Question

# Set up the integral for the divergence theorem both ways. Then find the flux.F(x,y,z) = 3x hat(i) + xy hat(j) + 2xz hat(k)E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.

Set up the integral for the divergence theorem both ways. Then find the flux.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}$$
E is the cube bounded by the planes $$x = 0, x = 3, y = 0, y = 3,\ and\ z = 0, z = 3$$.

2021-02-21

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 $$\displaystyle\frac{{405}}{{2}}$$.