I want to compute a surface-integral, but using cylindrical coordinates.
I'll denote the Cartesian coordinate system on $\mathbb{R^3}$ by $\mathbb{R}^3_{xyz}$, and the cylindrical by $\mathbb{R}^3_{\rho \theta z}$. Let $$\textbf{F}(x,y,z) = \begin{bmatrix}F_x(x,y,z) \\ F_y(x,y,z) \\ F_z(x,y,z) \end{bmatrix}$$ be a vector-field. Let $S \subset \mathbb{R}^3_{xyz}$ be the surface I've marked in black on the right below. I wish to compute the flux of $\textbf{F}$ out of $S$, i.e. $\int_S \textbf{F} \cdot \hat{n} \, dS $, where $\hat{n}$ is a unit-vector normal to $S$, pointing outward. Clearly, this calls for cylindrical coordinates. Denote by
$$ \phi(\rho, \theta, z) =\begin{bmatrix} \rho \cos \theta \\ \rho \sin \theta \\ z\end{bmatrix} \quad , \quad \phi^{-1}(x,y,z) =\begin{bmatrix} \sqrt{x^2+y^2} \\ \arctan(y,x) \\ z\end{bmatrix} $$ the standard change-of-variables. In orange I've drawn $\phi^{-1}(S)$.
My approach (which I'm not sure of):
I can write $\textbf{F}$ as a two-form: $\alpha = F_x \, dy \wedge dz \ + \ F_y \, dz \wedge dx \ +\ F_z \, dx \wedge dy$. And then try to pull it back, i.e. compute $\phi^{\star} \alpha$. However I don't know how to compute $\phi^{\star}(dy \wedge dz)$ (or any other one, for that matter). Even then - How do I treat the $\hat{n}$ part?
Somehow I feel there must be some nice way to compute the surface-integral. This is all in part for eventually computing the divergence of $\textbf{F}$ using cylindrical coordinates.
First Edit:
Following the suggestion, I'm instead using a simpler parametrization for $S$. Let $f(\rho, z)=(\rho \cos \theta, \rho \sin \theta, z)$. Then we can compute and get that the normal pointing out of the black surface $S$ is:
$$\tag{1} \hat{n} = \textbf{-} \big(\frac{\partial f}{\partial \rho} \times \frac{\partial f}{\partial z}\big) = - \begin{bmatrix} \sin \theta \\ \cos \theta \\ 0 \end{bmatrix}$$
if we write down
$$\tag{2} \textbf{F} = F_\rho \hat{\rho} + F_\theta \hat{\theta} + F_z \hat{z} = \begin{bmatrix} F_\rho \cos \theta - F_{\theta} \sin \theta \\ F_\rho \sin \theta + F_\theta \cos \theta \\ F_z \end{bmatrix} $$ then
$$\tag{3} \textbf{F} \cdot \hat{n} = \begin{bmatrix} F_\rho \cos \theta - F_{\theta} \sin \theta \\ F_\rho \sin \theta + F_\theta \cos \theta \\ F_z \end{bmatrix} \cdot \begin{bmatrix} -\sin \theta \\ -\cos \theta \\ 0 \end{bmatrix} = -2 F_{\rho} \sin \theta \cos \theta + F_\theta (\sin ^2 \theta - \cos^2 \theta)$$
And I don't see how that helps me. I wish to compute sums of surfaces integrals, and then finally diving by the volume, in order to compute the divergence.
Second edit:
Somewhere in my computation of the div formula I have a mistake, since I don't get the right result:
I'm given $\textbf{F} = F_{\rho} \hat{\rho} + F_{\theta} \hat{\theta} + F_z \hat{z}$. Let $V \subset \mathbb{R}^3_{\rho \theta z}$ be the box depiected on the left, with sides length $\Delta \rho, \Delta \theta, \Delta z$. Let $\Delta V$ denote the volume of V, then $\Delta V = \Delta \rho \cdot \Delta \theta \cdot \Delta z$. Let $(\rho_0, \theta_0, z_0)$ be a point right in the center of the box. I wish to compute
$$\tag{4} \mathop{div} \textbf{F} = \lim_{\Delta V \to 0} \frac{1}{\Delta V} \int_{\partial V} \textbf{F} \cdot \hat{n} \mathop{dS}$$ Let $S_1, \ldots, S_6$ denote the six faces of the box. Then $$ \int_{\partial V} \textbf{F} \cdot \hat{n} \mathop{dS} = \int_{S_1} \textbf{F} \cdot \hat{n}_{S_1} \mathop{dS} + \ldots + \int_{S_6} \textbf{F} \cdot \hat{n}_{S_6} \mathop{dS} $$
Let $S_1$ be the face drawn in orange. Then $\hat{n}_{S_1} = \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}_{(\hat{\rho}, \hat{\theta},\hat{z})}$. We can approximate the flux out of $S_1$ by evaluating $\textbf{F} \cdot \hat{n}_{S_1}$ at some point on $S_1$, and multiplying by the area of $S_1$:
\begin{align} \tag{5} \int_{S_1} \textbf{F} \cdot \hat{n}_{S_1} \mathop{dS} &\approx \textbf{F}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0) \cdot \hat{n}_{S_1} \Delta \rho \cdot \Delta z \\ &= F_{\theta}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0) \cdot \Delta \rho \cdot \Delta z \end{align} Similarly, we can compute for $S_2$, the face across it: \begin{align} \tag{6} \int_{S_2} \textbf{F} \cdot \hat{n}_{S_2} \mathop{dS} \approx F_{\theta}(\rho_0, \theta_0 + \frac{\Delta \theta}{2}, z_0) \cdot \Delta \rho \cdot \Delta z \end{align}
Summing, we have \begin{align} \tag{7} \int_{S_1 + S_2} \textbf{F} \cdot \hat{n} \mathop{dS} &\approx \Big( F_{\theta}(\rho_0, \theta_0 + \frac{\Delta \theta}{2}, z_0) - F_{\theta}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0) \Big) \cdot \Delta \rho \cdot \Delta z \\ &= \frac{F_{\theta}(\rho_0, \theta_0 + \frac{\Delta \theta}{2}, z_0) - F_{\theta}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0)}{\Delta \theta} \cdot \Delta \rho \cdot \Delta \theta \cdot \Delta z \end{align} and then we get that \begin{align} \tag{8} \lim_{\Delta V \to 0 }\frac{1}{\Delta V} \int_{S_1 + S_2} \textbf{F} \cdot \hat{n} \mathop{dS} &= \lim_{\Delta V \to 0 } \frac{F_{\theta}(\rho_0, \theta_0 + \frac{\Delta \theta}{2}, z_0) - F_{\theta}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0)}{\Delta \theta} \\ &= \lim_{\Delta \theta \to 0 } \frac{F_{\theta}(\rho_0, \theta_0 + \frac{\Delta \theta}{2}, z_0) - F_{\theta}(\rho_0, \theta_0 - \frac{\Delta \theta}{2}, z_0)}{\Delta \theta} \\ &= \frac{\partial F_{\theta}}{\partial \theta} \big|_{(\rho_0, \theta_0, z_0)} \end{align}
Repeating this argument for the rest of the faces, I eventually get
\begin{align} \tag{9} \mathop{div} \textbf{F} = \frac{\partial F_{\rho}}{\partial \rho} + \frac{\partial F_{\theta}}{\partial \theta} + \frac{\partial F_z}{\partial z} \end{align}
which is incorrect.
