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- Thread starter breez
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- #2

HallsofIvy

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A vector field [itex]\vec{f}[/itex] is said to be "conservative" (that's physics terminology; I prefer "is an exact derivative") if there exist a scalar function F, having continuous partials, such that [itex]\vec{f}= \nabla F[/itex]. IF there exist such an F, then yes, [itex]\vec{f}[/itex] is "conservative"!

- #3

nicksauce

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[tex] \int_{\partial S}\vec{F}\cdot d \vec{l} = \int_{S} \nabla \times \vec{F} \cdot d \vec{S}[/tex]

and because

[tex] \int_{\partial V}\vec{F} \cdot d \vec{S} = \int_{V} \nabla \cdot \vec{F} dV[/tex]

it follows that

[tex] \int_{\partial V}\vec{F}\cdot d \vec{l} = \int_{V} \nabla \cdot \nabla \times \vec{F} dV = 0 [/tex]

assuming F is smooth enough? The problem is that you can't go from the first theorem to the second. In the first S is an open surface (which has a boundary curve) and in the second S = delV is a closed surface (which does not have a boundary curve)

- #4

nicksauce

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The first integral in the third equation should be around delS not delV.

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