logo

Definition and Properties of Vector Areas 📂Mathematical Physics

Definition and Properties of Vector Areas

Definition

vector_area.png

For a given surface SS, the following integral is called the vector area of SS.

a:=∫Sda \mathbf{a} := \int_{\mathcal{S}} d \mathbf{a}

Description

hemisphere.png

As an example, let’s calculate the vector area of a hemisphere with a radius of RR. It is da=R2sin⁡θdθdϕr^d \mathbf{a} = R^{2}\sin\theta d\theta d\phi \hat{\mathbf{r}}. Here,

r^=cos⁡ϕsin⁡θx^+sin⁡ϕsin⁡θy^+cos⁡θz^ \hat{\mathbf{r}} = \cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}}

when integrated over the region of the northern hemisphere, both x^\hat{\mathbf{x}} and y^\hat{\mathbf{y}} components cancel out, leaving only the z^\hat{\mathbf{z}} component. Thus, we obtain the following.

a=∫ϕ=02π∫θ=0π/2R2sin⁡θcos⁡θdθdϕz^=2πR2∫θ=0π/2sin⁡θcos⁡θdθz^=2πR212z^=πR2z^ \begin{align*} \mathbf{a} &= \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} R^{2}\sin\theta \cos\theta d\theta d\phi \hat{\mathbf{z}} \\ &= 2\pi R^{2} \int_{\theta=0}^{\pi/2} \sin\theta \cos\theta d\theta \hat{\mathbf{z}} \\ &= 2\pi R^{2} \dfrac{1}{2} \hat{\mathbf{z}} \\ &= \pi R^{2} \hat{\mathbf{z}} \end{align*}

The integral on θ\theta is justified by the table of integrals of trigonometric functions at (1)(1).

Properties

  1. The vector area of a closed surface is always a=0\mathbf{a} = \mathbf{0}.

  2. The vector area of surfaces with the same boundary is always the same.

  3. The following integral holds. a=12∮r×dl \mathbf{a} = \dfrac{1}{2}\oint \mathbf{r} \times d \mathbf{l}

  4. For any constant vector c\mathbf{c}, the following is true. ∮(c⋅r)dl=a×c \oint (\mathbf{c} \cdot \mathbf{r}) d \mathbf{l} = \mathbf{a} \times \mathbf{c}