- #1

Laudator

- 18

- 1

$$\int (\nabla \times \mathbf v) \cdot d \mathbf a = \oint \mathbf v \cdot d \mathbf l$$ Now, if for a specific situation, I can work out the RHS and it's equal to zero, does it necessarily mean that ##\nabla \times \mathbf v = 0##? I mean all that tells me is that the surface integral on the LHS is zero and integrals are like infinite summations, could all those tiny ##(\nabla \times \mathbf v) \cdot d \mathbf a## magically cancel each other (for any arbitrary surfaces, not just one specially chosen) yet left ##\nabla \times \mathbf v## none zero? And if this possibility doesn't exist, how to prove it?