Trouble understanding a common vector calculus example I have difficulty understanding the following vector calculus example. Text can be found here. It is the 5th Q&A -- starting with equation (31.1035).It concerns finding the vector potential of a current loop.
I get the argument up to $$\vec A=\frac{\mu I}{4\pi r^3}\oint \vec r'\cdot \vec r \,\,\,d\vec l'$$.
But I don't understand why $$\oint \vec r'\cdot \vec r \,\,\,d\vec l'=\left(\int d\vec a\right)\times \vec r$$
[ADDED]: Definitions of the terms:
This describes a current loop, with a current $I$ flowing in it.
$\vec r$ is the position vector of an arbitrary point in space. $\vec r'$ is a position vector of a point on the loop. $d\vec l'$ is an infinitesimal line element along the loop.
$\vec A(\vec r)$ is the  vector potential, given by $$\frac{\mu I}{4\pi}\oint \frac{d\vec l'}{|\vec r'-\vec r|}$$
The above expression is obtained by Taylor expanding this definition.
$d\vec a$ is an infinitesimal area element of the enclosed area in the loop with direction pointig out of the surface enclosed by the loop. So that $$\int d\vec a$$ is the area enclosed by the loop times the unit vector $\hat n$ pointing out of the plane (supposing the loop is planar...)

The first thing I tried is the vector triple product identity 
$$(\vec r'\cdot \vec r )\,\,\,d\vec l'=\vec r\times(d\vec l'\times \vec r')+(\vec r\cdot d\vec l')\vec r'$$
The first term looks like it's doing the right sort of thing, but I don't see why the second term vanishes (or fits in) ...Maybe I have missed out something quite obvious?

This expression relates a loop integral with an area integral, so it has to be related to Green's theorem.
