91Ó°ÊÓ

Consider the vector field below:

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

The goal is to find the integral ${âˆ\neg }_S\mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS$, where$S$is the surface of the region $W$, which is limited by the planes$y=0,y=z,z=3,x=0,x=4$, 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 $F(x,y,z)$an open region containing has a vector field , then

${âˆ\neg }_{s}F(x,y,z).ndS=∫{âˆ\neg }_{w}divF(x,y,z)dV$

Where $n$is the normal vector pointing outwards

${âˆ\neg }_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={∭}_W \mathrm{di}v\mathbf{F}(x,y,z)dV.$

First, determine the vector field's divergence$F(x,y,z)=x{y}^{2}i+y(z-3x)j+4xyzk$

A vector field's divergence $F(x,y,x)=F_1(x,y,z)i+F_2(x,y,z)j+F_3(x,y,z)k$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 $F(x,y,z)=x{y}^{2}i+y(z-3x)j+4xyzk)$will be,

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

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

$$\begin{gathered} =y^2+z-3x+4xy \\ =y^2+4xy+(z-3x) \end{gathered}$$

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

The surface $S$, which is the surface of the region $W$ bordered by the planes, defines the region $y=0,y=z,z=3,x=0,x=4$

The integration region will be here,

$R=\left\{\right(x,y,z\left)\right|0≤x4,0≤y≤z,0≤z≤3\}$

Now evaluate the integral using the divergence theorem ${âˆ\neg }_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS$as follows

${âˆ\neg }_{s}F.ndS=∫{âˆ\neg }_{R}divFdV$

localid="1650908686064" $={∫}_0^4{∫}_0^3{∫}_0^z(y^2+4xy+(z-3x\left)\right)dydzdx$

localid="1650908367876" $={∫}_0^4{∫}_0^3\left[{∫}_0^z\right(y^2+4xy+z-3x\left)\right)dy]dzdx$

$={∫}_0^4 {∫}_0^3 {\left[\frac{y^3}{3}+2x{y}^2+(z−3x)y\right]}_0^{z}dzdx$

$={∫}_0^4{∫}_0^3[\frac{y^3}{3}+2x{y}^2+(z-3x)y-(\frac{0^3}{3}+2x.0^2+(z-3x).0]dzdx$

$={∫}_0^4{∫}_0^3(\frac{z^3}{3}+2x{z}^2+z^2-3xz)dzdx$

$={∫}_3^4\left[{∫}_0^3\right(\frac{z^3}{3}+2x{z}^2+z^2-3xz\left)dz\right]dx$

$={∫}_0^4 {\left[\frac{z^4}{12}+\frac{2x{z}^3}{3}+\frac{z^3}{3}−\frac{3x{z}^2}{2}\right]}_0^{3}dx$

$={∫}_0^4\left[\right(\frac{3^4}{12}+\frac{\left(2x\right)\left(3^3\right)}{3}+\frac{3^3}{3}+\frac{\left(3x\right)\left(3^2\right)}{2})-(\frac{0^4}{12}+\frac{\left(2x\right)\left(0^3\right)}{3}+\frac{0^3}{3}-\frac{\left(3x\right)(0^{2)}}{2}\left)\right]dx$

$={∫}_0^4(\frac{63}{4}+\frac{9x}{2})dx$

$={\left[\frac{63x}{4}+\frac{9x^2}{4}\right]}_0^4$

$$\begin{gathered} =(\frac{\left(63\right)\left(4\right)}{4}+\frac{\left(9\right)\left(4^2\right)}{4})-(\frac{\left(63\right)\left(0\right)}{4}+\frac{\left(9\right)\left(0^2\right)}{4}) \\ =99 \end{gathered}$$

$={∫}_0^4 {∫}_0^3 {\left[\frac{y^3}{3}+2x{y}^2+(z−3x)y\right]}_0^{z}dzdx$