Flux integral

For an oriented smooth surface $S$ with unit normal $\hat{\mathbf{N}}$ and a continuous vector field $\mathbf{F}$ on $S$, the flux integral (or vector surface integral) is

${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_S\mathbf{F}\cdot \hat{\mathbf{N}}dS.$

Using a parameterization $\mathbf{r}(u,v)$ where ${\mathbf{r}}_u\times {\mathbf{r}}_v$ matches the chosen orientation:

${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_D\mathbf{F}(\mathbf{r}(u,v))\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)dudv.$

The magnitudes cancel: $\hat{\mathbf{N}}dS=({\mathbf{r}}_u\times {\mathbf{r}}_v)dudv$. The same parameterization derivative that gives a magnitude factor for scalar integrals gives a vector factor for flux integrals.

Physical meaning

If $\mathbf{F}$ is the velocity of a fluid, then ${\iint }_S\mathbf{F}\cdot d\mathbf{S}$ is the rate of volume flow across $S$ in the direction of $\hat{\mathbf{N}}$ (positive if more goes that way than the opposite way).

If $\mathbf{F}$ is the electric field, then ${\iint }_S\mathbf{F}\cdot d\mathbf{S}$ is the electric flux — proportional, by Gauss’s law, to the enclosed charge.

Orientation matters

Reversing orientation flips the sign:

${\iint }_{-S}\mathbf{F}\cdot d\mathbf{S}=-{\iint }_S\mathbf{F}\cdot d\mathbf{S}.$

Unlike scalar surface integrals.

The graph shortcut

For a surface $z=f(x,y)$ over a region $D$ in the $xy$-plane, with upward orientation:

$d\mathbf{S}=⟨-f_x,-f_y,1⟩dxdy.$

The flux integral becomes

${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\iint }_D\mathbf{F}(x,y,f(x,y))\cdot ⟨-f_x,-f_y,1⟩dxdy.$

In words: replace $z$ by $f(x,y)$ in $\mathbf{F}$, dot with $⟨-f_x,-f_y,1⟩$, integrate over the planar domain. Workhorse for graph-shaped surfaces. Flip the sign for downward orientation.

Worked example: flux through a hemisphere

$\mathbf{F}=⟨0,0,z⟩$ outward through the upper unit hemisphere $x^2+y^2+z^2=1$, $z\ge 0$.

Sphere parameterization with ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }$ pointing outward at radius $1$: ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }=⟨{\sin }^2\theta \cos \varphi ,{\sin }^2\theta \sin \varphi ,\sin \theta \cos \theta ⟩$.

On the surface, $z=\cos \theta$, so $\mathbf{F}=⟨0,0,\cos \theta ⟩$. Dot with ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }$: $\cos \theta \cdot \sin \theta \cos \theta =\sin \theta {\cos }^2\theta$.

${\iint }_S\mathbf{F}\cdot d\mathbf{S}={\int }_0^{2\pi }{\int }_0^{\pi /2}\sin \theta {\cos }^2\theta d\theta d\varphi =2\pi \cdot \frac{1}{3}=\frac{2\pi }{3}.$

Worked example: graph version

$\mathbf{F}=⟨x,y,1⟩$ through $z=x^2+y^2$ for $z\le 4$, upward.

Surface is a graph with $f(x,y)=x^2+y^2$. $d\mathbf{S}=⟨-2x,-2y,1⟩dA$. Dot with $\mathbf{F}$: $-2x^2-2y^2+1$. Domain $D$ is the disk $x^2+y^2\le 4$.

In polar:

${\iint }_D(1-2r^2)rdrd\theta =2\pi {\int }_0^2(r-2r^3)dr=2\pi (2-8)=-12\pi .$

Negative: more goes downward than upward.

Symmetry shortcuts

Build the habit of looking for symmetry before grinding.

• Constant field through a flat surface: $\iint \mathbf{F}\cdot d\mathbf{S}=(\mathbf{F}\cdot \hat{\mathbf{N}})\mathrm{Area}(S)$.
• Radial field through a sphere centered at origin: if $\mathbf{F}(\mathbf{r})=g(r)\hat{\mathbf{r}}$, the flux is $g(a)\cdot 4\pi a^2$.
• Antisymmetry: if $\mathbf{F}$ is odd in some direction and $S$ is symmetric across the perpendicular plane, fluxes cancel.

These can collapse a tedious integral to a one-line answer.

Closed surfaces

A closed surface has no boundary — it bounds a 3D region. Sphere, surface of a cube, surface of a cylinder (top + bottom + lateral). The integral with outward orientation,

$\iint \mathbf{F}\cdot d\mathbf{S},$

is the total flux: net outflow from the enclosed volume. The Divergence theorem relates it to the integrated divergence over the volume.

In context

The flux integral is the 3D analog of the 2D vector line integral. The line integral asks “how much of $\mathbf{F}$ points along the curve?” via dot with tangent; the flux integral asks “how much of $\mathbf{F}$ points through the surface?” via dot with normal. Both reduce to scalar integrals after taking a dot product.

The flux integral is the right side of Stokes’ theorem and the left side of the Divergence theorem.