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