I have read in a few places that the change in direction a vector experiences when parallel transported along a closed circuit $\partial D$ (holonomy) is equal to the surface integral of the gaussian curvature inside the circuit, $\iint_D K dS=\Delta\phi(\partial D)$.
An example I understand is this: take a vector on the north pole, transport down to the equator, go around a quarter circle, go back to the north pole. This will give $\Delta\phi=\pi/2$. The integral of the curvature equals 1/8 of the area of the sphere, so also $\pi/2$. Fine.
However, what if I do almost the same thing, but don't go all the way down to the equator? I go down some amount, then go around a quarter circle along a fixed latitude and go back. Won't the angle still be $\pi/2$ in this case? But the area is smaller...
Also, books notice that this theorem only holds when the domain $D$ lies in an orthogonal parametrization or is covered by a single orhonormal frame. I am not sure what this means and how restrictive it is.