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

Using the Divergence Theorem,

${}_S$vec$(F)*dvec(S)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}{\displaystyle \underset{p}{\overset{q}{}}},dzdydx=$

where $a=$dots,$b=,c=,d=,$ and $q=$

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

On $x=3,{}_S$vec$(F)*dvec(S)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}},dzdy=$

where $a=$dots,$b=$dots and $d=$

On $x=2,{}_S$vec$(F)*dvec(S)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}|,dzdy||=$

where $a=,b=,c=$ and $d=$

On $y=2,{}_S$vec$(F)*dvec(S)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}{|}_{d}zdx||=$ where $a=,b=$ and $d=$

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

On $z=3,{}_S$vec$(F)*dvec(S)={\displaystyle \underset{a}{\overset{b}{}}}{\displaystyle \underset{c}{\overset{d}{}}}{?}_{,}dydx=$ where $a=b=,c=,$ and $d=$

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

Thus, summing these, we have ${}_S$vec$(F)*dvec(S)=$