flux of vector field

Let

$\displaystyle \vec{U} \;=\; U_x\vec{i}+U_y\vec{j}+U_z\vec{k}$

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

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

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

$\displaystyle \int_a\vec{U} \cdot d\vec{a}.$


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:

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

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

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

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$ ;

$\displaystyle dA = \cos\alpha\, da$

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

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

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

$\displaystyle \varphi \;=\; \frac{1}{\sqrt{3}}\int_A(x+2y+3(1-x-y))\,\sqrt{3}\,dA
\;=\; \int_0^1\left(\int_0^{1-x}(3-2x-y)\,dy\right)dx \;=\; 1$



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As of this snapshot date, this entry was owned by pahio.