Edit 2: the OP's question is in fact equivalent to a question about single-variable integrals, namely: does $f(x)dx$ denote multiplication of $f(x)$ by $dx$? Perhaps the right answer is that it denotes a memory of multiplication. Namely, the multiplication is still there at the level of the hyperfinite Riemann sum. To pass from this to the integral one applies the standard part function, after which we have only a "memory" left. Similarly, one can form the differential quotient Δy/Δx which is still a ratio, but one doesn't get the derivative until one applies the standard part function. Here also there is only a memory of a division left. The advantage of the hyperreal framework is that one has a direct procedure for passing from the ratio to the derivative which isn't the case in the traditional real-based framework where one must appeal to an indirect notion of an epsilon, delta limit.

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What do you mean by "memory of multiplication"? –  Emin May 24 at 17:32
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@Emin, the multiplication is still there at the level of the hyperfinite Riemann sum. To pass from this to the integral one applies the standard part function, after which we have only a "memory" left. Similarly, one can form the differential quotient $\Delta y/\Delta x$ which is still a ratio, but one doesn't get the derivative until one applies the standard part function. Here also there is only a memory of a division left. The advantage of the hyperreal framework is that one has a direct procedure for passing from the ratio to the derivative... –  user72694 Jul 28 at 14:24
    
... which isn't the case in the traditional real-based framework where one must appeal to an indirect notion of an epsilon, delta limit. –  user72694 Jul 28 at 14:25
    
I think that comment would be the real answer to my question! You may include it on your answer so I can accept the answer. –  Emin Jul 29 at 13:18
    
As a physicist, I find that infinitesimals just are ordinary real numbers. And they don't even have to be small. With certain astronomical applications, they can be quite large, maybe the size of a galaxy. I've been working with a Fluid Tube Continuum where infinitesimals are the distance between two tubes in a tube bundle. A physicist works with different scales. Quantities that are small on one scale can be large on another scale. A distinction which apparently tends to disappear when going to exact, by "taking the limit". –  Han de Bruijn 12 hours ago

As others have said $dx\, dy$ does not represent a product of differentials. But it represents a product of measures. We have the "natural" Lebesgue measure $\lambda$ on the $x$-axis, and integration with respect to this measure is signalled by writing ${\rm d}x$ as right parenthesis of the integral. Similarly we have the "natural" Lebesgue measure $\lambda$ on the $y$-axis, and integration with respect to this measure is signalled by writing ${\rm d}y$ as right parenthesis of an integral involving the variable $y$. The individual measures $\lambda$ on the $x$-axis ${\mathbb R}$ and the $y$-axis ${\mathbb R}$ define a product measure $\lambda\otimes\lambda$ on the cartesian product ${\mathbb R}^2$, again called Lebesgue measure on ${\mathbb R}^2$. Integration with respect to this product measure is is signalled by writing ${\rm d}(x,y)$, ${\rm d}x\otimes {\rm d}y$, or simply $dx\,dy$, as right parenthesis of an integral over some subset $A\subset{\mathbb R}^2$. Fubini's theorem then tells us that $$\int\nolimits_A f(x,y)\>{\rm d}(x,y)=\int\nolimits_{A'}\left(\int\nolimits_{A_x} f(x,y)\> {\rm d}y\right)\ {\rm d}x\ ,$$ where $A'$ denotes the projection of $A$ onto the $x$-axis and $A_x:=\{y\mid (x,y)\in A\}$ collects the $y$-values to be weighted in for given $x\in A'$.

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From all answers this is to me more convinced, but still there are some gaps I think. From what I understood, you're saying that dxdy is just a notation that tells us that we are integrating with respect to $\lambda\otimes\lambda$ measure, but I'm not convinced that it's just a notation. –  Emin Mar 15 at 18:23
    
    
Improper integrals exist for functions that are not Lebesgue integrable: see Wikipedia –  Han de Bruijn 12 hours ago

Let's say $f(x,y)$ is in $\dfrac{\mathbf{kg}\cdot \mathbf{m}}{\mathbf{sec}\cdot\mathbf{dollar}}$ and $x$ is in $\mathbf{sec}$ and $\mathbf{dollar}$. Then $f(x,y)\,dx\,dy$ is in $\mathbf{kg}\cdot \mathbf{m}$, just as if we are multiplying.

