# flux of vector field

Let

$$\overrightarrow{U}={U}_{x}\overrightarrow{i}+{U}_{y}\overrightarrow{j}+{U}_{z}\overrightarrow{k}$$ |

be a vector field in ${\mathbb{R}}^{3}$ 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

$${\int}_{a}\overrightarrow{U}\cdot \mathit{d}\overrightarrow{a}.$$ |

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

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

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 $dA$;

$$dA=\mathrm{cos}\alpha da$$ |

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

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

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

$$\phi =\frac{1}{\sqrt{3}}{\int}_{A}(x+2y+3(1-x-y))\sqrt{3}\mathit{d}A={\int}_{0}^{1}\left({\int}_{0}^{1-x}(3-2x-y)\mathit{d}y\right)\mathit{d}x=\mathrm{\hspace{0.33em}1}$$ |

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 |