Why is curl zero for a closed surface?

Question

Why does $\iint_S \operatorname{curl}F \cdot dS = 0$ for any $C^1$ vector field if $S$ is a closed surface?

Is it because the flows cancel out?

Answer 1

By Stokes' theorem, $$\int \bigtriangledown \times \mathbf{v} \cdot \boldsymbol{da} = \oint \mathbf{v} \cdot \boldsymbol{dl}$$

i.e. We choose a closed path over whatever surface we are given and integrate its divergence with the vector field to get the left hand side of our equation(dot product of curl of v).

Think of a disc made of clay. It is its circumference that forms the boundary.

Now let me remodel the disc into a bowl with a wide open mouth. You can see that the portion of the disc which was previously the circumference has now become the rim of the bowl. Let us choose its rim as a closed path for our line integral and find the value for $$ \oint \mathbf{v} \cdot \boldsymbol{dl} $$ This will give us our $\iint (\nabla\times F)\cdot da=0$ .

Now, if I further attempt to turn that clay bowl(an open surface) into a sphere(a closed surface) by trying to close its mouth entirely, I would have to reduce the size of rim to an infinitesimally small point.

This leaves me with no line integral to choose on the surface. Therefore my $$ \oint \mathbf{v} \cdot \boldsymbol{dl} $$ will be $0$. Hence, we'll be left with $\iint (\nabla\times F)\cdot dS=0$ as per Stokes' theorem.