Compute the flux integral ${\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)$ in two ways, directly and using the Divergence Theorem. $S$ is the surface of the box with faces $x=3,x=6,y=0,y=2,z=0,z=3,$ closed and oriented outward, and vec$(F)=3x^2$vec$(i)+y^2$vec$(j)+4z^2$vec$(k).$

Next, calculating directly, we have ${\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)=($the sum of the flux through each of the six faces of the box$).$ Calculating the flux through each face separately, we have:

On $y=2,{\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}dzdx=$ where $a=2$ and where $a=0,b=c=2,$

$d=.$

On $x=3,{\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}2,dzdy=2$ where $a=0,b=2,c=$ and $d=$

$$

$,b=$

$$

$d=.$

On $y=0,{\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}dzdx=$ where $a=2$

$q,$ and $d=$

$,b=,c=$ and

On $z=3,{\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}dydx=$ where $a=2,b=$

$,c=$ and $d=$ And on $z=0,{\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}dydx=$ where $a=,b=,c=$ and $d=$

Thus, summing these, we have ${\displaystyle \underset{S}{\overset{\phantom{}}{}}}$vec$(F)*dvec(A)==$