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Differential Lengths (Cylinder & Sphere)

  1. Sep 30, 2012 #1
    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.
     
  2. jcsd
  3. Sep 30, 2012 #2
    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!
     
    Last edited: Sep 30, 2012
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