If an infinitely small rectangle has length and width respectively $dx$ and $dy$, then its area is $dx\,dy$; if $f(x,y)$ is density of something (mass, probability, energy$\ldots$) with respect to area, then $f(x,y)\,dx\,dy$ is measure in the same units as that "something". Why does $dx\,dy$ become $r\,dr\,d\theta$. Sometimes people say "because you multiply by a Jacobian". That's Ok as far as it goes, maybe. I regard it as $(dr)(r\,d\theta)$. If $r$ is in meters and $\theta$ is dimensionless and in radians, then $r\,d\theta$ is in meters. A length of an arc of a circle is the radius times the radian measure of the arc. The radius is $r$; the radian measure of the arc is $d\theta$, so $r\,d\theta$ is the length, and $dr$ is disance in a direction orthogonal to that so $(dr)(r\,d\theta)$ is area of that rectangle. It's a rectangle because an infinitely small arc is a straight line.

If you say that this is not rigorous, I agree.

If you object that this is not rigorous, I disagree.

Intuitive ideas can be made rigorous [comment inspired by comments below: The following should be obvious, but apparently there was one person to whom it wasn't, so maybe there are others. I do not condone making infinitesimals rigorous in most first-year calculus courses.]. What is the right way to do that may be subject to philosophical disagreements. But one should not assert that intuition to be made rigorous is the rigorous end-product. Just which way of making something rigorous is the right one depends on the context. Some other way of making something rigorous that will be discovered 100 years from now may have its place. But the idea that is to be made rigorous exists independently of the ways of making it rigorous.

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That something can be made rigorous (and of course infinitessimals and things like «infinitely small rectangle» can) does not imply that it is made rigorous. These infinitely small quantities you speak of are usually not made rigorous in most courses, which instead focus on another way of setting things up. In that setting, $dx\,dy$ is certainly nothing more than a notational device. –  Mariano Suárez-Alvarez Jul 29 at 18:21
    
@MarianoSuárez-Alvarez : I think that is nonsense. Yes, there are lots of lousy calculus courses that don't explain the intuition. But a good introductory calculus course does that and largely eschews rigor. They shouldn't be just used as a notational device. I don't think we should tell people who ask questions here that they shouldn't want to know the answers simply because the answers won't be explained in most courses. –  Michael Hardy Jul 29 at 19:17
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Something in your comment is very strange: you see to think that only a course that makes intuitive ideas rigorous should speak of them at all. Rigor is a bad thing in most introductory calculus courses. Intuitive explanation of infinitesimals is a good thing in such courses. –  Michael Hardy Jul 29 at 19:18
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@MarianoSuárez-Alvarez : You really know how to miss a point and drop context. Looking at my answer, I see that I wrote "Intuitive ideas can be made rigorous", and from your comments it looks as if you construed that as meaning one should make them rigorous so that one can say that $dx\,dy$ is an instance of multiplication. If you read the WHOLE paragraph, you will see that my point was nearly the opposite of that. I wrote that those intuitions exist INDEPENDENTLY of the ways of making them rigorous. –  Michael Hardy Jul 29 at 19:28
    
I find it rather insulting that you assume I did not read the whole paragraph, so Iwill just ignore the whole thing. –  Mariano Suárez-Alvarez Jul 29 at 21:37

$dx$ and $dy$ aren't real numbers; they are things called differential forms. Thus, you can't use the real number multiplication operation to multiply them.

However, $dx \, dy$ is a thing, and it is not terribly unreasonable to define "the multiplication of $dx$ and $dy$ to be $dx \, dy$. The trick is that you have to make the inferences in the opposite direction from what you're used to -- to work out the first properties, it's not because you are understanding $dx \, dy$ in terms of multiplication, it's because you are using your understanding of $dx \, dy$ to figure out what 'multiplication' means.

