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Surface integral

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate [double integral]f.n ds where f=xi+yj-2zk and S is the surface of the sphere x^2+y^2+z^2=a^2 above x-y plane.


    3. The attempt at a solution

    I know that the sphere's orthogonal projection has to be taken on the x-y plane,but I'm having trouble with the integration.
     
  2. jcsd
  3. Mar 14, 2007 #2

    HallsofIvy

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    Of course, you will have to do upper and lower hemispheres separately. One way to get the projection into the xy-plane is to find the gradient of x2+ y2+z2, 2xi+ 2yj+ 2zk, and "normalize by dividing by 2z: (x/z)i+ (y/z)j+ k. Then n dS is (x/z)i+ (y/z)j+ k dxdy.
    f.n dS is ((x2/z)+ (y2/z)- 2z) dxdy. I think I would rewrite that as ((x2/z)+ (y2/z)+ z- 3z) dxdy= ((x2+ y2+ z2)/z- 3z) dxdy= (a^2/z- 3z)dxdy. Now, for the upper hemisphere, [itex]z= \sqrt{a^2- x^2- y^2}[/itex] while for the negative hemisphere it is the negative of that. Because your integrand is an odd function of z, I think the symmetry of the sphere makes this obvious.

    Finally, do you know the divergence theorem?
    [tex]\int\int_T\int (\nabla \cdot \vec{v}) dV= \int\int_S (\vec{v} \cdot \vec{n}) dS[/tex]
    where S is the surface of the three dimensional region T. Here [itex]\nabla\cdot f[/itex] is very simple and, in fact, you don't have to do an integral at all! I wouldn't be surprized to see this as an exercise in a section on the divergence theorm.
     
  4. Mar 14, 2007 #3
    I know this is a bit embarrassing for me,but how d'you integrate
    (a^2/z- 3z)dxdy .After having substituted for z,and converted to polar co-ordinates,I get zero in the denominator!

    This is the expression:
    [double integral]a^2/(sqrt(a^2-x^2-y^2)) dx dy.
    For conversion to polar co-ords,if I substitute x=a cos(theta) and
    y=a sin(theta),the denominator becomes zero.

    (Thanks a lot for the help anyway:smile: )
     
  5. Mar 14, 2007 #4

    HallsofIvy

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    Why bother to find an anti-derivative? The function is odd in z and the region of integration is symmetric about the origin- the integral is 0.

    My point about the divergence theorem is that [itex]\nabla \cdot (x\vec{i}+y\cdot\vec{j}-2z\vec{k}) = 1+ 1- 2= 0[/tex]! The integral of that over any region is 0!
     
  6. Mar 15, 2007 #5
    Thanks a lot.:smile:
    (The question did ask not to use divergence theorem,by the way)
     
  7. Mar 15, 2007 #6

    HallsofIvy

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    Did it say you couldn't?:rolleyes:
     
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