Begin by calculating

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

Express $S$ as $z\=\sqrt{\left(1-x^2-4 y^2\right)\/6}$  and obtain $\mathrm{d\σ}\=\sqrt{1\+z_x^2\+z_y^2} \mathrm{dA}$ as

$\mathrm{d\σ}\=\sqrt{1\+\frac{x^2}{-6x^2-24y^2\+6}\+\frac{16y^2}{-6x^2-24y^2\+6}} \mathrm{dA}\=\frac{1}{\sqrt{6}}\sqrt{\frac{5x^2\+8y^2-6}{x^2\+4y^2-1}} \mathrm{dA}$ 

An upward normal on $S$ is given by $\left[\begin{array}{c}-z_x\\ -z_y\\ 1\end{array}\right]$ = $\left[\begin{array}{c}\frac{x}{\sqrt{-6x^2-24y^2\+6}}\\ \frac{4y}{\sqrt{-6x^2-24y^2\+6}}\\ 1\end{array}\right]$. Normalized, this is

$\mathbf{N}\=\frac{\sqrt{6}}{\sqrt{\frac{5x^2\+8y^2-6}{x^2\+4y^2-1}}}\left[\begin{array}{c}\frac{x}{\sqrt{-6x^2-24y^2\+6}}\\ \frac{4y}{\sqrt{-6x^2-24y^2\+6}}\\ 1\end{array}\right]$ 

Now $\mathbf{N} \mathrm{d\σ}\=$ $\left[\begin{array}{c}\frac{x}{\sqrt{-6x^2-24y^2\+6}}\\ \frac{4y}{\sqrt{-6x^2-24y^2\+6}}\\ 1\end{array}\right] \mathrm{dA}$ and

 $\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=\left(-\frac{yx}{\sqrt{-6x^2-24y^2\+6}}-\frac{4zy}{\sqrt{-6x^2-24y^2\+6}}-x^2\right) \mathrm{dA}$ 

On $S$, where $z\=z\left(x\,y\right)$, this becomes

$\left(-\frac{yx}{\sqrt{-6x^2-24y^2\+6}}-\frac{2}{3}y-x^2\right) \mathrm{dA}$

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 $\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ the integral

${\int }_0^{2 \mathrm{\π}}{\int }_0^E-r \left(\frac{r^2\sin \left(t\right)\cos \left(t\right)}{\sqrt{-6r^2{\cos \left(t\right)}^2-24r^2{\sin \left(t\right)}^2\+6}}\+\frac{2}{3}r\sin \left(t\right)\+r^2{\cos \left(t\right)}^2\right) \mathrm{dr} \mathrm{d\θ}\=-\frac{\mathrm{\π}}{8}$ 

To obtain ${∳}_C\mathbf{F}·\mathbf{dr}$, note that if $r\=x i\+y j\+0 k$, then $\mathbf{F}·\mathbf{dr}\=x^{2}y \mathrm{dx}\+y z \mathrm{dy}$, which, in the plane $z\=0$, becomes just $x^{2}y \mathrm{dx}$. Taking $y\=\pm \sqrt{1-x^2}\/2$ to describe the ellipse $C$ leads to the line integral

 ${∳}_C\mathbf{F}·\mathbf{dr}$ = ${\int }_1^{-1}{x}^2\sqrt{1-x^2}\/2 \mathrm{dx}-{\int }_{-1}^1{x}^2\sqrt{1-x^2}\/2 \mathrm{dx}\=-\frac{\mathrm{\π}}{8}$

To evaluate $\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, use the task template in Table 9.9.6(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.

Table 9.9.6(c) accesses the LineInt command through the Context Panel. Express the ellipse $C$ in polar coordinates $r$ and $t$ and give the resulting expression for $r\left(t\right)$ the name $E$. (See the Mathematical Solution for a derivation of the expression $r\left(t\right)$.)

Table 9.9.6(d) uses the LineInt command to obtain the value of ${∳}_C\mathbf{F}·\mathbf{dr}$, where $C$ is the ellipse $x^2\+4 y^2\=1$. In polar coordinates, this ellipse is given by $r\left(\mathrm{\θ}\right)\=1\/\sqrt{{\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)}$. (See the Mathematical Solution for a derivation of this expression for $r\left(\mathrm{\θ}\right)$.)