Answer
The average value: $\bar f = 2{\rm{e}} - 4$.
Work Step by Step
We have $f\left( {x,y,z} \right) = {{\rm{e}}^y}$ and the region ${\cal W}:0 \le y \le 1 - {x^2}{\rm{,}}0 \le z \le x$.
We sketch the solid region and choose to project ${\cal W}$ onto the $xy$-plane and obtain the domain ${\cal D}$. From the figure attached, we see that ${\cal D}$ is bounded by the curve $y = 1 - {x^2}$. It is both a vertically and horizontally simple region. Using $0 \le y \le 1 - {x^2}$, we choose to describe ${\cal D}$ as a vertically simple region. So, the description of ${\cal D}$ is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1 - {x^2}} \right\}$
Thus, ${\cal W}$ is a $z$-simple region consisting of all points lying between ${\cal D}$ and the upper face, which is the plane $z=x$. So, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1,0 \le y \le 1 - {x^2},0 \le z \le x} \right\}$
We evaluate the volume of ${\cal W}$:
$Volume\left( {\cal W} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} 1{\rm{d}}V$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} \mathop \smallint \limits_{z = 0}^x {\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} \left( {z|_0^x} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} x{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 x\left( {y|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {x - {x^3}} \right){\rm{d}}x$
$ = \left( {\frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right)|_0^1$
$ = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
To find the average value of $f$, we use Eq. (5):
$\bar f = \frac{1}{{Volume\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {{\rm{e}}^y}{\rm{d}}V$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} \mathop \smallint \limits_{z = 0}^x {{\rm{e}}^y}{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} {{\rm{e}}^y}\left( {z|_0^x} \right){\rm{d}}y{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} x{{\rm{e}}^y}{\rm{d}}y{\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 x\left( {{{\rm{e}}^y}|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 x\left( {{{\rm{e}}^{1 - {x^2}}} - 1} \right){\rm{d}}x$
$ = 4\mathop \smallint \limits_{x = 0}^1 \left( {x{{\rm{e}}^{1 - {x^2}}} - x} \right){\rm{d}}x$
$ = 4\left( { - \frac{1}{2}{{\rm{e}}^{1 - {x^2}}} - \frac{1}{2}{x^2}} \right)|_0^1$
$ = 4\left( { - \frac{1}{2} - \frac{1}{2} + \frac{1}{2}{\rm{e}}} \right)$
$ = 2{\rm{e}} - 4$
Thus, the average value: $\bar f = 2{\rm{e}} - 4$.
