The closed surface $S$ consists of two parts: $S_a$, the upper half of the ellipsoid; and $S_b$, the interior and boundary of the ellipse in the plane $z\=0$.

The calculation $\int {\int }_{S_a}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=-\mathrm{\π}\/8$ was implemented in Example 9.9.6, where $S_a$ was called $S$.

All that is needed here is to show that $\int {\int }_{S_b}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=\mathrm{\π}\/8$, so begin by calculating

$\nabla \times \mathbf{F}\=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ x^2 y& y z& x z\end{array}\right|$ = $\left[\begin{array}{c}-y\\ -z\\ -x^2\end{array}\right]$ 

A downward (and hence outward) unit normal on $S_b$ is clearly $-\mathbf{k}$. Consequently,

 $\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ} \= x^2 \mathrm{dA}$ 

To integrate this over the interior of $C$, express $C$ in polar coordinates by the calculation

${\left(r \cos \left(\mathrm{\θ}\right)\right)}^2\+4{\left(r \sin \left(\mathrm{\θ}\right)\right)}^2\=1$ 

from which it follows that $r\=r\left(\mathrm{\θ}\right)\=1\/\sqrt{{\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)}\equiv E$. Change to polar coordinates to obtain for $\int {\int }_{S_b}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ the integral

${\int }_0^{2 \mathrm{\π}}{\int }_0^{E}r \left(r^2{\cos \left(t\right)}^2\right) \mathrm{dr} \mathrm{d\θ}\=\frac{\mathrm{\π}}{8}$ 

The closed surface $S$ consists of two parts: $S_a$, the upper half of the ellipsoid; and $S_b$, the interior and boundary of the ellipse in the plane $z\=0$.

The calculation $\int {\int }_{S_a}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=-\mathrm{\π}\/8$ was implemented in Example 9.9.6, where $S_a$ was called $S$.

All that is needed here is to show that $\int {\int }_{S_b}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=\mathrm{\π}\/8$.

To evaluate $\int {\int }_{S_b}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, use the task template in Table 9.9.7(b). 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.

Maple obtains $-\mathrm{\π}\/8$ because it uses the upward normal ${\mathbf{R}}_x\times {\mathbf{R}}_y\=\mathbf{k}$, where $\mathbf{R}\=x \mathbf{i}\+y \mathbf{j}\+0 \mathbf{k}$. That this normal points upward, instead of downward (and hence outward) is established by the following calculation.

${\mathbf{R}}_x\times {\mathbf{R}}_y$ = $\left[\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$ = $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ = $\mathbf{k}$ 

Changing to the appropriate downward normal changes the sign of the integral.

The closed surface $S$ consists of two parts: $S_a$, the upper half of the ellipsoid; and $S_b$, the interior and boundary of the ellipse in the plane $z\=0$.

The calculation $\int {\int }_{S_a}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=-\mathrm{\π}\/8$ was implemented in Example 9.9.6, where $S_a$ was called $S$.

All that is needed here is to show that $\int {\int }_{S_b}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=\mathrm{\π}\/8$.

Maple obtains $-\mathrm{\π}\/8$ because it uses the upward normal ${\mathbf{R}}_x\times {\mathbf{R}}_y\=\mathbf{k}$, where $\mathbf{R}\=x \mathbf{i}\+y \mathbf{j}\+0 \mathbf{k}$. That this normal points upward, instead of downward (and hence outward) is established by the following calculation.

${\mathbf{R}}_x\times {\mathbf{R}}_y$ = $\left[\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$ = $\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ = $\mathbf{k}$ 

Changing to the appropriate downward normal changes the sign of the integral.