Flux Integral

For a particle moving counterclockwise around the closed path f $\vec{F}$ is a vector field containing surface S, then $\vec{F}$ describes the velocity of the flow or fluid at any point across the surface.

The volume of fluid (amount of flow) crossing the surface S per unit of time is called flux of F across S.

Flux:

$${\iint }_S\vec{F}\cdot \vec{N}dS$$

where $\vec{F}=<P,Q,R>$ and $\vec{N}$ is unit normal across $S$.

If a fluid has mass density $\rho (x,y,z)$ at a point then flux is`

$${\iint }_S\rho \cdot \vec{F}\cdot \vec{N}dS$$

For $\vec{F}=<P,Q,R>$ and $S$ be surface $z=g(x,y)$, $R$ is its projection on xy plane.

Flux: $\iint \vec{F}\cdot \vec{N}dS=\vec{F}$

$z=g(x,y)$

upward oriented: $z-g(x,y)=0$ => Level curve $G(x,y,z)$

$$\begin{array}{rl}\vec{N}=\nabla G=<-g_x,-g_y,1>& \\ \iint \vec{F}\cdot \frac{<-g_x,-g_y,1>}{\sqrt{g_x^2+g_y^2+1}}\cdot \sqrt{g_x^2+g_y^2+1}dA& =\iint \vec{F}\cdot <-g_x,-g_y,1>dA\end{array}$$

downward oriented: $z-g(x,y)=0$ => Level curve $G(x,y,z)$

$$\begin{array}{rl}\vec{N}=\nabla G=<-g_x,-g_y,-1>& \\ \iint \vec{F}\cdot \frac{<-g_x,-g_y,-1>}{\sqrt{g_x^2+g_y^2+1}}\cdot \sqrt{g_x^2+g_y^2+1}dA& =\iint \vec{F}\cdot <-g_x,-g_y,-1>dA\end{array}$$

Examples §

Example 1 §

Find the flux of $\vec{F}$ across S where $\vec{N}$ is upward-oriented.

$$\begin{array}{c}\vec{F}=<x,2y>\\ s:z=6-3x-y\end{array}$$

Solution §

$$\begin{array}{rl}\iint \vec{F}\cdot <-g_x,-g_y,1>dA& =\iint <x,2y>\cdot <3,1,1>dxdy\\ y=6-3x& ⟹0\le x\le 2,0\le y\le 6-3x\\ & ={\int }_{x=0}^{x=2}{\int }_{y=0}^{y-6-3x}3x+2ydydx\\ & =36\end{array}$$