# 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.

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

### Relevant Questions

Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
S is the surface of the solid bounded by the cylinder $$\displaystyle{y}^{{2}}+{z}^{{2}}={9}$$ and the planes x = −3 and x = 1.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. $$\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}$$
C: The triangle bounded by y = 0, x = 3, and y = x
Use the Divergence Theorem to find the flux of $$\displaystyle{F}={x}{y}^{{2}}{i}+{x}^{{2}}{y}{j}+{y}{k}$$ outward through the surface of the region enclosed by the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the planes z = 1 and z =-1.
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. F = (x - y)i + ( y - x)j
C: The square bounded by x = 0, x = 1, y = 0, and y = 1
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
Use the divergence theorem to evaluate $$\displaystyle\int\int_{{S}}{F}\cdot{N}{d}{S}$$, where $$\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}$$ and S is the boundary of the cube defined by $$\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}$$.