# The flux of a vector field through a cylinder [closed]

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

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

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