Show that the area vectors for a general $n$-sided closed shape sum to zero It is possible to show that the sum of the area vectors for a general, closed, $n$-sided figure in $\mathbb{R}^3$ (3-space) is zero. 
Hint: it may be easiest to consider orientable and non-orientable surfaces separately.
 A: A three-dimensional "body" $B\subset{\mathbb R}^3$, e.g., a ball, an octahedron, or a cylinder, has a boundary $\partial B$ which is an oriented $2$-chain, i.e., a union of smooth oriented pieces of surfaces, such that the "positive normal" ${\bf n}$ points to the outside at all points ${\bf p}\in\partial B$, where it is defined. For this situation one has Gauss' divergence theorem which says that
$$\int_{\partial B}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_B{\rm div}({\bf v})\ {\rm d}{\bf x}$$
for all $C^1$ vector fields ${\bf v}$ on $B$. When ${\bf v}$ is a constant vector field then ${\rm div}({\bf v})\equiv0$, and we can write
$${\bf v}\cdot\int_{\partial B}{\bf n}\ {\rm d}\omega=\int_{\partial B}{\bf v}\cdot{\bf n}\ {\rm d}\omega=\int_B{\rm div}({\bf v})\ {\rm d}{\bf x}=0\ .$$
Since this is true for arbitrary ${\bf v}\in{\mathbb R}^3$ it follows that
$$\int_{\partial B}{\bf n}\ {\rm d}\omega=0\ .$$
A: The dot product between an area vector and an arbitrary unit length direction vector $n$ is proportional to the area you obtain by projecting the area onto a plane orthogonal to $n$. So now consider your whole closed shape projected onto that plane. For every point in the plane, you can examine the ray originating from that point an in direction $n$. That ray must intersect the faces of the shape an equal number of times, half of them coming into the shape and half going out of it. So these whill exactly cancel, and the net effect is that there is no contribution of these areas to the point from which your ray originated. Since the same argument holds for all points (with some care for those intersecting edges, vertices and so on), you can tell that there will be no net contribution to any point of the projection. So the sum of the projected area vectors has to be zero. Since this is trie for every possible choice of $n$, you can tell that the sum of area vectors has to be zero, since the sum of dot products equals the dot product with the sum, and the only vector which has zero dot product with every unit-length vector is the zero vector.
