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    What does this mean when f is a scalar function:

    \int f(\mathbf{r})\ d \mathbf{r}

    I cannot find anything in my notes that refers to integrating a scalar with respect to a vector.

    All I can think of is that it means d \mathbf{r}=(dx,dy,dz) so the integral is a vector of 3 integrals, the integrand integrated with respect to dx, dy, dz in order. But then what does that mean? i.e. what does


    \int f(\mathbf{r})\ dx

    mean?
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    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
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    (Original post by jb444)
    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
    But my answer needs to be a vector for teh question to work.
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    Anyone?
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    (Original post by james22)
    Anyone?
    Can you post an actual question so we have some context for the problem?
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    AS jb444 wrote you have to integrate a
    scalar valued function which have vector as variable.
    The line is discribed by a vector
    \displaystyle \vec r=\vec r(s)
    where s is the arc length of the line, so
    \displaystyle \int_C F(\vec r) d\vec r =\int_S F(\vec r(s)) ds

    Or when you write down the line with t parameter
    \displaystyle \vec r=\vec r(t) and you paramerize the F with t, then
    \displaystyle \int F(\vec r(s) ds =\int F(x(t),y(t),z(t))\sqrt{x'^2(t)+y  '^2(t)+z'^2(t)}dt
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    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)
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    (Original post by james22)
    All I can think of is that it means d \mathbf{r}=(dx,dy,dz) so the integral is a vector of 3 integrals, the integrand integrated with respect to dx, dy, dz in order. But then what does that mean? i.e. what does


    \int f(\mathbf{r})\ dx
    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
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    (Original post by davros)
    Can you post an actual question so we have some context for the problem?
    (Original post by Mark85)
    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)
    (Original post by DFranklin)
    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.
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    (Original post by beelz)
    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...
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    (Original post by Farhan.Hanif93)
    See here.

    EDIT: Although, I can't see any reason why this question specifically is what the OP is after...
    Nor can I but that does happen to be the question.
 
 
 
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