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Line Integral Notation wrt Scalar Value function

  1. Jul 25, 2017 #1
    I'm getting a bit confused by the specific notation in the question and am unsure what exactly it is asking here/how to proceed.
    1. The problem statement, all variables and given/known data
    Given a scalar function ##f## find (a) ##∫f \vec {dl}## and (b) ##∫fdl##
    along a straight line from ##(0, 0, 0)## to ##(1, 1, 0)##.


    2. Relevant equations
    ##f(x,y,z) = 12xy + z##
    ##\vec {dl} = (\vec {dx},\vec {dy},\vec {dz})##
    3. The attempt at a solution

    So what i'm mainly confused about is with part a, I can't seem to understand what it's referring to thus don't know how to go about starting the integral.

    Is it implying that I need to parametricize the scalar function and then take the integral wrt to dl, treating it like a vector valued function approaching it with the dot product like this $$u = f(x,y,z) = 12xy + z , v = y , w = z $$ resulting in the vector function $$\vec {r}(u,v,w) = (12xy + z) \hat {\mathbf i} + y \hat {\mathbf j} + z \hat {\mathbf k} $$ with the parametrization of $$\vec {r}(t) = (t,t,0)$$ resulting in this integral $$\int (12t^2,t,0)⋅(1,1,0) dt$$

    or
    am I overthinking this and i'm simply looking find the line integral over each of the differential components like this: $$\int f \vec{dl} = \int f {\mathbf dx} \hat {\mathbf i} + \int f {\mathbf dy} \hat {\mathbf j} + \int f {\mathbf dz} \hat {\mathbf k}$$
     
  2. jcsd
  3. Jul 25, 2017 #2

    haruspex

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    Neither.
    I can rule out your first option because f is a scalar, so the integral must be a vector.
    Your second option overlooks the interaction between x and y in the computation of f.

    Parameterise the path from (0,0,0) to (1,1,0) and write f and ##\vec{dl}## in terms of that parameter.
     
  4. Jul 26, 2017 #3
    So since the parametrized path is ##\vec r(t) = t \hat i + t \hat j + 0 \hat k##, is the correct path to take then to evaluate the integral like so
    $$\int f(x,y,z) \vec {dl} = \int_0^1 f(x(t),y(t),z(t))(\vec {dx} \hat i + \vec {dy} \hat j +\vec {dz} \hat k)$$
    $$= \int_0^1 (12t^2)x'(t)dt \hat i + \int_0^1 (12t^2)y'(t)dt \hat j + \int_0^1 (12t^2)z'(t)dt \hat k ,$$ where $$(x'(t),y'(t),z'(t)) = (1,1,0)$$ ?
     
  5. Jul 26, 2017 #4

    haruspex

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    Yes.
     
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