There are some subtleties in what $dx \, dy$ means that I'm not up to fully explaining at the moment: e.g. it needs to talk about the orientation of a region, so that the fact that $dx \, dy = -dy \, dx$ can be properly explained. (you don't notice this fact when you do ordinary iterated integrals, since you flip the orientation of your region whenever you swap the order of $x$ and $y$, which cancels out the sign change)

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To consider: why is it that $\int \mathrm dx \wedge \mathrm dy = \int dx \, dy$ but $\int \mathrm dy \wedge \mathrm dx = - \int dx \, dy$? How can this be reconciled with the notion that order of integration doesn't usually matter? Because what's really going on is the 2-form is eating the 2-vector, and changing the sign of the 2-form changes the sign of the result. This also answers why $\int \mathrm dx \wedge \mathrm dy = \int dx \, dy$: it's conventional. People usually use the 2-vector $e_x \wedge e_y$ as the orientation, but if you use $e_y \wedge e_x$, you get the opposite. –  Muphrid Mar 9 at 20:49
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@Muphrid: That's not what's "really going on" -- it's merely one way to understand the integral. And is easy enough to grok orientation once you see the initial idea and, e.g. see how if $R = [a,b]\times[c,d]$, then $\int_a^b \int_c^d dy dx$ should be $\int_{-R} dy \wedge dx$ since you're tracing the rectangle out in the opposite orientation than standard. –  Hurkyl Mar 9 at 21:43
    
@Muphrid: And it's worth noting that, even when interpreting $dx$ and $dy$ as differential forms and all integrals as path integrals, that $$\int_I \left( \int_J dx \right) \, dy = \int_J \left( \int_I dy \right) \, dx$$ corresponding to two different ways to write a surface as a curve of curves. –  Hurkyl Mar 9 at 22:00
    
dx and dy aren't real numbers; they are things called differential forms. Well, one way of interpreting them is as forms. But the question explicitly points out that this is not the only way of interpreting them; in the context of NSA, they're infinitesimals. –  Ben Crowell May 19 at 22:18
    
@Ben: In the single-variable case, we can indeed interpret differential forms as infinitesimals. But they're still differential forms, and we can make semantic errors if we forget that. Note, however, that here we're in the multi-variable case. –  Hurkyl May 19 at 22:48

I find it very helpful in this case, and with multi-dimensional integrals in general, to write the $dx$ and $dy$ after the $\int$ sign. I.e.

$$ \int dy \int dx\, f(x,y) \;\;\;\text{ rather than }\;\;\; \int\int f(x,y)\, dx\,dy $$

This makes it clear that the $dx$ and $dy$, far from being infinitessimally small $\Delta x$ and $\Delta y$s, are part of the integral operators. Mathematical physicists often follow this convention, not just because it makes multi-dimensional integrals easier to understand, but also because it emphasizes integration as an operator -- and a linear operator at that.

$\int dy$ and $\int dx$ each operate on $f(x,y)$ as "atomic units" that can't be split up. $\int dy$ and $\int dx$ apply to $y$ and $x$ respectively just as $\sum_y$ and $\sum_x$ are units that apply to $y$ and $x$. It still makes perfect sense to see that when $f(x,y)$ factorizes to $f(x)f(y)$ you can rearrange $$\int dy \int dx\, f(x)f(y) \;\;\;\text{ to }\;\;\; \int dy \,f(y) \int dx\, f(x)$$ just as you can $$\sum_y \sum_x\, f(x)f(y) \;\;\;\text{ to }\;\;\; \sum_y \,f(y) \sum_x f(x)$$ However if you write $dxdy$ following the integrand it appears as though you are also factorizing $dxdy$ into $dx\times dy$ when in fact the $dx$ and $dy$ are part of the $\int$ operators.

