74dcc9…: “Amid the quick chord”

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Consider the cube with vertices: $(1,1,1) \quad (-1,1,1)\quad (-1,-1,1)\quad (1,-1,1)$ $(1,1,-1) \quad (-1,1,-1)\quad (-1,-1,-1)\quad (1,-1,-1)$ Let this cube be oriented with outward pointing normal vectors. Compute the flux of the field $\vec {F}=\vector {x,y,z}$ across the surface of the cube. $\text {Flux} = \answer {24}$

You will need to compute: $\iint _C \vec {F}\dotp \uvec {n}\d S$ where $C$ is the cube above.

One way to proceed is to break the integral above into $6$ different integrals, one for each face of the cube.

Start by considering the top face of the cube, call it $C_{\text {Top}}$ . In this case $\uvec {n} = \vector {0,0,1}$ .

Here $\vec {F}\dotp \uvec {n} = z$ .

Moreover on this side, $z = 1$ .

$\begin{align*} \iint_{C_{\text{Top}}} \vec{F}\dotp\uvec{n}\d S &= \iint_{C_{\text{Top}}}1 \d S\\ &= 4 \end{align*}$

The calculations for the other five sides are similar.