In my textbook, a question asks to find the curl $\int\int \nabla \times F \cdot \vec{dS}$ through all faces of a cube sitting on the $xy$ plane but through the base (the base has no face). The clever answer is to realize that the integration of curl on these 5 regions of interest is the same as the line integral ($\int F \cdot \vec{dr}$) around the perimeter of the base. This line integral of the base's perimeter is also equal to the curl through this one face (as if it were there).
1) I cannot convince myself that this boundary "covers" all 5 sides of the square (ie, that the line integral of the boundary equals the integration of curl on the five higher faces). I would think that each of these faces is only describable, with a line integral, using its own perimeter.
2) If this is true, I am curious to know whether the curl through any solid with edges (and with one empty face) can be found simply by finding the line integral of the vector field on the empty face's perimeter. For example, can taking the line integral around the pentagon shown in the buckeyball give me the flux through all other faces?
Recall that $\iint_{S} {\rm curl} \> \vec{F} \cdot \vec{n} \> dS = 0$ if $S$ is a closed surface. So, the flux of curl($\vec{F}$) through all the faces except the base should be equal to the negative of the curl of $\vec{F}$ through the base, making sure that you parametrize the base with the normal vector pointing downwards.
So, we have $\iint_{S'} {\rm curl} \> \vec{F} \cdot \vec{n} \> dS = -\iint_{B} {\rm curl} \> \vec{F} \cdot \vec{n} \> dS$, where $S'$ are the non-base faces and $B$ is the base. This is probably easier to compute than a line integral since the normal vector is just $(0, 0, -1)$.
To see that it is in fact equal to the line integral around the base, recall that Stokes' Theorem requires the surface and the boundary to be piecewise-smooth, which they are.