Begin by calculating

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

A unit (upward) normal on $S_1$ is $\mathbf{N}\=\frac{1}{\sqrt{29}}\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]$ where $z\=6-x\/2-3 y\/4$. Hence,

$\mathrm{d\σ}\=\sqrt{1\+z_x^2\+z_y^2} \mathrm{dA}\=\sqrt{1\+{\left(1\/2\right)}^2\+{\left(3\/4\right)}^2} \mathrm{dA}\=\sqrt{29}\/4 \mathrm{dA}$ 

and$$

which, on $S_1$ becomes

The projection of $S_1$ onto the plane $z\=0$ is a right triangle whose hypotenuse is $y\=8-2 x\/3$, from which it follows that

$\int {\int }_{S_1}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\={\int }_0^{12}{\int }_0^{8-2 x\/3}\left(6\+5 x\/2-y\/4\right) \mathrm{dy} \mathrm{dx}$ = $-328$ 

An outward normal on $S_2$ where $y\=0$ is the unit vector $-\mathbf{j}$. Hence, $\left(\nabla \times \mathbf{F}\right)·\mathbf{N}\=z$ on $S_2$, a right triangle with hypotenuse $z\=6-x\/2$ in the $\mathrm{xz}$-plane. This gives the integral

$\int {\int }_{S_2}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ = ${\int }_0^{12}{\int }_0^{6-x\/2}z \mathrm{dz} \mathrm{dx}$ = 72

An outward normal on $S_3$ where $x\=0$ is the unit vector $-\mathbf{i}$. Hence, $\left(\nabla \times \mathbf{F}\right)·\mathbf{N}\=z$ on $S_3$, a right triangle with hypotenuse $z\=6-3 y\/4$ in the $\mathrm{yz}$-plane. This gives the integral

$\int {\int }_{S_3}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ = ${\int }_0^8{\int }_0^{6-3 y\/4}y \mathrm{dz} \mathrm{dy}$ = 64

From these results it follows that $\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ = $-328\+72\+64\=-192$.$$

The surface $S$ caps $C$, the right triangle with vertices $\left(0\,0\,0\right)\,\left(12\,0\,0\right)\,\left(0\,8\,0\right)$. Since this triangle lies in the plane $z\=0$, $\mathbf{F}·\mathbf{dr}\=0$ on each leg, and is $x \left(8-2 x\/3\right)$ on the hypotenuse. See Table 9.9.5(b) for the details.$$

From Table 9.9.5(b), it follows that ${∳}_C\mathbf{F}·\mathbf{dr}$ = ${\int }_{12}^{0}x \left(8-2 x\/3\right) \mathrm{dx}\=-192$.

To evaluate $\int {\int }_{S_1}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, use the task template in Table 9.9.5(c). 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.

To evaluate $\int {\int }_{S_2}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, use the task template in Table 9.9.5(d). 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.

To evaluate $\int {\int }_{S_3}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, use the task template in Table 9.9.5(e). 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.

Unfortunately, Maple takes $\mathbf{N}\=\mathbf{i}$, not $\mathbf{N}\=-\mathbf{i}$ on $S_3$. Maple's default normal on the open surface described by the position vector $\mathbf{R}\=\mathbf{R}\left(y\,z\right)$ is the normalized version of ${\mathbf{R}}_y\times {\mathbf{R}}_z$. Since $\mathbf{R}\left(y\,z\right)\=y \mathbf{j}\+z \mathbf{k}$,

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

Hence, the value of $\int {\int }_{S_3}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ is actually 64, not $-64$, and

$\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=-328\+72\+64\= -192$ 

Table 9.9.4(f) accesses the LineInt command through the Context Panel.

Note that in order for Maple to trace the edges of the triangle $C$, the starting vertex must be repeated as the last node to be reached.

Unfortunately, Maple takes $\mathbf{N}\=\mathbf{i}$, not $\mathbf{N}\=-\mathbf{i}$ on $S_3$. Maple's default normal on the open surface described by the position vector $\mathbf{R}\=\mathbf{R}\left(y\,z\right)$ is the normalized version of ${\mathbf{R}}_y\times {\mathbf{R}}_z$. Since $\mathbf{R}\left(y\,z\right)\=y \mathbf{j}\+z \mathbf{k}$,

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

Hence, the value of $\int {\int }_{S_3}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$ is actually 64, not $-64$, and

$\int {\int }_S\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}\=-328\+72\+64\= -192$ 

Table 9.9.5(g) uses the LineInt command to obtain the value of ${∳}_C\mathbf{F}·\mathbf{dr}$, where $C$ is the triangle whose vertices are $\left(0\,0\,20\right)\,\left(30\,0\,0\right)\,\left(0\,24\,0\right)$.

Note that in order for Maple to trace the edges of the triangle $C$, the starting vertex must be repeated as the last node to be reached.