$\mathbf{F}(x,y,z)=15x{z}^2\mathbf{i}+15y{x}^2\mathbf{j}+15y^{2}z\mathbf{k}$, and $s$ is the surface of the lower half of the unit sphere, along with the unit circle in the plane.

Consider the vector field below:

$\mathbf{F}(x,y,z)=15x{z}^2\mathbf{i}+15y{x}^2\mathbf{j}+15y^{2}z\mathbf{k}$

The goal is to find the integral ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$, where $S$is the surface of the lower half of the unit sphere, as well as the unit circle in the plane, and $\mathbf{n}$is the normal vector heading outwards.

To evaluate this integral, use the Divergence Theorem.

Divergence According to the theorem,

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

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

Where $\mathbf{n}$ is the normal vector pointing outwards."

First, determine the vector field's divergence. $\mathbf{F}(x,y,z)=15x{z}^2\mathbf{i}+15y{x}^2\mathbf{j}+15y^{2}z\mathbf{k}$.

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

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

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

Then there's the vector field's divergence. $\mathbf{F}(x,y,z)=15x{z}^2\mathbf{i}+15y{x}^2\mathbf{j}+15y^{2}z\mathbf{k}$is going to be

$div\mathbf{F}(x,y,z)=\left(\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}\right)·\left(15x{z}^2\mathbf{i}+15y{x}^2\mathbf{j}+15y^{2}z\mathbf{k}\right)$

$=\frac{\partial }{\partial x}\left(15x{z}^2\right)+\frac{\partial }{\partial y}\left(15y{x}^2\right)+\frac{\partial }{\partial z}\left(15y^{2}z\right)$

$=15z^2+15x^2+15y^2$

$=15\left(x^2+y^2+z^2\right)$

Now evaluate the integral using the Divergence Theorem (1). ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS$in the following manner:

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

$={\iiint }_{R}15\left(x^2+y^2+z^2\right)dV$

$=15{\iiint }_R\left(x^2+y^2+z^2\right)dV$...........2

The region is defined by the surface $S$, where $S$which is the lower half of the unit sphere's surface, as well as the unit circle in the plane.

When represented in spherical coordinates $(\rho ,\phi ,\theta )$. Both the integrand and the region are substantially simplified. Change the coordinates of this integral (2) to spherical and integrate it.

The zone of integration is defined as follows in spherical coordinates:

$R=\left\{(\rho ,\phi ,\theta ):0\le \rho \le 1,\frac{\pi }{2}\le \phi \le \pi ,0\le \theta \le 2\pi \right\}.$

When expressed in spherical coordinates,

$x=\rho \sin \phi \cos \theta ,y=\rho \sin \phi \sin \theta ,z=\rho \cos \phi$, and

$x^2+y^2+z^2={\rho }^2$, and

$dV={\rho }^2\sin \phi d\rho d\phi d\theta$

Then, as follows, assess the integral (2):

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=15{\iiint }_R\left(x^2+y^2+z^2\right)dV$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }{\int }_0^1\left({\rho }^2\right)\left({\rho }^2\sin \phi d\rho d\phi d\theta \right)$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }{\int }_0^1{\rho }^4\sin \phi d\rho d\phi d\theta$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\sin \phi \left[{\int }_0^1{\rho }^{4}d\rho \right]d\phi d\theta$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\sin \phi {\left[\frac{{\rho }^5}{5}\right]}_0^{1}d\phi d\theta$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\sin \phi \left[\frac{1^5}{5}-\frac{0^5}{5}\right]d\phi d\theta$

$=15{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\frac{1}{5}\sin \phi d\phi d\theta$

$=3{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\sin \phi d\phi d\theta$

Reduce the last integral to the following:

${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=3{\int }_0^{2\pi }{\int }_{\pi /2}^{\pi }\sin \phi d\phi d\theta$

$=3{\int }_0^{2\pi }\left[{\int }_{\pi /2}^{\pi }\sin \phi d\phi \right]d\theta$

$=3{\int }_0^{2\pi }[-\cos \phi {]}_{\pi /2}^{\pi }d\theta$

$=3{\int }_0^{2\pi }\left[(-\cos \pi )-\left(-\cos \frac{\pi }{2}\right)\right]d\theta$

$=3{\int }_0^{2\pi }\left[\right(-(-1))-(-0\left)\right]d\theta$

$=3{\int }_0^{2\pi }1d\theta$

$=3[\theta {]}_0^{2\pi }$

$=3[2\pi -0]$

$=6\pi$

As a result, the necessary integral is ${\iint }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=6\pi$.