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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    How does Green's Theorem change from the original formula if we consider a clockwise line integral instead of counter-clockwise??

  2. anonymous
    • one year ago
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    Original form \[\oint_\limits C \phi dx+\psi dy=\iint_\limits R(\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial x})dxdy\]

  3. anonymous
    • one year ago
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    @IrishBoy123

  4. IrishBoy123
    • one year ago
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    you get a minus answer for the integral, ie a minus area!

  5. anonymous
    • one year ago
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    \[-\iint_\limits R (\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial y})dxdy?\]

  6. anonymous
    • one year ago
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    so \[\iint_\limits R (\frac{\partial \phi}{\partial y}-\frac{\partial \psi}{\partial x})dxdy\] If we go clockwise ?

  7. IrishBoy123
    • one year ago
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    yes, its all definitional if you go the other way you get a negative relationship between the area integral and the line integral. so we could all agree from now on that we'll go clockwise, but we could just swicth round Greens Theorem, and life would go on

  8. anonymous
    • one year ago
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    Is it like switching the positive and negative charges and things would still work out(it's just a convention)

  9. IrishBoy123
    • one year ago
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    indeed

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