How to find the area vector of a shape? If I have a simple $3-d$ shape like a square plate connected to an identical square plate at one edge and they are at an angle of $90\, ^{\circ}$, how would I find an area vector that describes it?
I think it might involve finding a unit vector which points in the direction of the faces of the two plates and multiplying it by the area. Is this correct?
 A: For a flat surface with area $A$, one can define its corresponding area vector as $\mathbf A = A\mathbf n$ where $A$ is the total area of the surface, and $\mathbf n$ is a normal unit vector to the surface.  
If you have two surfaces joined together as in your example, then one can define area vectors $\mathbf A_1$ and $\mathbf A_2$ for each surface independently.  However, since the area vectors of these surfaces will be orthogonal to one another, I don't personally know a reasonable physical interpretation of combining them into a single area vector, say by adding them.  Of course you can certainly make whatever definition you want, but it wouldn't be a standard one as far as I am aware.
A: The vector areas are not very hard to define. As joshphysics mentioned, the vector area is defined as perpendicular unit vector to the plane multiplied by the area of the plate itself. Hence, for a simple rectangular plate defined by two vectors $\vec{a}$ and $\vec{b}$ the vector area is described as:
$$
\vec{A} = \vec{a} \times \vec{b}
$$
And contrary to what joshphysics said, you can add vector areas, but it very much depends on the context whether the added vector areas are meaningful in a physical sense or not. Also, vector areas follow the same addition rules as simple vectors.
Just to extend the answer a bit more, you can find a vector area of any arbitrary 3D surface by using the formula below:
$$
\vec{A} = \int_S \operatorname{d}\vec{S}
$$
Where $S$ defines some surface. Bear in mind, that the vector will always point outwards of the surface. Also, an interesting thing is that, for a hemisphere, the vector area will be the same as the one from a plane circle, which can be understood by considering the way you add the infinitesimal vector areas and how some components get canceled. Also, it is easy to show that:
$$
\oint_S \operatorname{d} \vec{S} = \vec{0}
$$
