Answer
The average value: $\bar f = \frac{1}{{2\pi }}$
Work Step by Step
We have $f\left( {x,y,z} \right) = xy\sin \left( {\pi z} \right)$ and ${\cal W} = \left[ {0,1} \right] \times \left[ {0,1} \right] \times \left[ {0,1} \right]$.
So, the volume of ${\cal W}$ is 1.
Using Eq. (5), the average value of $f$ is
$\bar f = \frac{1}{{Volume\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \mathop \smallint \limits_{z = 0}^1 xy\sin \left( {\pi z} \right){\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = - \frac{1}{\pi }\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 xy\left( {\cos \left( {\pi z} \right)|_0^1} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{2}{\pi }\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 xy{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{\pi }\mathop \smallint \limits_{x = 0}^1 x\left( {{y^2}|_0^1} \right){\rm{d}}x$
$ = \frac{1}{\pi }\mathop \smallint \limits_{x = 0}^1 x{\rm{d}}x$
$ = \frac{1}{{2\pi }}\left( {{x^2}|_0^1} \right) = \frac{1}{{2\pi }}$
Thus, the average value: $\bar f = \frac{1}{{2\pi }}$.
