Flux in three dimensional vector field

1. Space vector field

At every point in space , There are $\overrightarrow{F}=P\hat{i}+Q\hat{j}+R\hat{k}$
among $P$,$Q$,$R$, Are all $x$,$y$,$z$ Function of

2. Traffic

stay 2 Weizhong , The vector field crosses the curve C The flow of is recorded as
$${\int }_C\overrightarrow{F}\cdot \hat{n}ds$$
stay 3 Weizhong , The vector field passes through the surface S The flow of is recorded as
$${\iint }_S\overrightarrow{F}\cdot \hat{n}dS$$
For the sake of representation , remember $d\overrightarrow{S}=\hat{n}dS$

3. take $d\overrightarrow{S}$ Convert to $dxdy$

（1） The surface display is represented as $z=f(x,y)$

As shown in the figure , Hypothetical surface $z=f(x,y)$ On a small piece $\Delta S$ stay xoy The projection on the axis is rectangular ABCD, Because the block is small enough , You can think of it as a parallelogram .

rectangular $AD$ The length of is $\Delta x$（x Amount of change ）,$AB$ The length of is $\Delta y$（y Amount of change ）
set up $E$ The coordinates of the points are $(x,y,f(x,y))$, be $F$ The coordinates of the points are $(x,y+\Delta y,f(x,y+\Delta y))$, among $f(x,y+\Delta y)$ Linear approximation can be used , namely $f(x,y+\Delta y)\approx f(x,y)+\Delta y\cdot f_y^{\prime }$, be $F$ The coordinates of the points are $(x,y+\Delta y,f(x,y)+\Delta y\cdot f_y^{\prime })$, The same can be $H$ The coordinates of the points are $(x+\Delta x,y,f(x,y)+\Delta x\cdot f_x^{\prime })$, be
$$\overrightarrow{EF}=\left(0,\Delta y,\Delta y\cdot f_y^{\prime }\right)$$
$$\overrightarrow{EH}=\left(\Delta x,0,\Delta x\cdot f_x^{\prime }\right)$$
$$\Delta \overrightarrow{S}=\overrightarrow{EF}\times \overrightarrow{EH}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& \Delta y& \Delta y\cdot f_y^{\prime }\\ \Delta x& 0& \Delta x\cdot f_x^{\prime }\end{array}\right|=\left(\Delta x\Delta y\cdot f_x^{\prime },\Delta x\Delta y\cdot f_y^{\prime },-\Delta x\Delta y\right)=\left(f_x^{\prime },f_y^{\prime },-1\right)\Delta x\Delta y$$
（ explain ：$\left|\overrightarrow{EF}\times \overrightarrow{EH}\right|$ It's a parallelogram EFGH The area of ,$\overrightarrow{EF}\times \overrightarrow{EH}$ The direction of is plane EFGH The direction of the normal vector of ）

therefore
$$d\overrightarrow{S}=\pm \left(f_x^{\prime },f_y^{\prime },-1\right)dxdy$$
The sign depends on which side is positive .

（2） Parametric surfaces

Let's say you can put a surface $S$ use $u$,$v$ Two variables represent , namely $\{\begin{array}{l}x=x\left(u,v\right)\\ y=y\left(u,v\right)\\ z=z\left(u,v\right)\end{array}$ Then the position vector $\overrightarrow{r}=\overrightarrow{r}\left(u,v\right)$.
Considerations and parameters $\Delta u$ and $\Delta v$ The surface corresponding to the change of .
because $\Delta u$ and $\Delta v$ smaller , So you can think of a surface as a parallelogram , The two sides are $\frac{\partial \overrightarrow{r}}{\partial u}\Delta u$ and $\frac{\partial \overrightarrow{r}}{\partial v}\Delta v$
therefore
$$\Delta \overrightarrow{S}=\pm \left(\frac{\partial \overrightarrow{r}}{\partial u}\Delta u\right)\times \left(\frac{\partial \overrightarrow{r}}{\partial v}\Delta v\right)$$
therefore
$$d\overrightarrow{S}=\pm \left(\frac{\partial \overrightarrow{r}}{\partial u}\times \frac{\partial \overrightarrow{r}}{\partial v}\right)dudv$$
The sign depends on which side is positive .

（3） The surface is implicitly expressed as $g(x,y,z)=0$

The normal vector of the surface $\overrightarrow{N}=\nabla g$.
Take a small slanted surface （ One of the edges is horizontal , And it is small enough to be regarded as a small plane ）, Set the surface and horizontal plane $xoy$ An Angle of $\alpha$, As shown in the figure .

Then the surface area $\Delta S$ And in $xoy$ The area of the projection on the axis $\Delta A$ The relationship is
$$\Delta S\cdot \cos \alpha =\Delta A$$
（ notes , It is not counted as $\Delta x\Delta y$ Because the two sides of the projection do not necessarily coincide with x Axis or y Axis parallel ）
because
$$\cos \alpha =\frac{\hat{k}\cdot \overrightarrow{N}}{\left|\hat{k}\right|\cdot \left|\overrightarrow{N}\right|}=\frac{\hat{k}\cdot \overrightarrow{N}}{\left|\overrightarrow{N}\right|}$$
be
$$dS=\frac{\left|\overrightarrow{N}\right|}{\hat{k}\cdot \overrightarrow{N}}dA$$
$$\hat{n}dS=\frac{\left|\overrightarrow{N}\right|\hat{n}}{\hat{k}\cdot \overrightarrow{N}}dA=\pm \frac{\overrightarrow{N}}{\hat{k}\cdot \overrightarrow{N}}dA=\pm \frac{\overrightarrow{N}}{\hat{k}\cdot \overrightarrow{N}}dxdy$$
The sign depends on which side is positive .

4. Divergence theorem （ Gauss formula ）

If $S$ It's a region $D$ Closed surface of , The direction of the curve normal vector is outward , And the vector $\vec{F}$ In area D It is internally defined and differentiable , be
$${\iint }_S\overrightarrow{F}\cdot d\overrightarrow{S}={\iiint }_{D}div\left(\overrightarrow{F}\right)dV,\text{among }div\left(\overrightarrow{F}\right)=\left(P_x+Q_y+R_z\right)$$