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Integration including unit vectors

  1. Apr 7, 2014 #1
    I have an integral of aΘ cos(Θ) dΘ

    a is the unit vector for Θ.

    I'm not sure what to do with it in the integration. I know the unit vector equals a/abs(a) but that would give a mess of an integral cause of the abs(a).
     
  2. jcsd
  3. Apr 7, 2014 #2

    jbunniii

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    Can you express the unit vector in terms of ##\theta##?
     
  4. Apr 7, 2014 #3
    Let this be the derivative of (A(θ)aθ(θ)+B(θ)ar(θ)) with respect to θ. Solve for A and B.

    Chet
     
  5. Apr 7, 2014 #4
    I don't get it.
     
  6. Apr 7, 2014 #5
    Yes i did. I can write it as Θ/abs(Θ)
     
  7. Apr 7, 2014 #6
    Take the derivative of the expression I gave with respect to θ, and set it equal to the function you are trying to integrate with respect to θ. Make use of the condition that the derivative of the unit vector in the θ direction with respect to θ is equal to minus the unit vector in the r direction, and the derivative of the unit vector in the r direction with respect to θ is equal to plus the unit vector in the θ direction. You should get two ordinary differential equations in the two unknowns A and B. Solve these equations for A and B.

    Chet
     
  8. Apr 8, 2014 #7

    pasmith

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    You need to express [itex]\mathbf{a}_\theta[/itex] in cartesian components in order to integrate it:

    [tex]
    \mathbf{a}_\theta = \frac1r \left(\frac{\partial x}{\partial \theta}\mathbf{e}_x +
    \frac{\partial y}{\partial \theta}\mathbf{e}_y + \frac{\partial z}{\partial \theta}\mathbf{e}_z
    \right) = \dots?
    [/tex]
     
  9. Apr 8, 2014 #8
    I like the method you are suggesting here much better than the approach I suggested in post #6. It is much more straightforward.

    Chet
     
  10. Apr 9, 2014 #9

    HallsofIvy

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    Is this in three dimensions or two? If it is in three dimensions, what angle is [itex]\theta[/itex]? If it is in two dimensions, then the unit vector in direction [itex]\theta[/itex] can be written [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex].
     
  11. Apr 9, 2014 #10
    Actually, it should be [itex]-sin(\theta)\vec{i}+ cos(\theta)\vec{j}[/itex]. The relation you gave is for the unit vector in the radial direction (assuming θ is the angle measured counterclockwise from the x axis).

    Chet
     
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