Step 1

Considering the given dunction is

\(\displaystyle{f{{\left({x},{y},{z}\right)}}}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}\).

Where E is the cube bounded by the planes \(x =0, x = 3, y = 0, y = 3\ and\ z = 0, z =3.\)

Find the flux.

Step 2

Consider

\(\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}\).

Find the div(f), then

grad \(F = 3+x+2x\)

\(=3+3x\)

By using the divergence theorem, then

\(\displaystyle{F}{l}{u}{x}=\int\int\int\nabla{F}{d}{v}\)

\(\displaystyle={\int_{{0}}^{{3}}}{\int_{{0}}^{{3}}}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\int_{{0}}^{{3}}}{{\left({z}\right)}_{{0}}^{{3}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{{\left({y}\right)}_{{0}}^{{3}}}{\left.{d}{x}\right.}\)

\(\displaystyle={9}{\int_{{0}}^{{3}}}{\left({3}+{3}{x}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={9}{{\left({3}{x}+\frac{{{3}{x}^{{2}}}}{{2}}\right)}_{{0}}^{{3}}}\)

\(\displaystyle=\frac{{405}}{{2}}\)

Hence, total flux is \(\displaystyle\frac{{405}}{{2}}\).