Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Integral of dot product

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