Flux & Surface Integrals
Contents
Flux & Surface Integrals#
Oriented Surfaces#
If at every point on a surface \(S\) there exist two unique unit normal vectors, \(\hat{\boldsymbol{n}}\) and \(-\hat{\boldsymbol{n}}\), with \(\hat{\boldsymbol{n}}\) varying continuously over \(S\), then \(S\) is called an oriented (or orientable or two-sided) surface.
If \(S\) is defined by \(\overrightarrow{\boldsymbol{r}}(u, v)\), the positive normal vector is in the direction of \(\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\).
For a closed surface bounding a solid region \(E\), the positive orientation is that for which normal vectors point outward from \(E\).
Flux#
The flux of \(\overrightarrow{\boldsymbol{F}}(x, y, z)\) across a surface \(S\) is given by \(\iint_{S} \overrightarrow{\boldsymbol{F}} \cdot \hat{\mathbf{n}} d S\) where \(\hat{\mathbf{n}}\) is a field of unit vectors perpendicular (‘normal’) to the surface.
Surface integrals of vector fields (flux integrals)#
If \(\overrightarrow{\boldsymbol{F}}\) is a continuous vector field defined on an oriented surface \(S\) with unit normal vector \(\hat{\boldsymbol{n}}\) then the surface integral of \(\overrightarrow{\boldsymbol{F}}\) over \(S\) is \(\iint_{S} \overrightarrow{\boldsymbol{F}} \cdot \hat{\boldsymbol{n}} d S=\iint_{S} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{S}} \quad\) where \(d \overrightarrow{\boldsymbol{S}}\) is a vector surface area element, \(d \overrightarrow{\boldsymbol{S}}=\hat{\boldsymbol{n}} d S\).
This integral is also called the flux of \(\overrightarrow{\boldsymbol{F}}\) over \(S\)
It is the integral of the normal component of \(\overrightarrow{\boldsymbol{F}}\) over \(S\)
For a closed surface, it is the flux of \(\overrightarrow{\boldsymbol{F}}\) out of \(S\) if \(\hat{\boldsymbol{n}}\) points outwards, or the flux into \(S\) if \(\hat{\boldsymbol{n}}\) points inwards
Evaluating flux integrals#
As for surface integrals of scalar fields, the method of finding \(d \overrightarrow{\boldsymbol{S}}\) and evaluating the integral depends on the surface given.
For a parametric surface \(\overrightarrow{\boldsymbol{r}}(u, v)\), we have \(d \overrightarrow{\boldsymbol{S}}=\hat{\boldsymbol{n}} d S=\overrightarrow{\boldsymbol{n}} d u d v=\frac{\partial \overrightarrow{\boldsymbol{r}}}{\partial u} \times \frac{\partial \overrightarrow{\boldsymbol{r}}}{\partial v} d u d v\)
For a surface \(z=g(x, y)\), using \(x, y\) as parameters, \(\frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y}=\left\langle-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y}, 1\right\rangle\).
So \(d \overrightarrow{\boldsymbol{S}}=\left\langle-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y}, 1\right\rangle d x d y\).
Sometimes \(d \overrightarrow{\boldsymbol{S}}\) can be deduced simply by geometry (e.g. on the faces of a rectangular box).
On the curved surface of a cylinder of radius \(a\), the direction of \(d \overrightarrow{\boldsymbol{S}}\) is radially outwards and the magnitude is \(d S=a d \theta d \phi\). On the curved surface of a sphere of radius \(a\), the direction of \(d \overrightarrow{\boldsymbol{S}}\) is radially outwards and the magnitude is \(d S=a^{2} \sin \phi d \theta d \phi\).
An application#
Gauss’s Law of electrostatics states that for a closed surface \(S\), the flux of the electric field \(\mathbf{E}\) through the surface is related to the charge \(Q\) enclosed:
This law can be used to find \(\mathbf{E}\) in many different situations.
Example
Let \(S\) be a spherical surface of radius \(r\) enclosing a single charge \(Q\) at its centre.
By symmetry, \(\mathbf{E}\) must be normal to the surface everywhere, hence