Surface Element Conversion for Flux Through Uncapped Cylinder

In summary, in this conversation the individual is discussing how to find the flux of a field through a cylindrical surface using surface integrals. They explain that the normal vector can be found by taking the gradient of the cylinder and that the integrand can be rewritten in terms of the angle and height of the cylinder. They then discuss how to perform the conversion for the directed surface element and surface area in a general case.
  • #1
RoyalFlush100
56
2

Homework Statement


In the attached image.

Homework Equations


Gradient(x, y, z) * <f, g, h> = <fx, gy, hz>

The Attempt at a Solution


Because the cylinder's not capped, I know that all the flux will be in the radial direction. So, I can find a normal vector by finding the gradient of the cylinder: n = <2x, 0, 2z>/(2sqrt(x^2+z^2)) = <x, 0, z>/sqrt(x^2+z^2)

Now, I want to put this in terms of t (the angle) and h (y):
r(t, y) = <acos(t), h, asin(t)>
Where: y: (-2, 2) and t: [0, 2pi)

Now we can rewrite the integrand:
<acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> * <acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> dS
=(a^2cos^2(t) + a^2sin^2(t))/a dS
=a dS

Now, the only thing I'm confused by (assuming everything else is right), is what to do with dS. I know it needs to be put in terms of dt and dh (where I already have the limits of integration), but I am unsure of how to perform this conversion.
 

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  • #2
Isn't the vector ## \vec{E }## of the field used in this flux equation simply a unit vector in the radially outward direction? If it is, (and I don't know that I should be giving you this much of the answer, but the problem is almost trivial), isn't the flux equal to the surface area?
 
  • #3
Charles Link said:
Isn't the vector ## \vec{E }## of the field used in this flux equation simply a unit vector in the radially outward direction? If it is, (and I don't know that I should be giving you this much of the answer, but the problem is almost trivial), isn't the flux equal to the surface area?
I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.
 
  • #4
RoyalFlush100 said:
I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.
You don't need to do any conversion. Surface area ## S=\pi a^2 h ##.
 
  • #5
Charles Link said:
You don't need to do any conversion. Surface area ## S=\pi a^2 h ##.
Okay. Just curious though, how would I perform that conversion in a less trivial case?
 
  • #6
Surface integrals in the general case can take much effort to evaluate. The problem, I think, was designed to keep it simple.
 
  • #7
In the general case, where you have a surface parametrised by the two variables ##t## and ##s##, the directed surface element is given by
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
d\vec S = \vec n \, dS = \dd{\vec x}{t} \times \dd{\vec x}{s} dt\, ds,
$$
where ##\vec x## is the position vector (which on the surface is parametrised by ##t## and ##s##.
 
  • Like
Likes Charles Link

Related to Surface Element Conversion for Flux Through Uncapped Cylinder

1. What is flux through a cylinder?

Flux through a cylinder is a measure of the flow of a vector field through the surface of a cylinder. It is calculated by taking the dot product of the vector field and the normal vector of the cylinder's surface, and then integrating this value over the surface.

2. How is the normal vector of a cylinder's surface determined?

The normal vector of a cylinder's surface is always perpendicular to the surface and points outward from the center of the cylinder. It can be found by taking the cross product of two tangent vectors on the surface.

3. What factors affect the flux through a cylinder?

The magnitude and direction of the vector field, as well as the orientation and size of the cylinder, can all affect the flux through a cylinder. Additionally, any obstacles or boundaries within the cylinder can also impact the flux.

4. How is the flux through a cylinder used in real-world applications?

Flux through a cylinder is a useful concept in fluid dynamics, electromagnetism, and other areas of physics and engineering. It can help determine the rate of fluid flow through a pipe, the strength of an electric field around a charged cylinder, and more.

5. Can the flux through a cylinder be negative?

Yes, the flux through a cylinder can be negative if the vector field and the normal vector of the cylinder's surface are in opposite directions. In this case, the flux represents the amount of flow exiting the cylinder rather than entering it.

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