What is the dot product of a function and a differential?

Question

In a few theorems with integrals it looks like they are taking the dot product of a vector valued function or vector field and the differential.

$\oint_C \textbf{F} ⋅ d \textbf{r}$ from here and $\int_C \textbf{F}(\textbf{r})⋅d\textbf{r}$ here for examples.

However, we are always given its simplified form so that we can compute it. Is it just written that way to give some indication of what is being calculated, or can this opertion be done if I do not know its simplified form?

Edit: by simplified I mean getting the integral to the point that you are no longer taking the dot product between a function and the differential. For example $\int_C \textbf{F}(\textbf{r}) ⋅ d \textbf{r} = \int_a^b \textbf{F}(\textbf{r}(t)) ⋅ \textbf{r}'(t) dt$

What do you mean by "simplified form" ? – bubba Dec 26 '15 at 15:03 — bubba, Dec 26 '15 at 15:03
I added what I mean to the question. Hopefully that helps. – Corey Dec 26 '15 at 15:28 — Corey, Dec 26 '15 at 15:28

score 0 · Accepted Answer · answered Dec 26 '15 at 15:23

0

Let's say that $\mathbf{F}:\mathbb{R}^3 \to \mathbb{R}^3$. Then $$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C \begin{bmatrix} \mathit{F}_0 \\\mathit{F}_1 \\\mathit{F}_2 \end{bmatrix} \cdot \begin{bmatrix} dx \\ dy \\ dz \end{bmatrix} = \oint_C (\mathit{F}_0 dx + \mathit{F}_1 dy + \mathit{F}_2 dz) $$ It may be seen as a short hand for writing a sum, but does have mathematical meanings.

answered Dec 26 '15 at 15:23

Henricus V.

18,694

That makes more sense, but why does that result in r'(t) in the line integral? – Corey Dec 26 '15 at 15:42
@Corey Using the chain rule, $\frac{dr}{dt} dt = dr$ – Henricus V. Dec 26 '15 at 15:43

score 0 · Answer 2 · answered Dec 26 '15 at 16:09

$\oint_C \mathbf{F} \cdot d\mathbf{r}$ is a shorthand notation for the following:

Let $\mathbf{F}:\mathbb{R}^3 \to \mathbb{R}^3$ and $\textbf r:[0,1]\to \mathbb R^3$ be a parameterization of $C$ given by $t\mapsto (x(t),y(t),z(t))$.

Then, by definition the line integral of $\textbf F$ along $C$ is $\oint_C \mathbf{F} \cdot d\mathbf{r}=\int_0^1\mathbf{F}(\textbf {r}(t))\cdot \textbf r'(t)dt$.

If we suppress the parameter $t$ in the above, we get

$\mathbf{F}((x,y,z))=(F_0,F_1,F_2)$ and $\textbf r'(t)dt=(dx,dy,dz)$ so that

$\oint_C \mathbf{F}\cdot d\mathbf{r}=\oint_C (F_0,F_1,F_2)\cdot (dx,dy,dz)=\oint_C F_0dx+F_1dy+F_2dz,\ $ which is how one usually writes these line integrals.

What is the dot product of a function and a differential?

2 Answers2