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Surface integral of half sphere

  1. Dec 21, 2008 #1
    1. The problem statement, all variables and given/known data

    I am trying to sort out surface integrals in my head, and have become more confused when attempting to calculate the surface integral of a hemisphere. I am getting confused about which values to use as boundaries.



    2. Relevant equations

    da=R^2 sinθdθdφ

    3. The attempt at a solution



    To get half a sphere I integrated dφ between 0 and 2pi, and integrated sinθdθ between 0 and pi/2.

    This gives R^2*0*2pi=0

    Whereas if I integrate dφ between 0 and pi, and integrate sinθdθ between 0 and pi i get

    R^2*1*pi

    which is the answer I am looking for but shouldn't I get the same using both sets of boundary values? Confused...
     
  2. jcsd
  3. Dec 21, 2008 #2

    HallsofIvy

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    ??? Not if you integrate correctly! [itex]d\phi[/itex], integrated from 0 to [itex]2\pi[/itex] is, of course, [itex]2\pi[/itex] and [itex]sin\theta d\theta[/itex] integrated from 0 to [itex]\pi/2[/itex] is [itex]-cos\theta[/itex] evaluated between 0 and [itex]2\pi[/itex]: -(0- (1))= 1, not 0. The half sphere area is [itex]2\pi R^2[/itex].

    If, instead, you integrate [itex]d\phi[/itex] from 0 to [itex]\pi[/itex] and [itex]sin\theta d\theta[/itex] from 0 to [itex]\pi[/itex] you get [itex]\pi[/itex] for the first integral and [itex]-cos(\theta)[/itex] evaluated from 0 to [itex]\pi[/itex], which is -(-1- (1))= 2 for the second, again getting [itex]2\pi R^2[/itex].
     
  4. Dec 21, 2008 #3
    Oh right, I always forget that cos(0) isn't just 0. You would think I'd have learnt that by now. Thanks!
     
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