Calculating flux through a square

Question

I did not quite understand the latest lecture I've been to and would like a thorough explanation if possible. A field vector is given by $F=(\cos(xyz), \sin(xyz), xyz)$. Calculate the flux through a square that's parallel to the $XY$ plane, its center is in $(0,0,a)$, and its side's length is $a$. Thanks and regards.

Answer 1

The flux is given by

$$\iint_{\text{square}} dA \, \mathbf{F} \cdot \hat{\mathbf{n}} $$

where $\hat{\mathbf{n}}$ is the unit normal to the square, which in this case is simply the $z$ direction. Thus, $\mathbf{F} \cdot \hat{\mathbf{n}}$ is the $z$ component of $\mathbf{F}$, or $x y z$. Thus, the flux is

$$a \int_{-a/2}^{a/2} dx \, x \, \int_{-a/2}^{a/2} dy \, y = 0$$