How Do You Calculate Outward Flux Through the Base of a Cylinder?

In summary, the student attempted to solve a homework problem, but was not entirely confident in the answer. They asked for help and the author looked at the problem and provided a solution.
  • #1
BOAS
552
19

Homework Statement


Calculate the outward flux of [itex]F = 3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}[/itex] through the base of a cylinder centered at the origin with radius [itex]\mathrm{R}[/itex] and height [itex]\mathrm{2H}[/itex].

Homework Equations

The Attempt at a Solution



I am unsure of how to tell if I have calculated the outward flux, and not entirely confident that it makes sense to have 'H' in my final answer. I'd really appreciate it if you could take a look.

[itex]\Phi = \int_{s} \mathbf{F}. \mathbf{n} dA[/itex]

[itex]\mathbf{F}. \mathbf{n} = (3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}).(\mathbf{e_{z}}) = 2z[/itex]

[itex]\Phi = \int_{s} 2z dA[/itex]

[itex]\Phi = \int_{s} 2z \rho d\rho d\phi[/itex]

[itex]\Phi = \int_{0}^{2\pi} \int_{0}^{R} 2z \rho d\rho d\phi[/itex]

[itex]\Phi = \int_{0}^{2\pi} z \rho^{2} d\phi = z \rho^{2} [\phi]^{2\pi}_{0}[/itex]

[itex]\Phi = 2 \pi z \rho^{2}[/itex]

[itex]\Phi = -2 \pi H R^{2}[/itex]

Thanks!
 
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  • #2
BOAS said:

Homework Statement


Calculate the outward flux of [itex]F = 3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}[/itex] through the base of a cylinder centered at the origin with radius [itex]\mathrm{R}[/itex] and height [itex]\mathrm{2H}[/itex].

Homework Equations

The Attempt at a Solution



I am unsure of how to tell if I have calculated the outward flux, and not entirely confident that it makes sense to have 'H' in my final answer. I'd really appreciate it if you could take a look.

[itex]\Phi = \int_{s} \mathbf{F}. \mathbf{n} dA[/itex]

[itex]\mathbf{F}. \mathbf{n} = (3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}).(\mathbf{e_{z}}) = 2z[/itex]

[itex]\Phi = \int_{s} 2z dA[/itex]

[itex]\Phi = \int_{s} 2z \rho d\rho d\phi[/itex]

[itex]\Phi = \int_{0}^{2\pi} \int_{0}^{R} 2z \rho d\rho d\phi[/itex]

[itex]\Phi = \int_{0}^{2\pi} z \rho^{2} d\phi = z \rho^{2} [\phi]^{2\pi}_{0}[/itex]

[itex]\Phi = 2 \pi z \rho^{2}[/itex]

[itex]\Phi = -2 \pi H R^{2}[/itex]

Thanks!

Seems ok. Except I would say the OUTWARD normal on the bottom of the cylinder is ##-\mathbf{e_z}##. You could actually have done this calculation in your head, since you are integrating over a disk and the integrand is constant.
 
  • #3
Dick said:
Seems ok. Except I would say the OUTWARD normal on the bottom of the cylinder is ##-\mathbf{e_z}##. You could actually have done this calculation in your head, since you are integrating over a disk and the integrand is constant.

Ah, of course! I should have picked up on that.

Thanks!
 

Related to How Do You Calculate Outward Flux Through the Base of a Cylinder?

1. What is outward flux through a cylinder?

The outward flux through a cylinder is a measure of the amount of fluid or energy flowing outward from the surface of a cylinder.

2. How is outward flux through a cylinder calculated?

The outward flux through a cylinder can be calculated using the formula Flux = (Pi * radius^2 * velocity), where pi is the mathematical constant, radius is the radius of the cylinder, and velocity is the speed of the fluid or energy.

3. What factors affect the outward flux through a cylinder?

The outward flux through a cylinder is affected by the size and shape of the cylinder, the velocity of the fluid or energy, and the angle at which the fluid or energy is flowing relative to the surface of the cylinder.

4. How is outward flux through a cylinder used in scientific research?

Outward flux through a cylinder is used in many scientific research fields, including fluid dynamics, electromagnetism, and thermodynamics. It helps scientists understand the flow of fluids and energy and how it affects different systems.

5. What are some real-world applications of outward flux through a cylinder?

Outward flux through a cylinder has many real-world applications, such as calculating the flow of air or water through pipes, determining the efficiency of heat exchangers, and understanding the flow of blood through blood vessels in the human body.

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