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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How does Green's Theorem change from the original formula if we consider a clockwise line integral instead of counterclockwise??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Original form \[\oint_\limits C \phi dx+\psi dy=\iint_\limits R(\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x})dxdy\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you get a minus answer for the integral, ie a minus area!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\iint_\limits R (\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial y})dxdy?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[\iint_\limits R (\frac{\partial \phi}{\partial y}\frac{\partial \psi}{\partial x})dxdy\] If we go clockwise ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it like switching the positive and negative charges and things would still work out(it's just a convention)
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