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Integral of dot product

  1. Apr 26, 2013 #1
    Hi all,

    I'm working on a math problem with a known answer - though I can't reproduce the maths.

    The problem is this: there is a random 3d vector of unit length with a uniform probability, [itex]\vec{v}[/itex], and a secondary unit vector [itex]\vec{u}[/itex]. It is stated that:
    [tex]f = \int_{S^2}{| \vec{v} \cdot \vec{u} | d\vec{v}} = 2\pi[/tex]

    Now, I've never worked with integrals of this kind, and I'm not even exactly sure what [itex]S^2[/itex] means in the integral subscript, but given the problem I attempted to take a vector for [itex]\vec{u} = [0, 0, 1][/itex] (which shouldn't matter, as the distribution is uniform), and I set
    [tex]\vec{v} = [cos(\alpha)sin(\sigma), sin(\alpha)sin(\sigma), cos(\sigma)][/tex]
    for [itex]0 \leq \alpha \leq 2\pi[/itex] and [itex]0 \leq \sigma \leq \pi[/itex].

    Now I expected that:
    [tex]f = \int_{S^2}{| \vec{v} \cdot \vec{u} |d\vec{v}} = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{\pi} | \vec{v} \cdot \vec{u} |d{\sigma}d\alpha[/tex]
    [tex]f = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{\pi} | cos(\sigma) |d{\sigma}d\alpha = \int_{\alpha = 0}^{2\pi}\int_{\sigma = 0}^{{1 \over 2} \pi} 2cos(\sigma)d{\sigma}d\alpha[/tex]
    [tex]f = \int_{\alpha = 0}^{2\pi} 2[sin({1 \over 2}\pi) - sin(0)]d\alpha = 4\pi[/tex]

    Which is not what the apparent solution should be. What is the error in my logic? And how can it be proven that [itex]f = 2\pi[/itex]?

    Thanks in advance
     
  2. jcsd
  3. Apr 26, 2013 #2

    mfb

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    S_2 is the unit sphere - the set of all points with distance 1 to the origin.

    Your parametrization of the sphere is not uniform, you favor points close to u (and on the opposite side) in the integral.
    This is easier to solve in spherical coordinates.
     
  4. Apr 26, 2013 #3
    Thanks for your answer.

    I did manage to reproduce the answer now. Though I didn't know integration in spherical coordinates. I actually did convert it to spherical coordinates (right?), with p = 1 (the distance from the origin), except I didn't take into account the "Jacobian determinant". I really only had to multiply it with [itex]p^2 sin(\sigma)[/itex], is that correct?
     
  5. Apr 26, 2013 #4

    mfb

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    Staff: Mentor

    I think so, yes.
     
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