Surface element in cylindrical coordinates

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SUMMARY

The discussion focuses on solving the integral of the bound current density \(\vec J_b = 3s \hat z\) in cylindrical coordinates for an infinite cylinder. The area element is debated, with two expressions presented: \(d \vec a = s d\phi dz \hat s\) and \(d \vec a = s d\phi \hat z\). The latter leads to the correct result of \(\frac{2 \pi}{3} s^3 \hat \phi\) when integrated, while the former results in a zero value due to the dot product with the \(\hat z\) unit vector. The integration variables indicate that the integral is performed over the surface of the cylinder.

PREREQUISITES
  • Understanding of cylindrical coordinates
  • Familiarity with vector calculus and dot products
  • Knowledge of current density concepts in electromagnetism
  • Ability to perform double integrals
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  • Study the derivation of area elements in cylindrical coordinates
  • Learn about the physical significance of bound currents in electromagnetism
  • Explore the application of the divergence theorem in cylindrical coordinates
  • Investigate the use of vector fields in electromagnetic theory
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Students and professionals in physics, particularly those studying electromagnetism, as well as anyone working with vector calculus in cylindrical coordinates.

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Homework Statement



\vec J_b = 3s \hat z
\int \vec J_b \, d\vec a

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:
d \vec a = s d\phi dz \hat s
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.

d \vec a = s d\phi \hat z

The Attempt at a Solution



\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

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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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.
 

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