The question is by using Gauss’ Theorem calculate the flux of the
vector field$\overrightarrow{F} = x \hat{i} + y \hat{j}+ z \hat{k}$
through the surface of a cylinder of radius A and height H, which has
its axis along the z-axis and the base of the cylinder is on the
xy-plane.
So, first of all I converted the vector field into cylindrical coordinates
$\overrightarrow{F}= \rho \cos^2 \phi \hat{e}_\rho + \rho \sin^2 \phi \hat{e}_\rho + z \hat{e}_z $
which can be further reduced to-
$\overrightarrow{F}= \rho \hat{e}_\rho + z \hat{e}_z$
The surface of the cylinder has three parts, $ \ S_1 $, $ \ S_2 $, and $ \ S_3 $. $ \ S_1 $ and $ \ S_2 $ are the top and bottom of surface of the cylinder and $ \ S_3 $ is the curved surface. We can write the surface integral over the surface of the cylinder as
$\unicode{x222F}_S \overrightarrow{F} . d\overrightarrow{S}=\iint_{S_1} \overrightarrow{F} . d\overrightarrow{S_1} +\iint_{S_2} \overrightarrow{F} . d\overrightarrow{S_2} + \iint_{S_3} \overrightarrow{F} . d\overrightarrow{S_3} $
As the area element is in $\rho \phi$ plane (for a constant value of z) has the value $\rho d \rho d \phi$. So an area element on $ \ S_1 $ and $ \ S_2 $ will have magnitude $\rho d \rho d \phi$, and the outward unit normals to $ \ S_1 $ and $ \ S_2 $ are then $ \hat{e}_z$ and $- \hat{e}_z$, respectively
$\therefore d\overrightarrow{S_1}= \rho d \rho d \phi \hat{e}_z$ and $d\overrightarrow{S_2}= -\rho d \rho d \phi \hat{e}_z$
And the area element for the $d\overrightarrow{S_3}= \rho dz d \phi \hat{e}_ \rho $
Now, keeping the conditions in mind-
$0 \le \rho \le A$ ; $0 \le \phi \le 2 \pi$; $0 \le z \le H$
$\unicode{x222F}_S \overrightarrow{F} . d\overrightarrow{S}=\iint_{S_1} [\rho \hat{e}_\rho + z \hat{e}_z].[\rho d \rho d \phi \hat{e}_z]+ \iint_{S_2} [\rho \hat{e}_\rho + z \hat{e}_z].[-\rho d \rho d \phi \hat{e}_z]+ \iint_{S_3} [\rho \hat{e}_\rho + z \hat{e}_z].[\rho dz d \phi \hat{e}_ \rho]$
The flux of $d\overrightarrow{S_1}$ and $ d\overrightarrow{S_2}$ will cancel out each other. Now, integrating $\iint_{S_3} \overrightarrow{F} . d\overrightarrow{S_3} $ as double integral-
$\int _{\phi =0}^{2\pi }\:\int _{z=0}^H\:\rho^2 dz d \phi$
$= 2 \pi A^2 H$ where $\rho = A$
So, the total flux is $= 2 \pi A^2 H$ which I think is wrong, as the flux should be the curved surface area of the cylinder,i.e., $= 2 \pi A H$
I am still learning this topic, so please mention any mistake that I've done while solving it
Best Answer
I think switching to cylindrical coordinates makes things way too complicated. It also seems to me you ignored the instructions to apply Gauss's Theorem. From the cartesian coordinates, we see immediately that $\text{div}\, \vec F = 3$, so the flux across the entire closed surface will be $3(\pi A^2H)$. The flux of $\vec F$ downwards across the bottom, $S_2$, is $0$ (since $z=0$); the flux of $\vec F$ upwards across the top, $S_1$, is $H\cdot(\pi A^2)$. Thus, the flux across the cylindrical surface $S_3$ is $2\pi A^2H$.
Your intuition is a bit off, because you need another factor of $A$ (since $\vec F$ is $A$ times the unit radial vector field). By the way, using $A$ for a radius is very confusing, as most of us would expect $A$ to denote area.