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Easy flux surface integral

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data
    find the flux of the field F(vector) across the portion of the sphere x^2+y^2+z^2=a^2 in the first octant directed away from the origin



    2. Relevant equations
    F(x,y,z)=zk(hat)



    3. The attempt at a solution
    i used Flux=double integral over x-y plane F.n(unit normal)dsigma
    where dsigma=abs(grad(g))/abs(grad(g).p(unit normal to surface))
    i let g=x^2+y^2+z^2-a^2=0
    and grad(g)=(2x,2y,2z)
    p=n=(x,y,z)/sqrt(x^2+y^2+z^2)

    and after some algebra i got:
    double integral( (a^2-x^2-y^2)/a dA)
    which i used polar co-ordinates
    to get 1/a double integral( a^2-r^2)r dr dtheta
    which gave me pi*a^3/8 yet the solutions in back of book is pi*a^3/6

    Looks like iv made a simple error somewhere(iv looked like 3 times but cant find one), but i just need to check im using the right method.
    sorry bout the way i worte out the maths but im unaware how to put it in nice maths form
    Can any1 help please?

    cheers,
    cos(e)
     
  2. jcsd
  3. Jun 6, 2009 #2
    ok i get the answer with p=k(hat)
    now im sure why im letting p=k(hat), is it because p is the normal to the area of the double intergral i take, in this case the x-y plane with theta varying from 0 to pi/2 adn r varying from 0 to a?
     
  4. Jun 6, 2009 #3
    If we call the integral I, then by symmetry, the integral over the entire surface of the sphere is 8 I. By the divergence theorem, we have:

    8 I = Integral of div F over volume = 4/3 pi a^3 -------->

    I = pi a^3/6
     
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