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Surface Integral

  1. Nov 11, 2007 #1
    May be this should have been in math section but since this came out while studying Electrodynamics i put it here
    we have
    [tex]\boxed{\int_{S} \nabla \times \vec{B}.d\vec{a}=\oint \vec{B}.d\vec{l}}[/tex]

    Q.well there are many areas with the same boundary which one to choose from?

    well if we know the area the boundary is fixed but not vice-versa does only the right side equal left but nor always the left side equals right.
    Can someone explain
     
  2. jcsd
  3. Nov 11, 2007 #2

    CompuChip

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    It doesn't matter which surface you take, that's the beauty of it. For example, if the curve on the right hand side is a circle, depending on the symmetry it may be easiest to take either a flat disk, or a half-sphere, or anything else.

    There's probably a nice proof for it too, but I wouldn't be able to give you that by heart.
     
  4. Nov 11, 2007 #3
    wow
    tat's gr8
     
  5. Nov 23, 2007 #4
    Stokes's Theorem.
     
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