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# flux of vector field

Let

$\vec{U}\;=\;U_{x}\vec{i}+U_{y}\vec{j}+U_{z}\vec{k}$ |

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

$d\vec{a}\;=\;\vec{n}\,da,$ |

where $\vec{n}$ is the unit normal vector on the positive side of $da$.

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

$\int_{a}\vec{U}\cdot d\vec{a}.$ |

Remark. One can imagine that $\vec{U}$ represents the velocity vector of a flowing liquid; suppose that the flow is stationary, i.e. the velocity $\vec{U}$ depends only on the location, not on the time. Then the scalar product $\vec{U}\cdot d\vec{a}$ 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 side to the positive side or contrarily.

Example. Let $\vec{U}=x\vec{i}+2y\vec{j}+3z\vec{k}$ 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 positive normal away from the origin.

One has the constant unit normal vector:

$\vec{n}\;=\;\frac{1}{\sqrt{3}}\vec{i}+\frac{1}{\sqrt{3}}\vec{j}+\frac{1}{\sqrt% {3}}\vec{k}.$ |

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

$\varphi\;=\;\int_{a}\vec{U}\cdot d\vec{a}\;=\;\frac{1}{\sqrt{3}}\int_{a}(x+2y+% 3z)\,da.$ |

However, this surface integral may be converted to one in which $a$ is replaced by its projection $A$ 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 $\vec{n}$ and $\vec{k}$:

$\cos\alpha\;=\;\vec{n}\cdot\vec{k}\;=\;\frac{1}{\sqrt{3}}.$ |

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

$\varphi\;=\;\frac{1}{\sqrt{3}}\int_{A}(x+2y+3(1-x-y))\,\sqrt{3}\,dA\;=\;\int_{% 0}^{1}\left(\int_{0}^{{1-x}}(3-2x-y)\,dy\right)dx\;=\;1$ |

## Mathematics Subject Classification

26B15*no label found*26B12

*no label found*

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