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 $y$:

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

Title	flux of vector field
Canonical name	FluxOfVectorField
Date of creation	2013-03-22 18:45:25
Last modified on	2013-03-22 18:45:25
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Definition
Classification	msc 26B15
Classification	msc 26B12
Synonym	flux of vector
Related topic	GaussGreenTheorem
Related topic	MutualPositionsOfVectors
Related topic	AngleBetweenTwoVectors
Defines	flux