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Quick question - Flux through a sphere

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫∫r.ndS where r=(x,y,z) and n is a normal unit vector to the surface S, which is a sphere of radius a centred on the origin.

    2. The attempt at a solution

    I decided to use polar coordinates. The radius of the sphere is clearly constant, a. So a surface element is dS = a2dθdø.

    r = a(sinθcosø, sinθsinø, cosθ)

    n = (sinθcosø, sinθsinø, cosθ)

    r.n = a

    therefore, ∫∫r.ndS = ∫∫a3dθdø

    where θ varies from 0 to 2π, and ø varies from 0 to π.

    This gives an answer of 2π2a3. Is this correct? I'm not so sure.
     
  2. jcsd
  3. Feb 13, 2014 #2

    vanhees71

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    Science Advisor
    2016 Award

    You forgot that
    [tex]\mathrm{d} S=a^2 \sin \theta \mathrm{d} \theta \mathrm{d} \phi.[/tex]
     
  4. Feb 13, 2014 #3
    Good point, thank you.

    But then I get ∫∫a3sinθdθdø

    Which evaluates to zero. Are the limits incorrect?
     
  5. Feb 13, 2014 #4

    pasmith

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    Homework Helper

    Yes: If [itex]z = r\cos\theta[/itex] (there are two exactly opposite conventions for the angle coordinates) then [itex]dS = r^2 \sin\theta\,d\theta\,d\phi[/itex] and the sphere is given by [itex]0 \leq \theta \leq \pi[/itex] and [itex]0 \leq \phi \leq 2\pi[/itex].

    Also: once you have that [itex]\mathbf{r} \cdot \mathbf{n} = a[/itex] you know that the answer must be [itex]a[/itex] times the surface area of a sphere of radius [itex]a[/itex].
     
  6. Feb 13, 2014 #5
    Oh, I had theta and phi mixed up. So that gives

    ∫∫r.ndS = ∫(2a3)dø

    from 0 to 2π, hence

    = 4πa3
     
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