What does the notation for a line integral of a vector field actually mean? I have been told that the line integral of a vector field, F(r) along a curve $C$ is:
$$I =\int_C\textbf{F}\cdot \text{d}\textbf{r}=\int_C(F_x,F_y)\cdot (\text{d}x,\text{d}y),$$
where $\text{d}\textbf{r}=(\text{d}x,\text{d}y)$ is the line element on $C$.
What does the bit at the end of the integral mean? I was always told that it was there to show the variable of integration but clearly it means more than that as here the dot product is being taken between it and a vector. Also, by extension, what is $(\text{d}x,\text{d}y)$?
 A: What you were told is not accurate, or rather, technically accurate if you consider it as abuse of notation, but very much unhelpful, and not a definition by any means. In the context of physics, a curve $C$ is merely a function $C:[t_0,t_1]\rightarrow\mathbb{R}^n$ satisfying some differentiability requirements. If $\mathbf{F}$ is some vector field, then $$\int_C\mathbf{F}(\mathbf{r})\cdot{\mathrm{d}\mathbf{r}}:=\int_{t_0}^{t_1}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)\,\mathrm{d}t$$ which is well defined, since $$\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)$$ is just some $f(t)$ for some function $f$, because dot products always result in scalar quantities. This is indeed, the correct definition. However, you can do some things here to arrive at the result that you were presented with. If $n=2$, and you are given an orthonormal basis $B=\{\mathbf{e_x},\mathbf{e_y}\}$, then $$\mathbf{F}=F_x\mathbf{e_x}+F_y\mathbf{e_y}=(F_x,F_y)_B$$ and $$\mathbf{r'}(t)=(x'(t),y'(t))_B,$$ implying $$\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)=(F_x(\mathbf{r}(t)),F_y(\mathbf{r}(t)))_B\cdot(x'(t),y'(t))_B.$$ Now, in an abuse of notation, one may consider $\mathrm{d}x=x'(t)\,\mathrm{d}t$ and $\mathrm{d}y=y'(t)\,\mathrm{d}t$. All in all, one would have $$\int_{t_0}^{t_1}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r'}(t)\,\mathrm{d}t=\int_{t_0}^{t_1}(F_x(\mathbf{r}(t)),F_y(\mathbf{r}(t)))_B\cdot(x'(t),y'(t))_B\,\mathrm{d}t=\int_C(F_x(\mathbf{r}),F_y(\mathbf{r}))_B\cdot(\mathrm{d}x,\mathrm{d}y)_B,$$ which in your post, is more shortly denoted as $$\int_C(F_x,F_y)\cdot(\mathrm{d}x,\mathrm{d}y),$$ not specifying the basis. And again, this is abuse of notation, because it treats the symbol $\mathrm{d}t$ in the notation of the Riemann integral as an actual mathematical object with its own objective, independent existence, which is not the case at all. And this all assumes a specific type of basis, which is inadequate, because in physics, these path-integrals are basis-independent, which is to say, they form tensors: they are invariants. So a specific basis for the dot product should not be chosen. It also assumes specifically that space is 2-dimensional. This last one is not a particularly big deal, but it still makes the formula you were presented with very awkward, since what you were presented with is presumably supposed to be a definition, not a circumstance-based theorem.
Hopefully this clarifies any doubts you have, and more importantly, you hopefully have a better understanding of what a path/line integral of a vector field actually is.
A: It's a use or abuse of notation. Just expand the scalar product $(F_x,F_y)\cdot(\text{d}x,\text{d}y)$ as $F_x\text{d}x+F_y\text{d}y$ to get
$$\int_CF_x\text{d}x+\int_CF_y\text{d}y$$
A: Starting off, you have your vector field $\mathbf{F}$ and a path over which you want to integrate it.  Now imagine that you divide your path up into "short" chunks (what "short" actually means in detail I don't have time or space to go into), so that the boundaries of the chunks are at $\mathbf{x}_0, \mathbf{x}_1, \mathbf{x}_2, $ etc.  In your first chunk, you have a displacement $\Delta\mathbf{x} = \mathbf{x}_1-\mathbf{x}_0$, or $\Delta\mathbf{x} =(x_1, y_1)-(x_0,y_0)=(\Delta x, \Delta y)$.  Now you want to know "how much" of $\mathbf{F}$ "points along" your displacement.  One measure of that would be $F\cdot\Delta\mathbf{x}=F_x \Delta x + F_y \Delta y$.  Calculate this for all chunks along your curve.  Now repeat this process for an ever increasing number of ever smaller chunks, and you get the line integral.
A: The $\textrm{d}\mathbf{r}$ represents an infinitesimal change in $\mathbf{r}$. The tuple $(\textrm{d}x,\textrm{d}y)$ just represents the coordinates of $\textrm{d}\mathbf{r}$, meaning that $\textrm{d}\mathbf{r}=\textrm{d}x \hat{\mathbf{i}} + \textrm{d}y\;\hat{\mathbf{j}}$. You have to make sure you take the dot product because, if you don't, you'll get a vector as your end result.