One reason why $dx$ is often treated as if it is a standalone object to be manipulated separately is the convenience of doing just that when substituting variables. For example $\int \sin^2 x \cos x \, dx = \int u^2 \,du$, substituting $u=\sin x$ and $dx = du / \cos x$. This is very convenient and everyone does it but strictly speaking there is a rule being applied which transforms the integration operation from one to another: $\int dx\, \frac{du}{dx} = \int du$. Similarly in differentiation it is very convenient to think about cancelling the numerator of one derivative with the denominator of another, e.g. when applying the chain rule, but this is not strictly speaking correct.

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So if I understood good, we don't need to say is dxdy a multiplication of dx and dy, but is $\iint dxdy=\int dx \int dy$ as operators, and the answer would be yes! Right? Than it means that dx, dy and dxdy are not differential forms, but they are just a notation (just as many of others said). Am I right? –  Emin Mar 16 at 10:14
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Yes, essentially it's about notation as at least one of the earlier answers said, but the notation I've given here makes it particular clear what's going on. So you can rewrite $\iint dxdy$ as $\int dy\int dx$ and then treat them as atomic operators. Also $\iint dA$ is common for differentating w.r.t. area. –  TooTone Mar 16 at 10:21
    
Can we see dx as a differential in the formula $\int f(x)dx$? –  Emin Mar 16 at 10:24
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please see my update –  TooTone Mar 16 at 10:32
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What you are talking about are iterated integrals rather than double integrals. Modulo suitable hypotheses on the function, these are equal, but one cannot separate dx from dy if one is talking about the double integral. The claim that $dx$ and $dx$ are "far from being infinitesimal" is merely persisting in denial and refusal to face reality. –  user72694 Mar 21 at 8:35

In Cartesian linear coordinates just as we sum up separate rectangle areas as $ \int y. dx $ , we sum up prismatic volumes as $ \int z.dx. dy $. Here we consider an infinitesimal(differential) area as a product of two infinitesimal(differential) lengths.

If we are considering curvilinear coordinates, we need to consider their Jacobian to multiply Cartesian differential area by the Jacobian in order to define relation between the old small area and the new distorted small area. The way product is looked upon as an area in the small is exactly the same as in Cartesian before it leads to the integrated in the large situation.While changing from Cartesian to polar,

$ dx dy = r * dr* d\theta $ where $ r$ is Jacobian $ J$ in the general case

$J = r= \partial(x,y)/\partial ( r,\theta). $

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No, because the differentials that appear in an integral are just notation; they only signify what variables are to be integrated over.

Consider a sum:

$$\sum_{i=1}^n i^2$$

The $i$ in the bottom of the summation symbol tells you that $i$ is the dummy variable here.

You could, in principle (though no one does this), notate integrals the same way:

$$\int_{x=a}^b x^2$$

And you could do it for a double integral:

$$\int_{y=c}^d \int_{x=a}^b e^{-x^2 - y^2}$$

The differentials being used in integrals are just empty notation.


Edit: Fly by Night suggests that differentials mean something because they transform. There is a transformation involved when changing coordinates, but it is absolutely incorrect to say this has to do with the differential.

This can be seen by considering the integral abstractly as a map, with its internals being the limit of a Riemann sum. Let $\mathcal I(f, \ell, d)$ be the integral of a function $f$ over a region described by the function $\ell$, with an interval $d$ describing the parameter region of $\ell$. That is, if $d = [a,b)$, then we integrate $f$ from $\ell(a)$ to $\ell(b)$.

The integral can then be written as a limit of a Riemann sum:

$$\mathcal I(f, \ell, d) = \lim_{N \to \infty} \sum_{n=0}^{N-1} [f \circ \ell]\left (a + n \frac{b-a}{N} \right) \ell'\left(a + n \frac{b-a}{N}\right) \frac{b-a}{N} $$

Consider the case $\ell(x) = x$. Then $\ell'(x) = 1$, and you get the usual, familiar form of a Riemannian integral.

