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Line Integral in E & M

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the line integral of A between to points x=1 and x=3 along the semicircular path with a center at x=2

    2. Relevant equations

    A=kx in the x hat direction.

    [tex]\int A \bullet dl [/tex]

    dl = ds s hat +s d[tex]\phi[/tex] [tex]\phi[/tex] hat +dz z hat


    3. The attempt at a solution

    My biggest problem seems to be getting that A into cylindrical coordinates so I can take the dot product with the dl.
     
    Last edited: Sep 12, 2009
  2. jcsd
  3. Sep 12, 2009 #2

    gabbagabbahey

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    HI msd213, welcome to PF!:smile:

    You can write this more clearly as follows (click on the [itex]\LaTeX[/itex] images to see the code used to generate them!):

    [tex]\textbf{A}=kx\mathbf{\hat{x}}[/tex]

    [tex]\int \textbf{A}\cdot d\textbf{l}[/tex]

    [tex]d\textbf{l}=ds\mathbf{\hat{s}}+s d\phi\mathbf{\hat{\phi}}+dz\mathbf{\hat{z}}[/tex]

    In cylindrical coordinates, centered at [itex]x=2[/itex] (since that is the center of your curve, you might as well choose the origin of your cylindrical coordinates to be there) [itex]x-2=s\cos\phi[/itex] and [itex]\mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}[/itex]
     
  4. Sep 13, 2009 #3
    Thanks for your response but I'm still confused. How did you derive that [itex]\mathbf{\hat{x}}[/itex]?
     
  5. Sep 13, 2009 #4

    gabbagabbahey

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    I didn't derive it...I looked it up!:smile:

    The derivation is done in most Multi-variable calculus texts, and I think is done in an Appendix in Griffiths' Introduction to Electrodynamics as-well. But it is really just a matter of drawing a picture and doing some very basic trig.
     
  6. Sep 13, 2009 #5
    I think I see now. But when you take the dot product between the A (now converted into cylindrical coordinates, and dl, aren't you going to have terms multiplied by differentials such as (some term)*[tex]ds[/tex] +(another term)*[tex]d\phi[/tex]...etc. How can you take the integrals with that? Where does the volume element fit in?
     
    Last edited: Sep 13, 2009
  7. Sep 13, 2009 #6

    gabbagabbahey

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    Well, for your particular curve (the semicircle), [itex]s=1[/itex] and [itex]z[/itex] are both constant, so [itex]ds=dz=0[/itex] and hence [itex]d\textbf{l}=d\phi\mathbf{\hat{\phi}}[/itex]

    In general, any 3D curve can be described by a single parameter. So, in cylindrical coordinates, for example, you would have something like [itex]s=s(u)[/itex], [itex]\phi=\phi(u)[/itex] and [itex]z=z(u)[/itex] and so all the differentials ([itex]ds[/itex], [itex]d\phi[/itex] and [itex]dz[/itex]) can be written in terms of a single differential [itex]du[/itex]. (For your particular curve, the easiest way is just to use [itex]\phi[/itex] as your parameter)

    If instead you wanted to integrate over a surface, you would express each coordinate in terms of two independent parameters, and your surface area element in terms of their differentials.
     
  8. Sep 13, 2009 #7
    I'm still a little confused about how this clears up the problem with the repeated differentials. For example, in my problem, isn't the integral going to look like.

    [tex]\int -k s^2\\\ cos\phi\\ sin\phi\\ d\phi\\\ s\\ ds\\ d\phi\\ dz[/tex]

    Using [itex]\mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}[/itex]
    and [tex]kx=kscos\phi[/tex]

    I'm not sure how I can write these in terms of one parameter.
     
  9. Sep 13, 2009 #8

    gabbagabbahey

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    No,

    [tex]\begin{aligned} \textbf{A} & = kx\mathbf{\hat{x}} \\ & = k(s\cos\phi+2)\left(\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}\right) \\ & = k(\cos^2\phi+2\cos\phi)\mathbf{\hat{s}}-k(\sin\phi\cos\phi+2\sin\phi)\mathbf{\hat{\phi}} \end{aligned}[/tex]

    (Since [itex]s=1[/itex] along your curve)

    And [itex]d\textbf{l}=d\phi\mathbf{\hat{\phi}}[/itex], so

    [tex]\begin{aligned} \textbf{A}\cdot d\textbf{l} & = \left(k(\cos^2\phi+2\cos\phi)\mathbf{\hat{s}}-k(\sin\phi\cos\phi+2\sin\phi)\mathbf{\hat{\phi}}\right)\cdot(d\phi\mathbf{\hat{\phi}}) \\ & = -k(\sin\phi\cos+2\sin\phi)\phi d\phi\end{aligned}[/tex]
     
    Last edited: Sep 13, 2009
  10. Sep 13, 2009 #9
    Thanks for your patience and prompt responses.

    Whenver I use cylindrical coordinates, I suppose I just reflexively put in that [tex]s\\ ds\\ d\phi\\ dz[/tex] whenever I convert to cylindrical coordinates and integrate. Why exactly isn't it needed here?
     
  11. Sep 13, 2009 #10

    gabbagabbahey

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    [itex]d\tau=s ds d\phi dz[/itex] is the differential volume element....you're not performing a volume integral here, so [itex]d\tau[/itex] is pretty useless to you.
     
  12. Sep 13, 2009 #11
    Thanks, I guess I've just been conditioned from calc to put that there.

    Also, I think the whole idea of this section is path independence.

    So, when you just take the path just along the x-axis, you have

    [tex]\int_1^3 k\\ x\\ dx[/tex]

    This integral yields 4k. But the integral you derive yields zero, unless I'm way off. The bounds should be 0 to [tex]\pi[/tex] correct? Shouldn't the integrals yield the same answer?
     
  13. Sep 13, 2009 #12

    gabbagabbahey

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    Right, sorry my post contained an error (substituted [itex]x=(s-2)\cos\phi[/itex] instead of [itex]x-2=s\cos\phi[/itex])...i'll edit it to fix the error...
     
  14. Sep 13, 2009 #13
    Thank you, that should clear everything up. You've been a great help.:smile:
     
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