Surface element in cylindrical coordinates

In summary, the conversation is about solving an integral in cylindrical coordinates for the bound current of an infinite cylinder. There are two different equations given for the area element, which leads to confusion about the correct way to find the unit vector in the answer. The first equation does not provide the desired unit vector, but the second one does. The main question is about determining which equation to use and understanding the reasoning behind it.
  • #1
millifarads
5
0

Homework Statement



[tex]\vec J_b = 3s \hat z [/tex]
[tex]\int \vec J_b \, d\vec a [/tex]

I need to solve this integral in cylindrical coordinates. It's the bound current of an infinite cylinder, with everything done in cylindrical coordinates and s is the radius of the cylinder. The answer should end up with a phi unit vector, but I just don't see how to get the actual unit vector.

Homework Equations

According to a pdf on MIT's site, the area element is:
[tex] d \vec a = s d\phi dz \hat s [/tex]
There's a caveat, though, as the page says there's a choice of direction. Using this way makes the most sense to me, but if I use this, the dot product between the s and z unit vectors will cause the whole thing to go to zero.
http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf

I've found another site that says to use the following, but I don't see the reasoning as to where the phi hat unit vector comes from.

[tex] d \vec a = s d\phi \hat z [/tex]

The Attempt at a Solution



[tex]\int \vec J_b \, d\vec a = \int_0^s \int_0^{2\pi} s \hat z \cdot sdsd\phi \hat z = \frac{2 \pi}{3} s^3 \hat \phi [/tex]

Using the first equation for the area element doesn't get me where I need to be, but the second one does.

Any help in figuring this out would be appreciated.
 
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  • #2
Over what surface are you integrating? Your expression ##\int \vec J_b \, d\vec a ## does not provide that crucial information.

In your first equation in 2, the integration variables are ##\phi## and ##z##, which suggests the integral is on the surface of the cylinder.

But in the equation in 3, the integration variables are ##\phi## and ##s##, which suggests the integral is a cross sectional disc bounded by the cylinder.
 

1. What is the surface element in cylindrical coordinates?

The surface element in cylindrical coordinates is a measure of the area of an infinitesimal surface in three-dimensional space. It is used to calculate the surface area of a curved surface in cylindrical coordinates, which is a coordinate system that uses a distance from a central axis and an angle from a reference direction.

2. How is the surface element calculated in cylindrical coordinates?

The surface element in cylindrical coordinates is calculated using the formula dS = r · dr · dθ, where r is the distance from the central axis and dθ is the change in angle. This formula takes into account the curvature of the surface and the change in coordinates.

3. What are the applications of the surface element in cylindrical coordinates?

The surface element in cylindrical coordinates is commonly used in various fields of science and engineering, such as fluid mechanics, electromagnetism, and heat transfer. It is also used in calculating the surface area of curved objects in real-world applications, such as designing pipes, cylinders, and other cylindrical shapes.

4. How does the surface element in cylindrical coordinates relate to the surface element in Cartesian coordinates?

The surface element in cylindrical coordinates is related to the surface element in Cartesian coordinates through a conversion factor. In cylindrical coordinates, the surface element is multiplied by a factor of r, which takes into account the curvature of the surface. In Cartesian coordinates, the surface element is simply the product of the two dimensions, dx · dy.

5. Can the surface element in cylindrical coordinates be generalized to other coordinate systems?

Yes, the surface element in cylindrical coordinates can be generalized to other coordinate systems, such as spherical coordinates. In spherical coordinates, the surface element is given by dS = r² · sin(θ) · dr · dθ, where r is the distance from the origin, θ is the angle from the z-axis, and φ is the angle from the x-axis. This formula takes into account the curvature of the surface in three dimensions.

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