# Question regarding Stokes' Theorem

• I
Stokes' Theorem states that:
$$\int (\nabla \times \mathbf v) \cdot d \mathbf a = \oint \mathbf v \cdot d \mathbf l$$ 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?

## Answers and Replies

Follow up: Found an answer on StackExchange says “... if the integral vanishes for all possible regions of integration, then the function is zero everywhere.", he also pointed out the formal proof of this is not easy. So I guess at my level, I should settle for this.

PeroK