PlanetPhysics/Flux of Vector Field

Let $\overrightarrow{U}=U_x\overrightarrow{i}+U_y\overrightarrow{j}+U_z\overrightarrow{k}$ be a [[../NeutrinoRestMass/|vector field]] in ${ℝ}^3$\, 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 $d\overrightarrow{a}=\overrightarrow{n}da,$ where $\overrightarrow{n}$ is the unit normal [[../Vectors/|vector]] on the positive side of $da$.

The [[../AbsoluteMagnitude/|flux]] of the vector $\overrightarrow{U}$ through the surface $a$ is the surface integral $\underset{a}{\int }\overrightarrow{U}\cdot d\overrightarrow{a}.$\\

Remark. \, One can imagine that $\overrightarrow{U}$ represents the [[../Velocity/|velocity]] vector of a flowing liquid; suppose that the flow is stationary, i.e. the velocity $\overrightarrow{U}$ depends only on the location, not on the time.\, Then the [[../DotProduct/|scalar product]] $\overrightarrow{U}\cdot d\overrightarrow{a}$ is the [[../Volume/|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\, $\overrightarrow{U}=x\overrightarrow{i}+2y\overrightarrow{j}+3z\overrightarrow{k}$\, 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 positive normal 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 $\phi =\underset{a}{\int }\overrightarrow{U}\cdot d\overrightarrow{a}=\frac{1}{\sqrt{3}}\underset{a}{\int }(x+2y+3z)da.$

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 $\overrightarrow{n}$ 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 $y$: $\phi =\frac{1}{\sqrt{3}}\underset{A}{\int }(x+2y+3(1-x-y))\sqrt{3}dA=\underset{0}{\overset{1}{\int }}(\underset{0}{\overset{1-x}{\int }}(3-2x-y)dy)dx=1$