a73652…: “Adjacent to a super cube”

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The average value of a function $F:\R ^2\to \R$ on some two-dimensional set $S$ is given by: $\frac {1}{\mbox {Area of }S} \iint _S F \d A$ Let $F : \R ^2 \to \R$ be the periodic function defined by the rule $F(x,y) = \cos ^2 x \cos ^2 y$ What is the average value of $F$ ? $\mbox {The average value is } \answer {1/4}.$

Consider just the average value on: $S=\{(x,y):0\le x \le 2\pi , 0 \le y \le 2\pi \}$

The area of $S$ is $4\pi ^2$ .

The function is periodic, i.e., $f(x,y) = f(x+2n\pi , y+2m\pi )$ for integers $n,m$ . So the average value on $S$ is actually the average value of the function.

$\iint _S F \d A = \pi ^2$ .

Therefore the average value is $\frac {\pi ^2}{4\pi ^2} = \frac {1}{4}$ .