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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

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  • $\begingroup$ @G. Smith Can you please explain where I went wrong? $\endgroup$ – Kliendester Sep 28 '19 at 16:58
  • $\begingroup$ Don’t you mean $\rho^2$ in your final integral? $\endgroup$ – G. Smith Sep 28 '19 at 17:02
  • $\begingroup$ You think the answer should be an area. Is this dimensionally consistent with the flux of $\mathbf{F}$? $\endgroup$ – G. Smith Sep 28 '19 at 17:04
  • $\begingroup$ yes,the final solution is $ 2 \pi \rho^2 H$, the maximum value of $\rho$ is A, the radius of the cylinder. $\endgroup$ – Kliendester Sep 28 '19 at 17:04
  • $\begingroup$ I’m talking about the typo $\rho\,dz\,d\phi$ in your final integral, not the result you got. $\endgroup$ – G. Smith Sep 28 '19 at 17:07
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The problem says: use Gauss's theorem. So the flux out is the total divergence inside the volume.

The divergence is:

$$ D(x, y, z) = \frac{dF_x}{dx} + \frac{dF_y}{dy} +\frac{dF_z}{dz} = 1+1+1=3$$

So it's:

$$ 3V $$

where V is the volume of the cylinder.

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