Find both orientations for the paraboloid defined by $z=9-x^{2}-y^{2},$ $z\geq 0.$

Solution (a) Since the paraboloid is given by $f(x,y) = 9-x^{2} - y^{2}, z \ge 0$, we can use the equations (3) and (4) to find the unit normal vectors. We begin by finding the partial derivatives $$f_{x}(x,y) =-2x \qquad f_{y}(x,y) =-2y$$

Then $$\begin{eqnarray*} \mathbf{n}&=&\dfrac{2x\,\mathbf{i}+2y\,\mathbf{j}+\mathbf{k}}{\sqrt{[-2x]^{2}+[-2y]^{2}+1}}=\dfrac{2x\,\mathbf{i}+2y\,\mathbf{j}+\mathbf{k}}{\sqrt{4x^{2}+4y^{2}+1}} \\[7pt] -\mathbf{n}&=&\dfrac{-2x\,\mathbf{i-}2y\,\mathbf{j}-\mathbf{k}}{\sqrt{[-2x]^{2}+[-2y]^{2}+1}}=\dfrac{ -2x\,\mathbf{i}-2y\,\mathbf{j}-\mathbf{k}}{\sqrt{4x^{2}+4y^{2}+1}} \end{eqnarray*}$$

Notice that the $\mathbf{k}$ component of $\mathbf{n}$ is positive, indicating $\mathbf{n}$ is the upward-pointing unit normal vector.

(b) We parametrize $z=9-x^{2}-y^{2},$ $z\geq 0$, using cylindrical coordinates, by defining the parametric equations $$\begin{equation*} x=r\cos \theta \qquad y=r\sin \theta \qquad z=9-(x^{2}+y^{2}) =9-r^{2} \end{equation*}$$

Then the parametrization is $$\begin{equation*} \mathbf{r}=\mathbf{r}( r,\theta ) =r\cos \theta \,\mathbf{i} +r\sin \theta \,\mathbf{j}+( 9-r^{2}) \mathbf{k} \end{equation*}$$

We seek $\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert$. The partial derivatives of $\mathbf{r}$ are $$\begin{eqnarray*} \mathbf{r}_{r}&=&\cos \theta \,\mathbf{i}+\sin \theta \,\mathbf{j}-2r\mathbf{k} \qquad \mathbf{r}_{\theta }=-r\sin \theta \,\mathbf{i}+r\cos \theta \,\mathbf{j}\\[4pt] \mathbf{r}_{r}\times \mathbf{r}_{\theta } &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \cos \theta & \sin \theta & -2r \\[3pt] -r\sin \theta & r\cos \theta & 0 \end{array}\right| =2r^{2}\cos \theta\, \mathbf{i}+2r^{2}\sin \theta\, \mathbf{j}+r\mathbf{k} \\[4pt] \left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert &=&\sqrt{ 4r^{4}\cos ^{2}\theta +4r^{4}\sin ^{2}\theta +r^{2}}=\sqrt{4r^{4}+r^{2}}=r \sqrt{4r^{2}+1} \end{eqnarray*}$$

The unit normal vectors are $$\begin{eqnarray*} \mathbf{n}&=&\dfrac{\mathbf{r}_{r}\times \mathbf{r}_{\theta }}{\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert }=\dfrac{2r^{2}\cos \theta\, \mathbf{i}+2r^{2}\sin \theta\, \mathbf{j}+r\mathbf{k}}{r\sqrt{4r^{2}+1}} =\dfrac{2r\cos \theta\, \mathbf{i}+2r\sin \theta\, \mathbf{j}+\mathbf{k}}{\sqrt{ 4r^{2}+1}}\\[4pt] -\mathbf{n}&=&\dfrac{-2r\cos \theta\, \mathbf{i} -2r\sin \theta\, \mathbf{j}-\mathbf{k}}{\sqrt{4r^{2}+1}} \end{eqnarray*}$$

Figure 58 The outer unit normal vectors of $z=9-x^{2}-y^{2},$ $z\geq 0.$

Notice that the unit vectors found in (a) and (b) are equivalent.