is there any history at all for this notation of partial anti-derivatives?

Question

i have searched but can not find examples of any published book or online articles that use this notation: $$\int f(x,y) \partial x$$ seems it would be useful for example here: $$\int_I\int_J f(x,y)dxdy = \int_I\color{blue}{\left(\color{black}{\int_J f(x,y)}\partial x\right)}dy$$

is there a history of such notation? are there problems with such notation? any thoughts/help would be much appreciated thx

edited-> some background for the question.. a student writes $$A=xy$$ then writes $$dA=xdy+ydx$$ then the student tries to recover the A by integrating $$\int dA=\int ydx+\int xdy$$ which 'yields' $$A=yx+c(y)+yx+c2(x)=2xy+c(y)+c2(x)$$ which is NOT the correct value of A, the teacher says $\int ydx\ne yx$ the students replies "sometimes it is" we routinely compute $\int ydx=yx$ when doing the inside of a double integral...so in $\int ydx$ sometimes $y$ is held constant and sometimes not.. yet the notation is indistinguishable...

Answer 1

Sometimes $x=3$ and sometimes $x=17$. You can't tell which is which if someone just writes $x$! The notation is indistinguishable!

Okay, the comment is a bit snarky, but it illustrates the point: notation is introduced to mean something, and if you ignore the meaning, you have a problem. e.g. if $x$ and $y$ are defined to be functionally related, then it is a flat out mistake to imagine that one can vary while the other is held constant.

(nitpick: there are degenerate cases like the relationship being $y = 0$ when $x \leq 0$ and $y = x$ when $x \geq 0$)

Partial derivative notation is already somewhat problematic; I don't think it's a good idea to try and extend it to integrals! e.g. if you are working on the unit sphere, you have three coordinates $x,y,z$ related by the equation $x^2 + y^2 + z^2 = 1$. Now suppose you have some function $f$ on the sphere. What does $\partial f/\partial x$ mean? Are you supposed to hold $y$ constant while $x$ varies? Or are you supposed to hold $z$ constant? Or something else entirely?

For integrals, there is no ambiguity: the path of integration tells you how the variables are related. When we write

$$ \int_0^1 \int_0^1 x y \, dx \, dy $$

what we really mean is that the integral

$$ \int_0^1 x y \, dx $$

is supposed to be taken over the vertical path that starts at $(x,y) = (0,y)$ and ends at $(1,y) = (0,y)$. (Note that each value of $y$ gives a different vertical path!) If you wanted to be pedantic, you could write it explicitly as a path integral; e.g. define $\gamma_y$ to be the path I mention above, and then write the integral as

$$ \int_0^1 \int_{\gamma_y} x y \, dx \, dy $$

In your example with the student's mistake, the problem is that the student never picked a path. When you break the integral apart across a sum, you have to use the same path on all of the integrals; but he chose two different paths on the right hand side. :(

Even worse, if you were in the univariate case (e.g. $x,y,A$ all functions of some variable $t$), then you have to pick the path defined by the relationship between the variables; it doesn't even make sense to imagine $y$ is held constant while $x$ varies!

Answer 2

It does not follow from

$$ dA = x \, dy + y\, dx. $$

that $$ \int dA = \int x \, dy + \int y \, dx $$

Indeed, you gave a proof by contradiction of the above statement. Instead,

$$ \int dA = \int x \, dy + y \, dx $$

This is a path integral. The teacher might instead respond that if we choose a curve $C$ in $\mathbb{R}^2$of the sort that is susceptible to Green's theorem (and that encloses the region $D$) where $D$ is the region enclosed by $C$, then we would instead be trying to aim for

$$ \oint_C y \, dx + x \, dy = \iint_D \left( \frac{\partial x}{\partial x} - \frac{\partial y}{\partial y} \right) \, dx \, dy \\ = \iint_D \, dx \, dy = \dotsm = xy. $$

$x \, dy + y \, dx$ is a 1-form; integrating a 1-form is not like Calculus I/II integration.