Introduction

We will start by considering a very special case of flux and then slowly generalize until we have the flux integral. For now we will be considering flowing water. We wish to find the rate at which water is flowing through the surface S, when the velocity of the water is $\vec{v}$.

1.

Consider a flat surface S, which has area $\Delta S$, and has been oriented with a direction of positive flow $\vec{n}$. If water is flowing through the surface in the direction of $\vec{n}$ with a constant speed $\left\Vert \vec{v}\right\Vert$, what is the rate at which water is flowing through S? The rate will be the total volume of water which flows past S in a time $\Delta t$ divided by $\Delta t$. First compute the total volume that flows past S. The water flowing through S forms a cylinder with a height of $\left\Vert \vec{v}\right\Vert \Delta t$ and a base with area $\Delta S$. The total volume of water will be

$$V=\left\Vert \vec{v}\right\Vert \Delta t\Delta S$$

and the rate (or flux) of water through the surface will be

$${\mbox{rate}}=\frac{V}{\Delta t}=\left\Vert \vec{v}\right\Vert \Delta S$$

2.

Next remove the condition that the direction of flow $\left( \vec{v}\right)$ is the same as $\vec{n}$. Now we have a skewed cylinder with base S and height $\left\Vert \vec{v}\right\Vert \cos \theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{n}$.

(illus of skewed cylinder)

The total volume is

$$V=\left\Vert \vec{v}\right\Vert \Delta t\Delta S\cos \theta$$

If we insist that $\vec{n}$ be a unit vector (I'll start using $\hat{n}$), then we can write the volume as

$$V =\left( \left\Vert \vec{v}\right\Vert \left\Vert \hat{n}\right\Vert \cos \theta \right) \Delta t\Delta S$$

$$=\left( \vec{v}\cdot \hat{n}\right) \Delta t\Delta S$$

and the rate is  

$${\mbox{rate}}=\left( \vec{v}\cdot \hat{n}\right) \Delta S$$ (1)

3.

Finally we can dispense with the assumptions that S is flat and that $\left\Vert \vec{v}\right\Vert$ is constant. In this case we divide the surface S into many small patches, each of which is nearly flat. Now we can assume $\left\Vert \vec{v}\right\Vert$ is nearly constant on each little patch and use case 2. To get the total flux we add up the flux over all of the small patches. Divide S into k patches $S_{1},S_{2},\cdots ,S_{k}$ each of which has area $\Delta S_{1},\Delta S_{2},\cdots ,\Delta S_{k}$ and normal vector $\hat{n}_{1},\hat{n}_{2},\cdots ,\hat{n}_{k}$. Take the velocity at the center of the j^th patch (or anywhere on the patch) to be $\vec{v}_{j}$. Using $\left( \ref{flux}\right)$the flux through the j^th patch is given by

$$\left( \vec{v}_{j}\cdot \hat{n}_{j}\right) \Delta S_{j}$$

and the total flux is the sum over all j:

$${\mbox{flux}}\simeq \sum_{j=1}^{k}\left( \vec{v}_{j}\cdot \hat{n}_{j}\right) \Delta S_{j}$$

In the limit that $k\rightarrow \infty$ and $\Delta S_{j}\rightarrow 0$ the sum is replaced with an integral and we write  

$${\mbox{flux}}=\int_{S}\vec{v}\cdot \hat{n}dS$$ (2)

Here dS is an infinitesimal piece of surface area. If we define the vector field $\bar{F}$ by

$$\bar{F}=\vec{v}$$

then (2) can be written as

$$\int_{S}\bar{F}\cdot \hat{n}dS$$

any integral of this form is called a flux integral. Sometime the quantity $\hat{n}dS$ is defined to be the area vector $d\vec{A}$ and () will be written as

$$\int_{S}\bar{F}\cdot d\vec{A}$$

Note:

$$\int_{S}dS={\mbox{Surface area of }}S$$

Examples:

1.

Compute the flux of the vector field $\vec{F}=4z\hat{\imath}+e^{z^{2}}\hat{j}+\sin \left( x^{2}z\right) \hat{k}$ through the triangle with vertices (0,0,0), (0,2,4), and (0,0,4) and oriented in the +x direction. This triangle lies in the yz-plane (since x is zero for all of the points) so the normal vector is $\hat{n}=\pm \hat{\imath}$.

The problem says that the surface is oriented in the positive x direction so we will use $\hat{n}=\hat{\imath}$. An infinitesimal piece of area in the yz-plane looks like dS=dydz (because the surface is flat). Therefore $d\vec{A}=\hat{n}dS=\hat{\imath}dydz$. The flux through the triangle is

$$\int_{S}\vec{F}\cdot d\vec{A} =\int_{S}\left[ 4z\hat{\imath}... ...\sin \left( x^{2}z\right) \hat{k}\right] \cdot \hat{\imath}dydz$$

$$=\int_{S}4zdydz=\int_{0}^{2}\int_{2y}^{4}4zdzdy=\frac{128}{3}$$

So the flux of $\vec{F}$ through the triangle is $\frac{128}{3}$.

2.

Find the flux of $\vec{F}=\frac{\vec{r}}{\left\Vert \vec{r}\right\Vert ^{3}}$ through the sphere x²+y²+z²=R². First we notice that the normal vector must point radially inward or outward. By convention, the normal vector of a closed surface is taken to be outward. So $\hat{n}=\frac{\vec{r}}{\left\Vert \vec{r}\right\Vert }$, and $\vec{F}\cdot \hat{n}=\frac{\vec{r}\cdot \vec{r}}{\left\Vert \vec{r}\right\Vert ^{4}}=\frac{1}{\left\Vert \vec{r}\right\Vert ^{2}}=\frac{1}{R^{2}}$ since $\left\Vert \vec{r}\right\Vert =R$ everywhere on the sphere. The flux integral is thus

$$\int_{S}\vec{F}\cdot \hat{n}dS=\int_{S}\frac{1}{R^{2}}dS=\frac{1}{R^{2}}\int_{S}dS$$

The last integral is the surface area of a sphere of radius R.

$$\int_{S}\vec{F}\cdot \hat{n}dS=\frac{1}{R^{2}}\left( 4\pi R^{2}\right) =\frac{1}{4\pi }$$