Now instead, consider a reparameterization. Let $x = g(u) = u^2$, as Fly by Night suggested. Then there is a new paremeterization function $m$:

$$m(u) = (\ell \circ g)(u) = u^2$$

There is also a transformed interval:

$$e = [g^{-1}(a), g^{-1}(b)) = [p, q)$$

Now then, since $m([p, q)) = \ell([a, b))$, the integral of $f$ should be the same for both. That is, $\mathcal I(f, \ldots)$ only really cares about the region $f$ is actually being integrated over, not how that region is parameterized. So we should conclude that

$$\mathcal I(f, \ell, d) = \mathcal I(f, m, e)$$

But we know that $m'(u) = {\color{red}{2u}}$, and we get

$$\mathcal I(f, m, e) = \lim_{N \to \infty} \sum_{n=0}^{N-1} [f \circ m]\left( p + n \frac{q - p}{N} \right) \color{red}{2\left (p + n \frac{q-p}{N} \right)} \frac{q-p}{N}$$

What does this look like in traditional integral notation? We started with

$$\int_a^b [f \circ \ell](t) \ell'(t) \, dt = \int_a^b f(t) (1) \, dt$$

And then we changed to

$$\int_{p = \sqrt{a}}^{q = \sqrt{b}} [f \circ m](t) m'(t) \, dt = \int_p^q [f \circ m](t) 2t \, dt$$

All of this follows from the transformation of the tangent vector along the parameterized curve. Writing it as a change of differentials is nothing more than misleading voodoo. It is a fundamental mistake to say that the differentials are 1-forms. They are not; they are totally vacuous and without meaning, and thus my answer to the question is that it's meaningless to talk about multiplying differentials because they are themselves meaningless. All of the supposed properties of differentials actually come from other, better defined geometric principles and concepts.

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I disagree with you! In the definition of the integral we have those $\delta x$ and $\delta y$ which becomes dx and dy when we act with limit. This tel us that they are part of the integral, and not just a notation! –  Emin Mar 7 at 17:08
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Just because a $\delta x$ would be there when you write the integral as the limit of a Riemann sum doesn't mean anything. The $dx$ by itself only signifies what variable is being integrated over; this can be done in other ways. You never use the $dx$ in an integral for anything other than this purpose. –  Muphrid Mar 7 at 17:11
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@Muphrid I think that $\mathrm{d}x$ is a little more than simply notation: it's a differential one-form. If we change coordinates on the $x$-axis then $\mathrm{d}x$ transforms. For example if $x=u^2$ then $\mathrm{d}x = 2u\, \mathrm{d}u$. –  Fly by Night Mar 7 at 17:33
    
@FlybyNight: No, this so-called property of differentials comes from the transformation of the tangent vector. Differentials themselves have no innate properties, and they do not obey the transformation laws of forms. –  Muphrid Mar 15 at 20:51
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There are multiple good reasons why we don't write $\int x^2 dx$ as $\int x^2$. (1) The units come out wrong. (2) It doesn't specify what we're integrating with respect to. –  Ben Crowell May 19 at 22:17

Let's see what happens to $dx dy$ under a change of variables $x = f(z,w)$, $ y = g(z,w)$.

$$ dx = \frac{df}{dz}dz + \frac{df}{dw}dw $$ $$ dy = \frac{dg}{dz}dz + \frac{dg}{dw}dw $$

So locally this change of variables is a linear transformation and the dilation factor is the determinant:

$$ \left| \begin{array}{cc} \frac{df}{dz} & \frac{df}{dw} \\ \frac{dg}{dz} & \frac{dg}{dw} \end{array} \right| =\frac{df}{dz}\frac{dg}{dw} -\frac{df}{dw} \frac{dg}{dz} $$

We can derive this using exterior algebra and the wedge product.

$$ dx \wedge dy = \left(\frac{df}{dz}dz + \frac{df}{dw}dw \right) \wedge \left(\frac{dg}{dz}dz + \frac{dg}{dw}dw \right) $$

Using the identities $dz \wedge dz = dw \wedge dw = 0$ and $dz \wedge dw = - dw \wedge dz$. We recover the Jacobian formula.


In our case, $dx \wedge dy $ behaves like $dx dy$ since they are in a sense perpendicular, "$dx \perp dy$".

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