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Line integral with the inverse square field

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Integrate F(x) = x / |x|^3 along the straight line from (1,0,0) to (2,-2,1).

    2. Relevant equations

    Line integral = int (F dot dx)

    3. The attempt at a solution

    I don't know where to start. Usually I do line integrals by parameterizing the line and the vector field but I'm not sure what to do with this. Also, is the field centered at (1,0,0)? I don't really understand the question that well. Is this a conservative vector field? How can I check? Can I use the fundamental theorem of calculus? Is there some sort of potential function I can calculate? I just need someone to bump me in the right direction towards those questions.
     
  2. jcsd
  3. May 11, 2010 #2

    lanedance

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    its a conservative vector field if it can be written in terms of some scalar potential, say [itex] \phi (x)[/itex] .

    then the vector field can if the gradient of the field [itex] \vec{F}(x) = \nabla \phi (x)[/itex]
     
  4. May 11, 2010 #3

    lanedance

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    if you can do that, the line intergal will just be the diffenerence in potential
     
    Last edited: May 12, 2010
  5. May 11, 2010 #4
    How would I go about figuring out what the potential function is?
     
  6. May 12, 2010 #5

    lanedance

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    well, that depends exactly what your vector field is, as its not very clear, but wirte the partial dereivatives in terms of the functions & see if anything jumps out at you

    otherwise, parameterise your line, in term of say t, then find the vector field at each point on teh line given by t and carry out the line intergal. dx will be in a constant dierction along the line

    so do you actually mean:

    [tex] \vec{F}(x)

    = \frac{\vec{x}} {|\vec{x}|^3}

    = \frac{1}{ (x^2+y^2+z^2)^{\frac{3}{2}} }

    \begin{pmatrix} x \\ y \\ z \end{pmatrix} [/tex]
     
  7. May 12, 2010 #6
    Yeah, that's what I meant...I wasn't sure how to write vector hats. What would the partial derivatives be? I don't know how to express that line without parameterizing it in terms of t, but if t is my only variable then how can I write a gradient field?
     
  8. May 12, 2010 #7

    lanedance

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    click on the tex if you want to see how its written

    as for doing the line integral, its fine to parametrise in terms of t. you will end up with something like:
    [tex] \vec{p}(t) = \vec{a} + t \vec{u}[/tex]

    now evaluate value of the gradient field at each point on the line, [itex] \vec{F}(\vec{p}(t)) [/itex].
     
  9. May 12, 2010 #8

    lanedance

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    as for finding the potential for, it doesn't always exist for an arbitrary vector field

    a vector field derived from a scalar potential always has zero curl, so that could be one check

    One way to find it is to equate the form of [tex] \nabla \phi(x) [/tex] with your vector field, and see if the resulting partial differntial equations are eay to solve.

    In this case, due to the symmetry of the field I would try first converting ther fiedl to spherical coordinates, then looking at the gradient in those coords.
     
    Last edited: May 13, 2010
  10. May 12, 2010 #9
    Ok I think I got it. Thanks so much for your help!
     
  11. May 13, 2010 #10

    lanedance

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    no worries - its worth seeing both methods & the potential will usually be quicker if its easy to find
     
    Last edited: May 13, 2010
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