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Hangup with vector integrals

  1. Oct 19, 2013 #1

    \sigma_{b} = \vec{P}\bullet\hat{n}

    Now, integrate both sides over a closed surface,

    \oint \sigma_{b} da = \oint (\vec{P}\bullet\hat{n}) da

    My math is fuzzy, and I don't really understand this next step.

    \oint \sigma_{b} da = \oint \vec{P} \bullet d\vec{a}

    What's going on here?

    Thank you for your time!
  2. jcsd
  3. Oct 19, 2013 #2


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    da is the SCALAR area element, a positive number.
    [itex]d\vec{a}\equiv\vec{n}da[/itex] is the ORIENTED area element, a vector in direction of the local normal vector, and with magnitude da.
  4. Oct 19, 2013 #3


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    It's simply a matter of notation. "da" is the "differential of area". "[tex]d\vec{a}[/tex]" is defined as the unit normal vector times da. So [itex]\vec{P}\cdot\vec{n}da= \vec{P}\cdot\ieft(\vec{n}da\right)= \vec{P}\cdot d\vec{a}[/itex].
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