Compute the outward flux of the vector field $\mathbf{F}(x, y, z) = 2\!x\mathbf{i} + 3\!y\mathbf{j} + 1\!z\mathbf{k}$ across the boundary of the right cylinder with radius 6 with bottom edge at height z = -3 and upper edge at z = 7.
Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the cylinder to be positive.

Part 1 - Using a Surface Integral
First we parameterize the three faces of the cylinder.

Note: in the following type $\theta$ as theta.

The bottom face: $\sigma_1(r,\theta)=(x_1(r,\theta),y_1(r,\theta),z_1(r,\theta)$ where $0\leq\theta\leq 2\pi$, $0\leq r\leq 6$ and
$x_1(r,\theta) =$
$y_1(r,\theta) =$
$z_1(r,\theta) = -3$

The top face: $\sigma_2(r,\theta)=(x_2(r,\theta),y_2(r,\theta),z_2(r,\theta)$ where $0\leq\theta\leq 2\pi$, $0\leq r\leq 6$ and
$x_2(r,\theta) =$
$y_2(r,\theta) =$
$z_2(r,\theta) = 7$

The lateral face: $\sigma_3(\theta,z)=(x_3(\theta,z),y_3(\theta,z),z_3(\theta,z)$ where $0\leq\theta\leq 2\pi$, $-3\leq z\leq 7$ and
$x_3(\theta) =$
$y_3(\theta) =$
$z_3(z)=$

Then (mind the orientation)
$\displaystyle\int\int_\sigma F\cdot n dS = \int_0^{2\pi}\int_0^{6}F(\sigma_1)\cdot\left(\frac{\partial \sigma_1}{\partial \theta}\times\frac{\partial \sigma_1}{\partial r}\right) dr\,d\theta$ $\displaystyle +\int_0^{2\pi}\int_0^{6} F(\sigma_2)\cdot\left(\frac{\partial \sigma_2}{\partial r}\times\frac{\partial \sigma_2}{\partial \theta}\right) dr\,d\theta$ $\displaystyle +\int_0^{2\pi}\int_{-3}^{7} F(\sigma_3)\cdot\left(\frac{\partial \sigma_3}{\partial z}\times\frac{\partial \sigma_3}{\partial \theta}\right) dz\,d\theta$

$\hskip 20pt =$ + +

$\hskip 20pt =$

Part 2 - Using the Divergence Theorem

$\displaystyle\int\int_\sigma F\cdot n \,dS =\int\int\int_G {\rm div } F\,dV$

$\hskip 20pt =$

$\displaystyle\int$

$\displaystyle\int$

$\displaystyle\int$

$dz\,dr\,d\theta$

$\hskip 20pt =$