flux of vector field

Let

$$\overrightarrow{U}=U_x\overrightarrow{i}+U_y\overrightarrow{j}+U_z\overrightarrow{k}$$

be a vector field in ${ℝ}^3$  and let $a$ be a portion of some surface in the vector field.  Define one ; if $a$ is a closed surface, then the of it.  For any surface element $da$ of $a$, the corresponding vectoral surface element is

$$d\overrightarrow{a}=\overrightarrow{n}da,$$

where $\overrightarrow{n}$ is the unit normal vector on the of $da$.

The flux of the vector $\overrightarrow{U}$ through the surface $a$ is the

$${\int }_a\overrightarrow{U}\cdot 𝑑\overrightarrow{a}.$$

Remark.  One can imagine that $\overrightarrow{U}$ represents the velocity vector of a flowing liquid; suppose that the flow is , i.e. the velocity $\overrightarrow{U}$ depends only on the location, not on the time.  Then the scalar product $\overrightarrow{U}\cdot d\overrightarrow{a}$ is the volume of the liquid flown per time-unit through the surface element $da$; it is positive or negative depending on whether the flow is from the negative to the positive or contrarily.

Example.  Let  $\overrightarrow{U}=x\overrightarrow{i}+2y\overrightarrow{j}+3z\overrightarrow{k}$  and $a$ be the portion of the plane  $x+y+x=1$  in the first octant ($x\geqq 0,y\geqq 0,z\geqq 0$) with the away from the origin.

One has the constant unit normal vector:

$$\overrightarrow{n}=\frac{1}{\sqrt{3}}\overrightarrow{i}+\frac{1}{\sqrt{3}}\overrightarrow{j}+\frac{1}{\sqrt{3}}\overrightarrow{k}.$$

The flux of $\overrightarrow{U}$ through $a$ is

$$\phi ={\int }_a\overrightarrow{U}\cdot 𝑑\overrightarrow{a}=\frac{1}{\sqrt{3}}{\int }_a(x+2y+3z)𝑑a.$$

However, this surface integral may be converted to one in which $a$ is replaced by its projection (http://planetmath.org/ProjectionOfPoint) $A$ on the $xy$-plane, and $da$ is then similarly replaced by its projection $dA$;

$$dA=\cos \alpha da$$

where $\alpha$ is the angle between the normals of both surface elements, i.e. the angle between $\overrightarrow{n}$ and $\overrightarrow{k}$:

$$\cos \alpha =\overrightarrow{n}\cdot \overrightarrow{k}=\frac{1}{\sqrt{3}}.$$

Then we also express $z$ on $a$ with the coordinates $x$ and $y$:

$$\phi =\frac{1}{\sqrt{3}}{\int }_A(x+2y+3(1-x-y))\sqrt{3}𝑑A={\int }_0^1\left({\int }_0^{1-x}(3-2x-y)𝑑y\right)𝑑x=\mathrm{\hspace{0.33em}1}$$