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?


4 Answers 4


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.

  • $\begingroup$ 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. $\endgroup$
    – sdebarun
    Jan 16, 2018 at 15:48
  • 1
    $\begingroup$ @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. $\endgroup$ 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.



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 $A = \int {\rm d} A$


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.


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