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Integration over all angles?

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose I have a relation [itex]S(\theta, \phi) = U \frac{c}{2} \cos{^{2} \, \theta} [/itex] and I want to integrate over [itex]\phi[/itex] from [itex] 0 [/itex] to [itex] 2 \pi [/itex] and [itex] \theta [/itex] from [itex] 0 [/itex] to [itex] \frac{\pi}{2} [/itex]. How do I do this double integral? Do I just do it normally (without any transformation), or do I use a total angle? I know that I'm supposed to get [itex] S = U \frac{c}{4} [/itex] but I'm not sure how it works.

    Here's an image I made: S9VStKK.png

    Angles are a little confusing for me so I would like some feedback.
     
  2. jcsd
  3. Jan 16, 2014 #2

    haruspex

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    What you have literally stated as the problem is to find ##\int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} U \frac{c}{2} \cos{^{2} \, \theta} d\theta d\phi##. But that isn't going to give the desired answer.
    What you mean, I suspect, is that you want to integrate over a surface parameterised by those angles. In that case you need to include the Jacobian. This represents the area of an element dA in terms of small changes in the two angles. See e.g. http://en.wikipedia.org/wiki/Spheri..._and_differentiation_in_spherical_coordinates.
    (But make sure you get your θ and ϕ the right way around. Not everyone uses the same notation.)
     
  4. Jan 16, 2014 #3
    What is the difference between integrating over the angles and integrating over the surface parameterised by these angles?

    I shouldn't have put S as a function. If I do an integral on each side, I get the desired result if it is an integral over the angles.
     
  5. Jan 16, 2014 #4

    haruspex

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    Integrating over the angles doesn't take any regard of the fact that they are angles. it just treats them as arbitrary variables of integration. ∫∫(function)dθdϕ.
    Integrating over a surface means ∫(function)dA, where dA is an element of area. In spherical polar, dA = r2 sin(θ)dθdϕ, where θ is the polar angle and ϕ is the azimuthal angle (which is the way you have used them).
    You mean ##\int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} U \frac{c}{2} \cos{^{2} \, \theta} d\theta d\phi##?
    I get a π2 in the value of that. Please post your working.
    OTOH, if I include the sin(θ), I still get a single factor of π, so still not the answer given in the OP.
    Please post the problem exactly as given.
     
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