What is an unit area vector?

Question

As the title suggests,

What is a unit area vector?

I've tried googling but unable to arrive at any satisfactory answers.

Any help is appreciated.

Answer 1

Consider the surface $S$ defined by the Ampèrian loop $\partial S$. Then, a vector normal to this surface and with magnitude equal to the area of the surface is $\vec a$. This is commonly referred to as the vector area of $S$. Then, the unit vector $\hat a$ is a vector parallel to $\vec a$, but with unit length.

The total current flowing through $S$ is proportional to the integral of the current density $\vec j$ over $S$:

$$ \iint_{S}\vec j\cdot d\vec a = \iint_S \vec j\cdot \hat a\,da. $$

Generally, to determine the orientation of $\hat a$ at a point on a surface $S$, we have to parametrize $S$. For example, our surface may be defined on an open region $D$ as $S(s,t) = (x(s,t), y(s,t), z(s,t))$. For curved surfaces, the vector $\hat a$ is a function of $s$ and $t$. Using properties of the cross product, we calculate $S_s$ and $S_t$, and evaluate the cross product at a point $(S_s\times S_t)(s_0,t_0)$, to determine $\vec a(s_0,t_0)$. Determining $\hat a$ is a matter of dividing out the length. Thus, in general we have the formula for a vector surface integral of a vector field $\vec F$ over a parametrized surface $S: D\to \Bbb R^3$: $$ \iint_S \vec F\cdot d\vec a=\iint_D\vec F(\vec r(s,t))\cdot\hat a(s,t)\,ds\,dt = \iint_D\vec F(\vec r(s,t))\cdot\frac{S_s(s,t)\times S_t(s,t)}{\Vert S_s(s,t)\times S_t(s,t)\Vert}\,ds\,dt.$$

Answer 2

It is a somewhat unnatural phrasing of the geometric idea.

The area "vector" is first and foremost an area bivector :the bilinear, orientation-sensitive, parallelogram area gadget denoted $v_1 \wedge v_2$, calculable by some determinants, and thought of as a slice of surface area that can be added to other such slices. For surfaces in $3$-dimensional space that bivector can be identified with a vector using the cross product instead, $v_1 \times v_2$, to get a normal vector to the surface of length equal to the oriented area of the parallelogram with sides $v_1, v_2$.

Given a definition of unit length there is a way to standardize area and volume measurements to match that, and unit area is determined in that way. In $3$ dimensions it is the same as the normal vector having length $1$.