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