$\mathbf{F}(x,y,z)=\mathbf{i}+\mathbf{j}+\mathbf{k}$, where S is the lower half of the unit sphere, with n pointing outwards.

Consider the given question,

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

If a surface S is the graph of $z=z\left(x,y\right)$, then the Flux of $\mathit{F}\left(x,y,z\right)$ through S is given below,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})∥z_x\times z_y∥dA\\ ={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}dA\dots ...\left(i\right)\end{array}$

Here, the surface S is the lower half of the unit sphere, so its equation will be,

$z=-\sqrt{1-x^2-y^2}$

Now first find $\frac{\mathrm{\partial }z}{\mathrm{\partial }x}$. The first partial derivates of z are given below,

$$\begin{gathered} \frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(-\sqrt{1-x^2-y^2}\right) \\ =\frac{x}{\sqrt{1-x^2-y^2}} \end{gathered}$$

On finding the value of $\frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}$,

role="math" localid="1650342662824" $$\begin{gathered} \frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}\left(-\sqrt{1-x^2-y^2}\right) \\ =\frac{y}{\sqrt{1-x^2-y^2}} \end{gathered}$$

Then,

$\begin{array}{r}\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}=\sqrt{{\left(\frac{x}{\sqrt{1-x^2-y^2}}\right)}^2+{\left(\frac{y}{\sqrt{1-x^2-y^2}}\right)}^2+1}\\ =\sqrt{\frac{x^2}{1-x^2-y^2}+\frac{y^2}{1-x^2-y^2}+1}\\ =\frac{\sqrt{1-x^2-y^2}}{\sqrt{1}}\end{array}$

The choice of n should have pointing upwards. Then the following vector, is normal to the surface.

Now, for $z=-\sqrt{1-x^2-y^2}$, gives the following vector perpendicular to this surface,

Then the desired normal vector will be,

The value of $\mathbf{F}(x,y,z)\cdot \mathbf{n}$will be,

Substituting these values in equation (i),

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}\right)dA\\ ={\iint }_D \left(\sqrt{1-x^2-y^2}\left(1-\frac{x+y}{\sqrt{1-x^2-y^2}}\right)\right)\left(\frac{1}{\sqrt{1-x^2-y^2}}\right)dA\\ ={\iint }_D \left(1-\frac{x+y}{\sqrt{1-x^2-y^2}}\right)dA\dots \dots .\left(ii\right)\end{array}$

The surface S is the lower half of the unit sphere, so the region of integration D is the unit disk in the xy-plane and centered at the origin.

Hence, in polar coordinates, the region of integration will be,

$D=\left\{\right(r,\theta )\mid 0\le r\le 1,0\le \theta \le 2\pi \}$

In this case,

localid="1650343222895" $\begin{array}{r}x=r\cos \theta ,\\ y=r\sin \theta \end{array}$

Using equation (ii), the flux of the vector field through the surface S,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(1-\frac{x+y}{\sqrt{1-x^2-y^2}}\right)dA\\ ={\int }_0^{2\pi } {\int }_0^1 \left(1-\frac{r\cos \theta +r\sin \theta }{\sqrt{1-r^2}}\right)rdrd\theta \\ ={\int }_0^{2\pi } \left[{\int }_0^1 1dr-(\cos \theta +\sin \theta ){\int }_0^1 \frac{r^2}{\sqrt{1-r^2}}dr\right]d\theta \\ ={\int }_0^{2\pi } \left[[1-0]-(\cos \theta +\sin \theta )\left[\frac{1}{2}\left(-1\sqrt{1-1^2}+{\sin }^{-1}1\right)-\frac{1}{2}\left(-0\sqrt{1-0^2}+{\sin }^{-1}0\right)\right]\right]d\theta \\ \begin{array}{r}\begin{array}{r}={\int }_0^{2\pi } \left[1-\frac{\pi }{4}(\cos \theta +\sin \theta )\right]d\theta \\ ={\left[\theta -\frac{\pi }{4}(\sin \theta -\cos \theta )\right]}_0^{2\pi }\\ =\left(2\pi -\frac{\pi }{4}(\sin 2\pi -\cos 2\pi )-\left(0-\frac{\pi }{4}(\sin 0-\cos 0)\right)\right.\end{array}\\ =2\pi \end{array}\end{array}$