Hangup with vector integrals

  • Thread starter tolove
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  • #1
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Main Question or Discussion Point

Given,

[itex]
\sigma_{b} = \vec{P}\bullet\hat{n}
[/itex]

Now, integrate both sides over a closed surface,

[itex]
\oint \sigma_{b} da = \oint (\vec{P}\bullet\hat{n}) da
[/itex]

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

[itex]
\oint \sigma_{b} da = \oint \vec{P} \bullet d\vec{a}
[/itex]

What's going on here?

Thank you for your time!
 

Answers and Replies

  • #2
arildno
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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.
 
  • #3
HallsofIvy
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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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