Answer
$\bar f = \frac{9}{2}$
Work Step by Step
We have $f\left( {x,y,z} \right) = x{y^2}{z^3}$ and the region $\left[ {0,1} \right] \times \left[ {0,2} \right] \times \left[ {0,3} \right]$.
Let ${\cal W}$ denote the region. So, the volume of ${\cal W}$ is $V = 1 \times 2 \times 3 = 6$.
By definition, the average value of $f\left( {x,y,z} \right)$ is given by (Eq. (5) in Section 16.3):
$\bar f = \frac{1}{{{\rm{Volume}}\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V$
$ = \frac{1}{6}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^3 x{y^2}{z^3}{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{6}\left( {\mathop \smallint \limits_{x = 0}^1 x{\rm{d}}x} \right)\left( {\mathop \smallint \limits_{y = 0}^2 {y^2}{\rm{d}}y} \right)\left( {\mathop \smallint \limits_{z = 0}^3 {z^3}{\rm{d}}z} \right)$
$ = \frac{1}{6}\left( {\frac{1}{2}{x^2}|_0^1} \right)\left( {\frac{1}{3}{y^3}|_0^2} \right)\left( {\frac{1}{4}{z^4}|_0^3} \right)$
$ = \frac{1}{6}\left( {\frac{1}{2}} \right)\left( {\frac{8}{3}} \right)\left( {\frac{{81}}{4}} \right) = \frac{9}{2}$
So, $\bar f = \frac{9}{2}$.
