Begin by calculating

$\nabla \times \mathbf{F}\=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ y z& x^{2}z& x y\end{array}\right|$ = $\left[\begin{array}{c}x-x^2\\ 0\\ z \(2 x-1\)\end{array}\right]$ 

The normal component of this vector must be integrated over two surfaces: $S_1$, the cylinder; and $S_2$, the unit disk in the plane $z\=a$. Using the parametrization $x\=\cos \left(\mathrm{\θ}\right)\,y\=\sin \left(\mathrm{\θ}\right)$ for the cylinder, an outward unit normal is ${\mathbf{N}}_1\=\cos \left(\mathrm{\θ}\right) \mathbf{i}\+\sin \left(\mathrm{\θ}\right) \mathbf{j}\+0 \mathbf{k}$. Hence, on this surface

$\left(\nabla \times \mathbf{F}\right)·{\mathbf{N}}_1\=\left(\cos \left(\mathrm{\θ}\right)-{\cos }^2\left(\mathrm{\θ}\right)\right) \cos \left(\mathrm{\θ}\right)$ 

Represent the cylinder by the position vector $\mathbf{R}\=\cos \left(\mathrm{\theta }\right) \mathbf{i}\+\sin \left(\mathrm{\theta }\right) \mathbf{j}\+z \mathbf{k}$ so $\mathrm{d\σ}\=∥{\mathbf{R}}_{\mathrm{\theta }}\times {\mathbf{R}}_z∥ d\mathrm{\theta } \mathrm{dz}$. Now

${\mathbf{R}}_{\mathrm{\theta }}\times {\mathbf{R}}_z$ = $\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -\sin \left(\mathrm{\θ}\right)& \cos \left(\mathrm{\θ}\right)& 0\\ 0& 0& 1\end{array}\right|$ = $\left[\begin{array}{c}\cos \(\mathrm{\θ}\)\\ \sin \(\mathrm{\θ}\)\\ 0\end{array}\right]$ 

so $\mathrm{d\σ}\=∥{\mathbf{R}}_{\mathrm{\theta }}\times {\mathbf{R}}_z∥ d\mathrm{\theta } \mathrm{dz}$ = $1\cdot d\mathrm{\θ} \mathrm{dz}$. Hence,

 $\int {\int }_{S_1}\left(\nabla \times \mathbf{F}\right)·{\mathbf{N}}_1 \mathrm{d\σ}$ = ${\int }_0^{2 \mathrm{\π}}{\int }_0^a\cos \left(\mathrm{\θ}\right)\left(\cos \left(\mathrm{\θ}\right)-{\cos }^2\left(\mathrm{\θ}\right)\right)\mathrm{dz} d\mathrm{\θ}$ = $\mathrm{\π} a$

On the surface $S_2$, ${\mathbf{N}}_2\=\mathbf{k}$, so $\left(\nabla \times \mathbf{F}\right)·{\mathbf{N}}_2\=z \left(2 x-1\right)$. Using cylindrical coordinates,

${\int }_0^1{\int }_0^{2 \mathrm{\π}}a \left(2 r \cos \left(\mathrm{\θ}\right)-1\right) r \mathit{\ⅆ}\mathrm{\θ} \mathit{\ⅆ}r\= -\mathrm{\π} a$  

The net flux of $\nabla \times \mathbf{F}$ through $S\=S_1\bigcup S_2$ is $\mathrm{\π} a-\mathrm{\π} a\=0$.

The line integral 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\\ 0\\ \sin \(t\)\cos \(t\)\end{array}\right]·\left[\begin{array}{c}-\sin \(t\) \mathrm{dt}\\ \cos \(t\) \mathrm{dt}\\ 0\end{array}\right]\= 0 \mathrm{dt}$ 

Clearly, then, ${\int }_C\mathbf{F}·\mathbf{dr}$ = ${\int }_0^{2 \mathrm{\π}}0 \mathit{\ⅆ}t$   = 0. 

To evaluate $\int {\int }_{S_1}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, where $S_1$ is the vertical wall of the cylinder, use the task template in Table 9.9.3(a) where $\nabla \times \mathbf{F}$ is given in Cartesian coordinates, but the cylinder $S_1$ is given in cylindrical coordinates.

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 $\int {\int }_{S_2}\left(\nabla \times \mathbf{F}\right)·\mathbf{N} \mathrm{d\σ}$, where $S_2$ is the "lid" for the cylinder, use the task template in Table 9.9.3(b) where $\nabla \times \mathbf{F}$ is given in Cartesian coordinates, but the cylinder $S_1$ is given in cylindrical coordinates.

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.

The net flux of $\nabla \times \mathbf{F}$ through $S\=S_1\bigcup S_2$ is $\mathrm{\π} a-\mathrm{\π} a\=0$.

Table 9.9.3(c) accesses the LineInt command through the Context Panel.

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

Clearly, then, ${\int }_C\mathbf{F}·\mathbf{dr}$ = ${\int }_0^{2 \mathrm{\π}}0 \mathit{\ⅆ}t$   = 0. 

Let $S$ be the union of two surfaces $S_1$ (the cylinder) and $S_2$ (the disk on top).

The net flux of $\nabla \times \mathbf{F}$ through $S\=S_1\bigcup S_2$ is $\mathrm{\π} a-\mathrm{\π} a\=0$. $$

Table 9.9.3(d) uses the LineInt command to obtain the value of ${∳}_C\mathbf{F}·\mathbf{dr}$, where $C$ is the unit circle centered at the origin, and lying in the plane $z\=0$.

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.