91Ó°ÊÓ

Consider the vector field below:

$\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$

The goal is to find the integral${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$ for the surface s, which is defined as follows:

The surface $S$is the first-octant cube's surface with side length $\mathrm{Ï€}$and one vertex at the origin, and $n$ is the normal vector heading outwards.

To evaluate this integral, use the Divergence Theorem.

"Let $W$be a bounded region in $R^3$whose border $S$is a smooth or piecewise-smooth closed oriented surface," says the Divergence Theorem. If an open region containing $W$has a vector field $F(x,y,z)$, then

${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={∭}_{W}div\mathbf{F}(x,y,z)dV........\left(1\right)$

where $n$is the normal vector pointing outwards

First, determine the vector field's divergence $\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$

A vector field's divergence $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$has the following definition:

$divF(x,y,z)=(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k).(F_{1}i+F_{2}j+F_{3}k)$

$=\frac{∂F_1}{∂x}+\frac{∂F_2}{∂y}+\frac{∂F_3}{∂z}$

Then there's the vector field's divergence $\mathbf{F}(x,y,z)=4x^{3}yz\mathbf{i}+6x^2{y}^{2}z\mathbf{j}+6x^{2}y{z}^2\mathbf{k}$will be,

$div\mathbf{F}(x,y,z)=\left(\frac{∂}{∂x}\mathbf{i}+\frac{∂}{∂y}\mathbf{j}+\frac{∂}{∂z}\mathbf{k}\right)·\left(F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}\right)$

$=\frac{∂}{∂x}\left(4x^{3}yz\right)+\frac{∂}{∂y}\left(6x^2{y}^{2}z\right)+\frac{∂}{∂z}\left(6x^{2}y{z}^2\right)$

$=12x^{2}yz+12x^{2}yz+12x^{2}yz$

$=36x^{2}yz$

$divF(x,y,z)=(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k).(4x^{3}yzi+6x^2{y}^{2}zj+6x^{2}y{z}^{2}k)$

$div\mathbf{F}(x,y,z)=\left(\frac{∂}{∂x}\mathbf{i}+\frac{∂}{∂y}\mathbf{j}+\frac{∂}{∂z}\mathbf{k}\right)·\left(F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}\right)$

The region is defined by the surface $S$, which is the surface of the first-octant cube with a side length of$\mathrm{Ï€}$and one vertex at the origin.

The integration region will be here,

$R=\left\{\right(x,y,z)âˆ\pounds 0≤x≤\mathrm{Ï€},0≤y≤\mathrm{Ï€},0≤z≤\mathrm{Ï€}\}$

Now evaluate the integral using the Divergence Theorem (1) ${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$as follows:

$âˆ\neg F.ndS=âˆ\neg {∫}_{R}divFdV$

$={∫}_0^{\mathrm{π}}{∫}_0^{\mathrm{π}}{∫}_0^{\mathrm{π}}\left(36x^{2}yz\right)dxdydz$

$={∫}_0^{\mathrm{π}}{∫}_0^{\mathrm{π}}\left[\right(36x^{2}yz\left)dx\right]dydz$

$={∫}_0^{\mathrm{π}}{∫}_0^{\mathrm{π}}{\left[12x^{3}yz\right]}_0^{\mathrm{π}}dydz$

$={∫}_0^{\mathrm{π}}{∫}_0^{\mathrm{π}}\left(12{\mathrm{π}}^{3}yz-12{\left(0\right)}^{3}yz\right)dydz$

$={∫}_0^{\mathrm{π}}\left[{∫}_0^{\mathrm{π}}12{\mathrm{π}}^{3}yzdy\right]dz$

$={∫}_0^{\mathrm{π}}{\left[6{\mathrm{π}}^3{y}^{2}z\right]}_0^{\mathrm{π}}dz$

role="math" localid="1650903789951" $={∫}_0^{\mathrm{π}}(6{\mathrm{π}}^3{\left(\mathrm{π}\right)}^{2}z-6{\mathrm{π}}^3{\left(0\right)}^{2}z)dz$

$={∫}_0^{\mathrm{π}}6{\mathrm{π}}^{5}zdz$

$={\left[3{\mathrm{Ï€}}^5{z}^2\right]}_0^{\mathrm{Ï€}}$

$=3{\mathrm{Ï€}}^5{\mathrm{Ï€}}^2-3{\mathrm{Ï€}}^5{\left(0\right)}^2$

$=3{\mathrm{Ï€}}^7$