Differential Lengths (Cylinder & Sphere)

  • Thread starter Meadman23
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  • #1
Meadman23
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This isn't a homework problem, but I was wondering if anyone could explain two things to me.

1. When you have the differential lengths of a cylinder:


dlr= dr dl[itex]\theta[/itex] = r d[itex]\theta[/itex] dlz = dz



Why is dl[itex]\theta[/itex] equal to r d[itex]\theta[/itex] and not just d[itex]\theta[/itex]?


2. When you have the differential lengths of a sphere:

dlR = dR dl[itex]\theta[/itex] = R d[itex]\theta[/itex] dl[itex]\varphi[/itex] = R sin[itex]\theta[/itex] d[itex]\varphi[/itex]

Why is dl[itex]\theta[/itex] equal to R d[itex]\theta[/itex] and why is [itex]\varphi[/itex] equal to R sin[itex]\theta[/itex] d[itex]\varphi[/itex]???


I really want to be able to see rather than memorize what each of these differential lengths are equal to.
 

Answers and Replies

  • #2
stephenkeiths
54
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As I recall, the rdθ has to do with radians. When you're on a circle, and you move through an angle θ, you have traveled a distance rθ (provided θ is measured in radians!). If you move through an infinitesimal angle dθ you have gone a distance rdθ.

As for spherical, notice that θ is the angle from the z-axis (I think?) then the radius in the X-Y plane is Rsinθ, so by the same radian argument, an infinitesmal movement in the phi direction (angle from the x-axis) should be Rsinθdψ (I can't believe they don't have phi!!).

And same goes for Rdθ in the θ direction (angle from the z-axis).

hope this helps!
 
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