This problem will illustrate the divergence theorem by computing the outward flux of the vector field $\mathbf{F}(x, y, z) = 2\!x\mathbf{i} + 3\!y\mathbf{j} + 1\!z\mathbf{k}$ across the boundary of the right rectangular prism: $-4 \leq x \leq 6, -2 \leq y \leq 7, -3 \leq z \leq 5$ oriented outwards using a surface integral and a triple integral over the solid bounded by rectangular prism.
Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the prism to be positive.

Part 1 - Using a Surface Integral
First we parameterize the six faces using $0\leq s \leq1$ and $0\leq t \leq 1$:

The face with z = -3 :  $\sigma_1=(x_1(s),y_1(t),z_1(s,t))$
$x_1(s) =$
$y_1(t) =$
$z_1(s,t) = -3$

The face with z = 5 :   $\sigma_2=(x_2(s),y_2(t),z_2(s,t))$
$x_2(s) =$
$y_2(t) =$
$z_2(s,t) = 5$

The face with x = -4 :   $\sigma_3=x_3(s,t),y_3(s),z_3(t))$
$x_3(s,t) = -4$
$y_3(s) =$
$z_3(t)=$

The face with x = 6 :   $\sigma_4=(x_4(s,t),y_4(s),z_4(t))$
$x_4(s,t) = 6$
$y_4(s) =$
$z_4(t) =$

The face with y = -2 :   $\sigma_5=(x_5(s),y_5(s,t),z_5(t))$
$x_5(s) =$
$y_5(s,t) = -2$
$z_5(t) =$

The face with y = 7 :   $\sigma_6=(x_6(s),y_6(s,t),z_6(t))$
$x_6(s) =$
$y_6(s,t) = 7$
$z_6(t) =$

Then (mind the orientation)
$\displaystyle\int\int_\sigma F\cdot n \,dS$ $\displaystyle =\int_0^1\int_0^1 F(\sigma_1)\cdot\left(\frac{\partial \sigma_1}{\partial t}\times\frac{\partial \sigma_1}{\partial s}\right) ds\,dt$ $\displaystyle +\int_0^1\int_0^1 F(\sigma_2)\cdot\left(\frac{\partial \sigma_2}{\partial s}\times\frac{\partial \sigma_2}{\partial t}\right) ds\,dt$ $\displaystyle +\int_0^1\int_0^1 F(\sigma_3)\cdot\left(\frac{\partial \sigma_3}{\partial t}\times\frac{\partial \sigma_3}{\partial s}\right) ds\,dt$ $\displaystyle +\int_0^1\int_0^1 F(\sigma_4)\cdot\left(\frac{\partial \sigma_4}{\partial s}\times\frac{\partial \sigma_4}{\partial t}\right) ds\,dt$ $\displaystyle +\int_0^1\int_0^1 F(\sigma_5)\cdot\left(\frac{\partial \sigma_5}{\partial s}\times\frac{\partial \sigma_5}{\partial t}\right) ds\,dt$ $\displaystyle +\int_0^1\int_0^1 F(\sigma_6)\cdot\left(\frac{\partial \sigma_6}{\partial t}\times\frac{\partial \sigma_6}{\partial s}\right) ds\,dt$

$\hskip 20pt =$ + + + + +

$\hskip 20pt =$

Part 2 - Using the Divergence Theorem

$\displaystyle\int\int_\sigma F\cdot n \, dS =\int\int\int_G {\rm div } F\,dV=$

$\hskip 20pt =$

$\displaystyle\int$

$\displaystyle\int$

$\displaystyle\int$

$dx\,dy\,dz$

$\hskip 20pt =$