Can we get area vector of any flat shape using the cross product?

It might sound lame, but can all the area be defined as vector quantity? I understand how the area of a parallelogram or a triangle is a vector. But when it comes to a circle, I don't understand. Say it is $$\pi r^2$$.

Is that something $$r\cdot r$$ = $$r^2\sin0$$ = r^2?

And also in a parallelogram, we define the area as a vector cross product. So can the cross product only define the area as a vector? And if yes, why can't dot products do the same?

Its merely the formalism of vector algebra that allows us to do this. This is only true in 3d, in higher dimensions this can't be maintained: there is no cross product for example in 4d or higher. So we can't take the product of vectors and get a vector in general.

This means areas, volumes and the like are not really represented by vectors.

Instead what are used are wedge products and these give the signed area of a parallelogram as a 'bivector' such as $u \wedge v$, (where $u$ and $v$ are vectors indicating the two sides) or the signed volume of a parallelopid as a 'trivector' $u \wedge v \wedge w$ (where $u$, $v$ and $w$ are vectors indicating the three sides). In a sense, this is more natural as it follows our intuition more closely rather than representing an area by a 1d vector.

You may be confusing two different aspects of a vector cross product.

The absolute value of a vector cross product is: $$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin(\vec{a},\vec{b}).$$

This "happens" to be the area of a parallelogram with edges $$\vec{a}$$ and $$\vec{b}$$. Proof is immediate, because $$|\vec{b}|\sin(\vec{a},\vec{b})$$ is height to the base $$\vec{a}$$. You can push this idea further and calculate area of a triangle as half the vector cross product of two edges. But that's about it. You cannot use vector cross product for calculating general areas such as a circle.

The second aspect relates to the fact that a vector cross product (not its absolute value) is a vector by itself with a direction of its own. The direction is perpendicular to both vectors ($$\vec{a}$$ and $$\vec{b}$$) or perpendicular to the plane containing these two vectors. This aspect is useful in many fields but not necessarily related to areas.

A dot product is defined as $$|\vec{a}||\vec{b}|\cos(\vec{a},\vec{b})$$ and this cannot be used for calculating an area.

• It is a very good explanation.. but i am thinking then can't we define "any" area as a vector? What if the direction of the area is important. Jan 16, 2018 at 15:48
• @D.Saha Yes you can define any plane area as a vector $A\hat{n}$ where $A$ is the magnitude of area and $\hat{n}$ is the 'outwards' unit normal. Jan 17, 2018 at 14:43

Any small segment of an area of any shape (in 3D) can be represented by a vector perpendicular to the surface. It may or may not be convenient to do this with a cross product. (The two vectors you start with must lie in the surface.) A dot product does not yield a vector result. If you are working with a closed surface, it may be customary to define the positive vectors as pointing outward.

Yes.

Area Parametrization

Consider the parametrization of a closed area where each interior point is defined by $$\vec{\rm pos}(u,v)$$ with free parameters $$u=0 \ldots 1$$, $$v=0 \ldots 1$$.

The differential area element at $$(u,v)$$ is

$${\rm d}A = \left| \frac{\partial \vec{\rm pos}}{\partial u} \times \frac{\partial \vec{\rm pos}}{\partial v} \right| \; {\rm d}u\, {\rm d}v \tag{1}$$

where $$\times$$ is the cross product and $$| \cdot |$$ is the vector magnitude.

The total area is then $$A = \int {\rm d} A$$

Example

Consider a centered ellipse with semi-major axis $$a$$ and semi-minor axis $$b$$. The interior area is defined by the following function

$$\vec{\rm pos} = \pmatrix{ u\, a \cos(2 \pi v) \\ u\, b \sin(2 \pi v) \\ 0 }$$

this makes the area element equal to

$${\rm d}A = \left| \pmatrix{ a \cos(2 \pi v) \\ b \sin(2 \pi v) \\ 0 } \times \pmatrix{ -2\pi u\, a \sin(2 \pi v) \\ 2\pi u\, b \cos(2 \pi v) \\ 0 } \right| {\rm d}u \,{\rm d}v = (2\pi\, a\, b\, u){\rm d}u \,{\rm d}v$$

And so the area of the ellipse is

$$A = \int \limits_0^1 \int \limits_0^1 (2\pi\,a\,b\,u)\,{\rm d}u {\rm d}v = (2\pi\,a\,b) \int \limits_0^1 \int \limits_0^1 u\,{\rm d}u {\rm d}v = (2\pi\,a\,b) \frac{1}{2} = \pi a b \;\checkmark$$

Volume Parametrization

The above is related to the parametrization of volume. Given the interior point of a solid as $$\vec{\rm pos}(u,v,w)$$ then the volume element is given by the vector triple product.

$${\rm d}V =\frac{\partial \vec{\rm pos}}{\partial w} \cdot \left( \frac{\partial \vec{\rm pos}}{\partial u} \times \frac{\partial \vec{\rm pos}}{\partial v} \right) \; {\rm d}u\, {\rm d}v\,{\rm d}w \tag{2}$$

which is used to define mass $$m = \int \rho {\rm d}V$$ and other volume properties.