My book, Mathematical Methods for Physics and Engineering - K. F. Riley,
explains that the 'Integral definition of curl' is:-
$$\nabla\times a = \lim_{V\to0}(\frac{1}{V}\int_S dS\times a) $$
and then, after a few pages, it mentions:-
$$\int_S dS \times \nabla\phi = \int_C\phi dr$$
Now, if we take $a = \nabla\phi$ in the first equation, then $\int_S dS\times a$ becomes $0$ (because the curl of the gradient of anything is $0$). But this contradicts with the second equation.
So, what is the mistake here?
Presumably, in the first definition $S$ is a closed surface (i.e. a surface with no boundary), and $V$ is the volume enclosed by the surface.
In the second definition, $S$ is not necessarily a closed surface. $S$ may have a boundary, and the boundary of $S$ is the curve $C$.
But in the special case where $S$ is closed, then $C$ is an empty curve, so $\int_C \phi dr = 0$. There is therefore no contradiction.