Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {{\rm{e}}^z}{\rm{d}}V \simeq 168.19$
Work Step by Step
Referring to Figure 12, the front face of the tetrahedron has vertices at $\left( {4,0,0} \right)$, $\left( {0,4,0} \right)$ and $\left( {0,0,6} \right)$. Referring to Exercise 35 in Section 15.8, the equation of the plane of the front face is given by
$\frac{x}{4} + \frac{y}{4} + \frac{z}{6} = 1$
Multiplying both sides by $12$ gives
$3x + 3y + 2z = 12$
$z = 6 - \frac{3}{2}x - \frac{3}{2}y$
We can consider the tetrahedron in Figure 12 as a $z$-simple region ${\cal W}$ lying over the triangle ${\cal D}$ in the $xy$-plane bounded by $x=0$, $y=0$, and the line $y=4-x$.
Thus, ${\cal W}$ is the region between the planes $z=0$ and $z = 6 - \frac{3}{2}x - \frac{3}{2}y$. The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$ given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 4,0 \le y \le 4 - x} \right\}$
So,
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 4,0 \le y \le 4 - x,0 \le z \le 6 - \frac{3}{2}x - \frac{3}{2}y} \right\}$
We evaluate the triple integral as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} f\left( {x,y,z} \right){\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^{6 - \frac{3}{2}x - \frac{3}{2}y} {{\rm{e}}^z}{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} \left( {\mathop \smallint \limits_{z = 0}^{6 - \frac{3}{2}x - \frac{3}{2}y} {{\rm{e}}^z}{\rm{d}}z} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} \left( {{{\rm{e}}^z}|_0^{6 - \frac{3}{2}x - \frac{3}{2}y}} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{4 - x} \left( {{{\rm{e}}^{6 - \frac{3}{2}x - \frac{3}{2}y}} - 1} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \left( {\left( { - \frac{2}{3}{{\rm{e}}^{6 - \frac{3}{2}x - \frac{3}{2}y}} - y} \right)|_0^{4 - x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \left( { - \frac{2}{3}{{\rm{e}}^{6 - \frac{3}{2}x - 6 + \frac{3}{2}x}} - 4 + x + \frac{2}{3}{{\rm{e}}^{6 - \frac{3}{2}x}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 \left( { - \frac{{14}}{3} + x + \frac{2}{3}{{\rm{e}}^{6 - \frac{3}{2}x}}} \right){\rm{d}}x$
$ = \left( {\left( { - \frac{{14}}{3}x + \frac{1}{2}{x^2} - \frac{4}{9}{{\rm{e}}^{6 - \frac{3}{2}x}}} \right)|_0^4} \right)$
$ = - \frac{{56}}{3} + 8 - \frac{4}{9} + \frac{4}{9}{{\rm{e}}^6}$
$ = - \frac{{100}}{9} + \frac{4}{9}{{\rm{e}}^6}$
$ \simeq 168.19$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {{\rm{e}}^z}{\rm{d}}V \simeq 168.19$.
