A Cartesian representation of the unit sphere is $f\left(x\,y\,z\right)\equiv x^2\+y^2\+z^2\=1$.

A unit outward normal on the sphere is $\genfrac{}{}{0ex}{}{\nabla f\/∥\nabla f∥}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{f\=1}$. The normalized gradient, evaluated on the sphere, is then

$\frac{2}{2\sqrt{x^2\+y^2\+z^2}}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ ⇒ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\equiv \mathbf{N}$ 

According to Table 9.6.1, the element of surface area can be taken as $\frac{∥\nabla f∥}{\|f_z\|}$$\mathrm{dA}$; so, on the sphere itself, $\frac{2\sqrt{x^2\+y^2\+z^2}}{\left|2 z\right|}$ becomes $1\/\left|z\right|$. On both hemispheres, $\left|z\right|\=\sqrt{1-x^2-y^2}$. Now, the projection of each hemisphere onto the plane $z\=0$ is the unit disk, so there are two integrals over this disk to consider, each the integral of

$\mathbf{F}·\mathbf{N} \mathrm{d\σ}$ = $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]·\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \frac{1}{\left|z\right|} \mathrm{dA}$ = $\frac{x^2\+y^2\+z^2}{\left|z\right|} \mathrm{dA}$ = $\frac{1}{\sqrt{1-x^2-y^2}} \mathrm{dA}$

The two integrals are therefore the same, so in polar coordinates, the flux integrals become

$2{\int }_0^{2 \mathrm{\π}}{\int }_0^1{\left(1-r^2\right)}^{-1\/2} r \mathrm{dr} \mathrm{d\θ}\=4 \mathrm{\π}$ 

The display in Table 9.6.12(a) is obtained by accessing the MultiInt command through the Context Panel system after loading the Student MultivariateCalculus package, and after making the choices indicated in the dialogs in Figures 9.6.12(a, b).