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