Calculating Power Dissipation on Cylinder Surface

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 9K views
likephysics
Messages
638
Reaction score
4

Homework Statement


I am trying to calculate power dissipated over a cylindrical surface using poynting vector -
[tex]\oint[/tex] ExH ds

I know ds for a sphere is r^2 sin [tex]\theta[/tex] d[tex]\theta[/tex] d[tex]\phi[/tex]

But now sure what ds is for a cylinder?

Homework Equations





The Attempt at a Solution


 
Physics news on Phys.org
likephysics said:
I know ds for a sphere is r^2 sin [tex]\theta[/tex] d[tex]\theta[/tex] d[tex]\phi[/tex]

But now sure what ds is for a cylinder?

It depends on which surface you are talking about. A closed cylinder has 3 surfaces; one curved surface and two flat circular end-caps. For the end-caps, [itex]dS=s ds d\phi[/itex]. While, for the curved surface, [itex]dS=s d\phi dz[/itex]. (Using [itex]\{s,\phi,z\}[/itex] for the cylindrical coordinates)

Griffiths' Introduction to Electrodynamics derives the infinitesimal displacements ([itex]dl_s=ds[/itex], [itex]dl_\phi=s d\phi[/itex], [itex]dl_z=dz[/itex]) in cylindrical coordinates in section 1.4.2. And the author gives a brief discussion of how to obtain area elements from these infinitesimal displacements at the end of page 40 (3rd edition).
 
Great. I am going to go take a look at Griffith's right now.