$\mathbf{F}(x,y,z)=4\cos z+2x\mathbf{i}+(3y-z)\mathbf{j}+12z\mathbf{k}$, and $S$is the surface of the region $W$bounded below by $z=x^2+y^2$and above by the sphere $x^2+y^2+z^2=2$centered at the origin.

The given is $\mathbf{F}(x,y,z)=4\cos z+2x\mathbf{i}+(3y-z)\mathbf{j}+12z\mathbf{k}$, and $S$the surface of the region $W$bounded below by $z=x^2+y^2$and above by the sphere $x^2+y^2+z^2=2$centered at the origin. The objective is to find the integration${\iint }_S\mathbf{F}(x,y,z)\times \mathbf{n}dS$

Consider the following vector field$\mathbf{F}(x,y,z)=(4\cos z+2x)\mathbf{i}+(3y-z)\mathbf{j}+12z\mathbf{k}\text{.}$

The objective is to evaluate the integral ${\iint }_S\mathbf{F}(x,y,z).\mathbf{n}dS$, which$S$is the surface of the region $W$bounded below by $z=x^2+y^2$and above by the sphere $x^2+y^2+z^2=2$centered at the origin, and $\mathbf{n}$is the outwards-pointing normal vector.

Use Divergence Theorem to evaluate this integral.

Divergence Theorem states that,

"Let $W$be a bounded region in ${\mathrm{ℝ}}^3$whose boundary $S$is a smooth or piecewise-smooth closed oriented surface. If a vector field $\mathbf{F}(x,y,z)$is defined on an open region containing $W$, then,

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

where $\mathbf{n}$ is the outwards unit normal vector. "

First find the divergence of the vector field $\mathbf{F}(x,y,z)=(4\cos z+2x)\mathbf{i}+(3y-z)\mathbf{j}+12z\mathbf{k}$.

The divergence of a vector field $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

$$\begin{gathered} divF(x,y,z)=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k)\times (F_{1}i+F_{2}j+F_{3}k) \\ =\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial f_3}{\partial z} \end{gathered}$$

Then, the divergence of the vector field $\mathbf{F}(x,y,z)=(4\cos z+2x)\mathbf{i}+(3y-z)\mathbf{j}+12z\mathbf{k}$will be,

$$\begin{gathered} divF(x,y,z)=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).\left(\right(4\cos z+2x)i+(3y-z)j+12zk) \\ =\frac{\partial }{\partial x}(4\cos z+2x)+\frac{\partial }{\partial y}(3y-z)+\frac{\partial }{\partial z}\left(12z\right) \\ =2+3+12 \\ =17 \end{gathered}$$

Here, the region is bounded by the surface S, where S is the surface bounded below by $z=x^2+y^2$and above by the sphere $x^2+y^2+z^2=2$.

The region is greatly simplified when expressed it in cylindrical coordinates $(r,\theta ,z)$.

Because$x^2+y^2=r^2$, the equation $z=x^2+y^2$becomes $z=r^2$, and the equation $x^2+y^2+z^2=2$becomes $r^2+z^2=2or{z}^2=2-r^2$

The inner integral in localid="1651145205713" $Z$runs fromlocalid="1651145200803" $z=r^2$(the lower surface) to$z=\sqrt{2-r^2}$(the upper surface). The boundary of the projection onto the $xy$-plane is determined by the intersection of the two surfaces. Equating the $Z$-coordinates in the equations of the two surfaces $z^2=2-r$and $z=r^2$it gives ,

$$\begin{gathered} {\left(r^2\right)}^2=2-r^2 \\ {r}^4=2-r^2 \\ {r}^4+r^2-2=0 \\ r=1 \end{gathered}$$

Hence, in cylindrical coordinates, the region of integration is described as follows,

$R=\left\{(r,\theta ,z):0\le r\le 1,0\le \theta \le 2\pi ,r^2\le z\le \sqrt{2-r^2}\right\}$

In cylindrical coordinates,

$$\begin{gathered} x=r\cos \theta ,y=r\sin \theta ,z=zand \\ {x}^2+y^2=r^{2}and \\ dV=dzrdrd\theta \end{gathered}$$

Now, use Divergence Theorem (1) to evaluate the integral ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$as follows: $$\begin{gathered} \int {\int }_{s}F(x,y,z).ndS=\int \int {\int }_{R}divF(x,y,z)dV \\ =\int \int {\int }_{R}17dV \\ =17\int \int {\int }_{R}dV \end{gathered}$$ $={\iiint }_{R}17dV$

$=17{\int }_0^{2\pi }{\int }_0^{1\sqrt{2-r^2}}{\int }_{r^2}dzdrd\theta$

$=17{\int }_0^{2\pi }{\int }_0^1\left[{\int }_{r^2}^{\sqrt{2-r^2}}dz\right]rdrd\theta$

$=17{\int }_0^{2\pi }{\int }_0^1[z{]}_{r^2}^{\sqrt{2-r^2}}rdrd\theta$

$=17{\int }_0^{2\pi }{\int }_0^1\left(\sqrt{2-r^2}-r^2\right)rdrd\theta$

Simplify the last integral as follows:

$$\begin{gathered} {\int }_{}{\int }_{s}F(x,y,z).ndS=17{\int }_0^{2\mathrm{\pi }}{\int }_0^1(\sqrt{2-r^2}-r^2)rdrd\theta \\ =17{\int }_0^{2\mathrm{\pi }}\left[{\int }_0^1\right(\sqrt{2-r^2}-r^2\left)rdr\right]d\theta \\ =17{\int }_0^{2\mathrm{\pi }}\left[{\int }_0^1\right(r\sqrt{2-r^2}-r^3\left)dr\right]d\theta \\ =17{\int }_0^{2\mathrm{\pi }}{[-\frac{{(2-r^2)}^{3/2}}{3}-\frac{r^4}{4}]}_0^{1}d\theta \\ =17{\int }_0^{2\mathrm{\pi }}\left[\right(-\frac{{(2-1^2)}^{3/2}}{3}-\frac{1^4}{4})-(-\frac{{(2-0^2)}^{3/2}}{3}-\frac{0^4}{4}\left)\right]d\theta \\ =17{\int }_0^{2\mathrm{\pi }}\left[\right(-\frac{1}{3}-\frac{1}{4})-(-\frac{2^{2/3}}{3}\left)\right]d\theta \\ =17{\int }_0^{2\mathrm{\pi }}\left(\frac{8\sqrt{2}-7}{12}\right)d\theta \\ =17\left(\frac{8\sqrt{2}-7}{12}\right){\left[\theta \right]}_0^{2\mathrm{\pi }} \\ =17\left(\frac{8\sqrt{2}-7}{12}\right)[2\mathrm{\pi }-0] \\ =\frac{17\mathrm{\pi }}{6}(8\sqrt{2}-7) \end{gathered}$$

Therefore, the required integral is .