Consider the surface $S$ with positive orientation that forms the boundary of the closed, bounded solid $E$ shown in Figure 61. Notice that the surface $S$ can be decomposed into three surfaces, $S_{1},$ $S_{2},$ and $S_{3}$, and that each of the outer unit normal vectors $\mathbf{n}_{1},$ $\mathbf{n}_{2}$, and $\mathbf{n}_{3}$ of the three surfaces points away from the solid $E$ enclosed by $S$.

Figure 61

Assume the surfaces $S_{1}$ and $S_{2}$ are defined by the equations $$\begin{equation*} S_{1}{:}\quad z=f_{1}(x,y)\qquad \hbox{and}\qquad S_{2}{:}\quad z=f_{2}(x,y) \end{equation*}$$

where $f_{1}(x,y)<f_{2}(x,y)$ on $D,$ the functions $f_{1}$ and $f_{2}$ are continuous on $D$, and both $f_{1}$ and $\ f_{2}$ have continuous partial derivatives on $D$. The surface $S_{3}$ is the portion of a cylindrical surface between $S_{1}$ and $S_{2}$ formed by lines parallel to the $z$-axis and along the boundary of $D$. The solid $E$ is enclosed by $S_{2}$ on the top, $S_{1}$ on the bottom, and the lateral surface $S_{3}$ on the sides. A surface $S$ with these properties is called $\boldsymbol x$$\boldsymbol y$-simple.

Find the $\mathbf{k}$ component of the outer unit normal vectors of $S$.

Solution For the top surface $S_{2}$: $z=f_{2}(x,y)$, the outer unit normal vector $\mathbf{n}_{2}$ is the same as the upward-pointing unit normal vector to $S_{2}$. $$\begin{equation*} \mathbf{n}_{2}=\dfrac{-(f_{2})_x(x,y)\mathbf{i}-(f_{2})_ y(x,y)\mathbf{j}+\mathbf{k} }{\sqrt{[(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}} \end{equation*}$$

The $\mathbf{k}$ component of $\mathbf{n}_{2}$ is given by the direction cosine $\cos \gamma _{2}$, where $$\cos \gamma _{2}=\mathbf{n}_{2}\,{\cdot}\, \mathbf{k}=\dfrac{1}{\sqrt{ [(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}}$$

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For the bottom surface $S_{1}$: $z=f_{1}(x,y)$, the outer unit normal vector $\mathbf{n}_{1}$ equals the downward-pointing unit normal vector $-\mathbf{n}$ to $S_{1}$. $$\begin{equation*} \mathbf{n}_{1}=-\mathbf{n}=\dfrac{(f_{1})_{x}(x,y)\mathbf{i}+(f_{1})_y(x,y)\mathbf{j}- \mathbf{k}}{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}} \end{equation*}$$

The $\mathbf{k}$ component of $\mathbf{n}_{1}$ is given by the direction cosine $\cos \gamma _{1}$, where $$\cos \gamma _{1}=\mathbf{n}_{1}\,{\cdot}\, \mathbf{k}=\dfrac{-1}{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}}$$

For the lateral surface $S_{3}$, the outer unit normal vector $\mathbf{n} _{3}$ is orthogonal to $\mathbf{k}$ at each point on $S_{3}$. The $\mathbf{k }$ component of $\mathbf{n}_{3},$ given by the direction cosine, $\cos \gamma _{3}$, is therefore $0$.