Let $Z\=\sqrt{1-x^2-y^2}$ be a Cartesian representation of the upper hemisphere $S$. It follows that

$\mathbf{R}\=\left[\begin{array}{c}x\\ y\\ Z\end{array}\right]$ $\Rightarrow$${\mathbf{R}}_x\times {\mathbf{R}}_y$ = $\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& -x\/Z\\ 0& 1& -y\/Z\end{array}\right|$ = $\left[\begin{array}{c}x\/Z\\ y\/Z\\ 1\end{array}\right]$  and  $\mathbf{N}\=\left[\begin{array}{c}x\\ y\\ Z\end{array}\right]$ 

The element of surface area can be obtained from

 $\sqrt{1\+Z_x^2\+Z_y^2}\=\sqrt{1\+{\left(x\/Z\right)}^2\+{\left(y\/Z\right)}^2}$ = $1\/Z$

or from $∥{\mathbf{R}}_x\times {\mathbf{R}}_y∥\=1\/Z$. In either event,

$\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=\left[\begin{array}{c}-1\\ 1\\ -1\end{array}\right]·\left[\begin{array}{c}x\\ y\\ Z\end{array}\right]\cdot \frac{1}{Z} dA\=\left(\frac{y}{Z}-\frac{x}{Z}-1\right)\mathrm{dA}$ 

so that $\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ can be implemented in Cartesian coordinates as

${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{{\sqrt{1-x^2}}_{}}\left(\frac{y}{\sqrt{1-x^2-y^2}}-\frac{x}{\sqrt{1-x^2-y^2}}-1\right) \mathrm{dy} \mathrm{dx}\=-\mathrm{\π}$ 

or in polar coordinates as

${\int }_0^1{\int }_0^{2 \mathrm{\π}}r\left(\frac{r \sin \left(\mathrm{\θ}\right)}{\sqrt{1-r^2}}-\frac{r \cos \left(\mathrm{\θ}\right)}{\sqrt{1-r^2}}-1\right) \mathrm{d\θ} \mathrm{dr}\= -\mathrm{\π}$ 

The line integral around $C$, the unit circle centered at the origin, given by ${∳}_C\mathbf{F}·\mathbf{dr}$, can be evaluated if  $C$ is parametrized by the position vector $\mathbf{r}\=\cos \left(t\right) \mathbf{i}\+\sin \left(t\right) \mathbf{j}\+0 \mathbf{k}$ so that on $C$

$\mathbf{F}·\mathbf{dr}\=\left[\begin{array}{c}0\\ -\cos \(t\)\\ -\sin \(t\)\end{array}\right]·\left[\begin{array}{c}-\sin \(t\) \mathrm{dt}\\ \cos \(t\) \mathrm{dt}\\ 0\end{array}\right]\= -{\cos }^2\left(t\right) \mathrm{dt}$ 

Consequently, ${∳}_C\mathbf{F}·\mathbf{dr}$ = $-{\int }_0^{2 \mathrm{\π}}{\cos }^2\left(t\right) \mathrm{dt}\= -\mathrm{\π}$.

The parametrization chosen for $C$ induces a counterclockwise traverse of the circle, an orientation consistent with the choice of an outward normal on $S$.$$

Table 9.9.1(a) contains a task template with which the flux of $\nabla \times \mathbf{F}$ through $S$ is computed. Should the "Clear All and Reset" button in the Task Template be pressed, all the data that has been input to the template will be lost. In that event, the reader should simply re-launch the example to recover the appropriate inputs to the template.

Table 9.9.1(b) contains the calculation of the line integral ${∳}_C\mathbf{F}·\mathbf{dr}$ , where $C$ is the circle capped by $S$.

The astute reader will realize that Maple has evaluated a line integral as an iterated double-integral by invoking Stokes' theorem! Consequently, a validation of Stokes' theorem demands that the line integral be evaluated from first principles. This is easily done if $C$ is parametrized by the position vector

 $\mathbf{r}\=\cos \left(t\right) \mathbf{i}\+\sin \left(t\right) \mathbf{j}\+0 \mathbf{k}$ so that on $C$

$\mathbf{F}·\mathbf{dr}\=\left[\begin{array}{c}0\\ -\cos \(t\)\\ -\sin \(t\)\end{array}\right]·\left[\begin{array}{c}-\sin \(t\) \mathrm{dt}\\ \cos \(t\) \mathrm{dt}\\ 0\end{array}\right]\= -{\cos }^2\left(t\right) \mathrm{dt}$ 

and hence, ${\int }_C\mathbf{F}·\mathbf{dr}$ = $-{\int }_0^{2 \mathrm{\π}}{\cos }^2\left(t\right) \mathit{\ⅆ}t$ = $-\mathrm{\pi }$.

The astute reader will realize that Maple has evaluated a line integral as an iterated double-integral by invoking Stokes' theorem! Consequently, a validation of Stokes' theorem demands that the line integral be evaluated from first principles. This has already been done twice, in the previous two sections.

Figure 9.9.1(a) can be obtained with the Flux command, provided the integration is implemented in Cartesian coordinates with the following syntax. The actual options applied in the figure can be seen in the code hidden in the table cell containing the figure.