Line Integral

None of the above -- $\mathbf r$ is a vector describing (e.g.) $\mathbb R^3$ and $f$ is a function mapping (e.g.) $\mathbb R^3 \to \mathbb R$. Therefore

$\int f(\mathbf r) d\mathbf r$

indicates the integral of $f$ (which will be a real number) over (e.g. )some line.


For example take a line $\ell : [0,1] \to \mathbb R^3$ given by $\ell(t) = (t,0,0)$ and $f(x,y,z) = x + y + z$. Then

$\int_\ell f(\mathbf r) d\mathbf r = \int_0^1 t dt = 1/2$

Anyone?

Can you post an actual question so we have some context for the problem?

We need the fuller context from where this question arose and you need to ask coherent questions in order that anyone provides further help.

The only thing that comes to mind is the fact that a scalar field can be thought of as a vector field onto a one dimensional subspace of your ambient space (which I am guessing here is $\mathbb{R}^3$) but such an association yields an infinite amount of choice (basically the choice of a vector spanning such a subspace)

If your answer is supposed to be a vector, then I would agree with your interpretation.

If your path is parameterized by t (i.e. r is a function of t), then $\int f(\mathbf{r})\ dx$ is going to be the same thing as

$\int f(\mathbf{r}) \left(\mathbf{i}.\dfrac{d\mathbf{r}}{dt}\right) \,dt$

or more generally,

$\int f({\bf r}) \,d{\bf r} = \int f({\bf r}) \dfrac{d\mathbf{r}}{dt} \,dt$

I believe this is the question to which he was referring. I am also stuck on this so would really appreciate some help.

See here.

EDIT: Although, I can't see any reason why this question specifically is what the OP is after...