Compute the flux integral $\int_S\vec F\cdot d\vec A$ in two ways, directly and using the Divergence Theorem. $S$ is the surface of the box with faces $x=0,x=2,y=1,y=4,z=0,z=3$, closed and oriented outward, and $\vec{F}=3x^{2}\vec i + 2y^{2}\vec j + 4z^{2}\vec k$.

Using the Divergence Theorem,
$\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d\int_p^q\,$ $\,dz\,dy\,dx =$ ,
where $a =$, $b =$, $c =$, $d =$, $p =$ and $q =$.

Next, calculating directly, we have $\int_S\vec F\cdot d\vec A =$ (the sum of the flux through each of the six faces of the box). Calculating the flux through each face separately, we have:
On $x = 2$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dz\,dy$ =
where $a =$, $b =$, $c =$ and $d =$.

On $x = 0$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dz\,dy$ =
where $a =$, $b =$, $c =$ and $d =$.

On $y = 4$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dz\,dx$ =
where $a =$, $b =$, $c =$ and $d =$.

On $y = 1$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dz\,dx$ =
where $a =$, $b =$, $c =$ and $d =$.

On $z = 3$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dy\,dx$ =
where $a =$, $b =$, $c =$ and $d =$.

And on $z = 0$, $\int_S\vec F\cdot d\vec A = \int_a^b\int_c^d$ $dy\,dx$ =
where $a =$, $b =$, $c =$ and $d =$.

Thus, summing these, we have $\int_S\vec F\cdot d\vec A =$