Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 14{{\rm{e}}^7} - 6{{\rm{e}}^4} + 2{{\rm{e}}^3} - 10$
Work Step by Step
Write $f\left( {x,y,z} \right) = z{{\rm{e}}^{2x + y}}$.
From Figure 20 we see that there are six faces that build up the total surface of the box. So, the integral of $z{{\rm{e}}^{2x + y}}$ over the surface of the box is equal to
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$ + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_3}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_4}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z$
$ + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_5}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_6}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z$
1. $z=0$ (the $xy$-plane), $||{\bf{N}}\left( {x,y} \right)|| = || - {\bf{k}}|| = 1$
Since $z=0$, so $f\left( {x,y,z} \right) = 0$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} f\left( {x,y,z} \right){\rm{d}}x{\rm{d}}y = 0$
2. $z=4$, $f\left( {x,y,4} \right) = 4{{\rm{e}}^{2x + y}}$, $||{\bf{N}}\left( {x,y} \right)|| = ||{\bf{k}}|| = 1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 4\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^3 {{\rm{e}}^{2x + y}}{\rm{d}}x{\rm{d}}y$
$ = 4\mathop \smallint \limits_{x = 0}^2 {{\rm{e}}^{2x}}{\rm{d}}x\mathop \smallint \limits_{y = 0}^3 {{\rm{e}}^y}{\rm{d}}x{\rm{d}}y = 4\cdot\frac{1}{2}\left( {{{\rm{e}}^{2x}}|_0^2} \right)\left( {{{\rm{e}}^y}|_0^3} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} f\left( {x,y,z} \right){\rm{d}}x{\rm{d}}y = 2\left( {{{\rm{e}}^4} - 1} \right)\left( {{{\rm{e}}^3} - 1} \right)$
3. $y=0$ (the $xz$-plane), $f\left( {x,0,z} \right) = z{{\rm{e}}^{2x}}$, $||{\bf{N}}\left( {x,z} \right)|| = || - {\bf{j}}|| = 1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_3}}^{} f\left( {x,y,z} \right){\rm{d}}x{\rm{d}}z = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{z = 0}^4 z{{\rm{e}}^{2x}}{\rm{d}}x{\rm{d}}z$
$ = \mathop \smallint \limits_{x = 0}^2 {{\rm{e}}^{2x}}{\rm{d}}x\mathop \smallint \limits_{z = 0}^4 z{\rm{d}}z = \frac{1}{2}\left( {{{\rm{e}}^{2x}}|_0^2} \right)\left( {\frac{1}{2}{z^2}|_0^4} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_3}}^{} f\left( {x,y,z} \right){\rm{d}}x{\rm{d}}z = 4\left( {{{\rm{e}}^4} - 1} \right)$
4. $y=3$, $f\left( {x,3,z} \right) = z{{\rm{e}}^{2x + 3}}$, $||{\bf{N}}\left( {x,z} \right)|| = ||{\bf{j}}|| = 1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_4}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{z = 0}^4 z{{\rm{e}}^{2x + 3}}{\rm{d}}x{\rm{d}}z$
$ = {{\rm{e}}^3}\mathop \smallint \limits_{x = 0}^2 {{\rm{e}}^{2x}}{\rm{d}}x\mathop \smallint \limits_{z = 0}^4 z{\rm{d}}z = \frac{{{{\rm{e}}^3}}}{2}\left( {{{\rm{e}}^{2x}}|_0^2} \right)\left( {\frac{1}{2}{z^2}|_0^4} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_4}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z = 4{{\rm{e}}^3}\left( {{{\rm{e}}^4} - 1} \right)$
5. $x=0$ (the $yz$-plane), $f\left( {0,y,z} \right) = z{{\rm{e}}^y}$, $||{\bf{N}}\left( {y,z} \right)|| = || - {\bf{i}}|| = 1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_5}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z = \mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{z = 0}^4 z{{\rm{e}}^y}{\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{y = 0}^3 {{\rm{e}}^y}{\rm{d}}y\mathop \smallint \limits_{z = 0}^4 z{\rm{d}}z = \left( {{{\rm{e}}^y}|_0^3} \right)\left( {\frac{1}{2}{z^2}|_0^4} \right) = 8\left( {{{\rm{e}}^3} - 1} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_5}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z = 8\left( {{{\rm{e}}^3} - 1} \right)$
6. $x=2$ (the $yz$-plane), $f\left( {2,y,z} \right) = z{{\rm{e}}^{4 + y}}$, $||{\bf{N}}\left( {y,z} \right)|| = ||{\bf{i}}|| = 1$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_6}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z = \mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{z = 0}^4 z{{\rm{e}}^{4 + y}}{\rm{d}}y{\rm{d}}z$
$ = {{\rm{e}}^4}\mathop \smallint \limits_{y = 0}^3 {{\rm{e}}^y}{\rm{d}}y\mathop \smallint \limits_{z = 0}^4 z{\rm{d}}z = {{\rm{e}}^4}\left( {{{\rm{e}}^y}|_0^3} \right)\left( {\frac{1}{2}{z^2}|_0^4} \right) = 8{{\rm{e}}^4}\left( {{{\rm{e}}^3} - 1} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_6}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {y,z} \right)||{\rm{d}}y{\rm{d}}z = 8{{\rm{e}}^4}\left( {{{\rm{e}}^3} - 1} \right)$
We add all these integral to get the integral of $z{{\rm{e}}^{2x + y}}$ over the surface of the box:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S$
$ = 0 + 2\left( {{{\rm{e}}^4} - 1} \right)\left( {{{\rm{e}}^3} - 1} \right) + 4\left( {{{\rm{e}}^4} - 1} \right) + 4{{\rm{e}}^3}\left( {{{\rm{e}}^4} - 1} \right) + 8\left( {{{\rm{e}}^3} - 1} \right) + 8{{\rm{e}}^4}\left( {{{\rm{e}}^3} - 1} \right)$
$ = 14{{\rm{e}}^7} - 6{{\rm{e}}^4} + 2{{\rm{e}}^3} - 10$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 14{{\rm{e}}^7} - 6{{\rm{e}}^4} + 2{{\rm{e}}^3} - 10$.
