Find the outer unit normal vectors to the solid $E$ enclosed by $$z=f(x,y)=\sqrt{R^{2}-x^{2}-y^{2}}\quad\hbox{and}\quad z=0\qquad 0\leq x^{2}+y^{2}\leq R^{2}$$

Solution The solid $E$ is the interior of a hemisphere with center at $(0,0,0)$ and radius $R$ as shown in Figure 62. The surface $S$ consists of two surfaces, $S_{1}$ and $S_{2}$. The bottom surface $S_{1}$ and top surface $S_{2}$ are defined by $$\begin{equation*} S_{1}{:}\quad z=0 \quad \hbox{and}\quad S_{2}{:}\quad z=f(x,y)=\sqrt{R^{2}-x^{2}-y^{2}}\qquad 0\leq x^{2}+y^{2}\leq R^{2} \end{equation*}$$

The outer unit normal vector $\mathbf{n}_{1}$ of $S_{1}$ is $-\mathbf{k}$.

To find the outer unit normal vector $\mathbf{n}_{2}$ of $S_{2}$, we find $f_{x}(x,y)$ and $f_{y}(x,y) .$ $$\begin{equation*} f_{x}(x,y)=\dfrac{-x}{\sqrt{R^{2}-x^{2}-y^{2}}}=\dfrac{-x}{z}\qquad \hbox{and} \qquad f_{y}(x,y)=\dfrac{-y}{\sqrt{R^{2}-x^{2}-y^{2}}}=\dfrac{-y}{z} \end{equation*}$$

Then $$\begin{eqnarray*} \mathbf{n}_{2}&=&\dfrac{-f_{x}(x,y)\mathbf{i}-f_{y}(x,y)\mathbf{j}+\mathbf{k}}{ \sqrt{[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}}=\dfrac{\dfrac{x}{z}\mathbf{i}+ \dfrac{y}{z}\mathbf{j}+\mathbf{k}}{\sqrt{\dfrac{x^{2}}{z^{2}}+\dfrac{y^{2}}{ z^{2}}+1}}\\ &=&\dfrac{x\mathbf{i}+y\mathbf{j}+z\,\mathbf{k}}{\sqrt{ x^{2}+y^{2}+z^{2}}}=\dfrac{x\mathbf{i}+y\mathbf{i}+z\mathbf{k}}{R} \end{eqnarray*}$$