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I was thinking about this while solving an electrostatics problem. If we have a vector ##\vec V## such that ##\oint \vec V \cdot d\vec A = 0## for any enclosed area, does it imply ##\vec V = \vec 0##?
Can you give us a wild such case?This is true only if V is assumed continuous, otherwise its values can vary pretty wildly even in small regions.
Sure, let me think a bit. If you want just a scalar field, consider any subset S of the interior and use its characteristic function, making sure S has measure 0, e.g., for scalar fields, the Char function of ( a finite collection of) Rationals. The integral will be 0 ( and will exist for a finite collection) but the function is not identically zero.Can you give us a wild such case?
I am not saying you're not correct, I am just not that good in finding examples.
No, not at all.I was thinking about this while solving an electrostatics problem. If we have a vector ##\vec V## such that ##\oint \vec V \cdot d\vec A = 0## for any enclosed area, does it imply ##\vec V = \vec 0##?
No. This is the other way around. A constant vector field is conservative. You must not conclude from free of rotations to vanishing!Yes because you can take your enclosed area to be small enough such that ##\vec{V}## is constant in it.
Even if! My example is simply connected and the area closed if you choose a disc around zero; or I misunderstood the condition. Stokes does not mean that the integrand is zero! If you run around in an electrical potential, you will not gain or lose energy. That doesn't mean there is no field.Dont we need simple-connectedness and a simple-closed curve? Maybe this is assumed?