91Ó°ÊÓ

Consider the vector field below:

$\mathbf{F}(x,y,z)=⟨xz,yz,xyzâŸ\copyright$

The goal is to assess the integral.${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, where $S$is the cylinder's surface with equation $x^2+y^2=9$for $-2≤z≤2$, and $\mathrm{n}$is the normal vector that points outwards.

To evaluate this integral, use the Divergence Theorem.

Divergence According to the theorem,

"Let be a bounded region in with a smooth or piecewise-smooth closed oriented surface as its boundary . If a vector field is defined on an open area containing , then,

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

Where is the unit normal vector inwards."

First, determine the vector field's divergence. $\mathbf{F}(x,y,z)=⟨xz,yz,xyzâŸ\copyright$.

A vector field's divergence. is defined in the following way:

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

Then there's the vector field's divergence. $\mathbf{F}(x,y,z)=⟨xz,yz,xyzâŸ\copyright$is going to be

$=\frac{∂}{∂x}\left(xz\right)+\frac{∂}{∂y}\left(yz\right)+\frac{∂}{∂z}\left(xyz\right)$

$=z+z+xy$

$=2z+xy$

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

${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={∭}_{R}div\mathbf{F}(x,y,z)dV$

$={∭}_R(2z+xy)dV$

Here, the region is bounded by the surface $S$, where $S$ is the surface of the cylinder with equation $x^2+y^2=9$ for $-2≤z≤2$.

When stated in cylindrical coordinates, the region is substantially simplified.$(r,\mathrm{θ},z)$. Change the coordinates of this integral (2) to cylindrical and integrate it.

The domain of integration in cylindrical coordinates is described as follows:

$R=\left\{\right(r,\mathrm{θ},z):0≤r≤3,0≤\mathrm{θ}≤2\mathrm{π},-2≤z≤2\}.$

In cylindrical coordinates,

$x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},z=z$, and

$x^2+y^2=r^2$, and

$dV=dzrdrd\mathrm{θ}$

Then, as follows, evaluate the integral $\left(2\right)$:

${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={∭}_R(2z+xy)dV$

$={∫}_0^{2\mathrm{π}}{∫}_0^3{∫}_{-2}^2(2z+r\cos \mathrm{θ}r\sin \mathrm{θ})dzrdrd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3{∫}_{-2}^2\left(2z+r^2\cos \mathrm{θ}\sin \mathrm{θ}\right)dzrdrd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3{∫}_{-2}^2\left(2zr+r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)dzdrd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3\left[{∫}_{-2}^2\left(2zr+r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)dz\right]drd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3{\left[z^{2}r+z{r}^3\cos \mathrm{θ}\sin \mathrm{θ}\right]}_{-2}^{2}drd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3\left[\left(2^{2}r+2r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)-\left((-2{)}^{2}r+(-2)r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)\right]drd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{∫}_0^3\left(4r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)drd\mathrm{θ}$

Reduce the last integral to the following:

${âˆ\neg }_S\mathbf{F}(x,y,z)·\mathbf{n}dS={∫}_0^{2\mathrm{Ï€}}{∫}_0^3\left(4r^3\cos \mathrm{θ}\sin \mathrm{θ}\right)drd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}\cos \mathrm{θ}\sin \mathrm{θ}\left[{∫}_0^{3}4r^{3}dr\right]d\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}\cos \mathrm{θ}\sin \mathrm{θ}{\left[r^4\right]}_0^{3}d\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}\cos \mathrm{θ}\sin \mathrm{θ}\left[3^4-0^4\right]d\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}81\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}$

$={\left[\frac{81{\cos }^2\mathrm{θ}}{2}\right]}_0^{2\mathrm{π}}$

$=\frac{81{\cos }^{2}2\mathrm{Ï€}}{2}-\frac{81{\cos }^{2}0}{2}$

$=\frac{81(1{)}^2}{2}-\frac{81(1{)}^2}{2}$

$=0$

As a result, the necessary integral is ${âˆ\neg }_S\mathrm{F}(x,y,z)·\mathbf{n}dS=0$.