1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The length element in cylindrical coordinates

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that in cylindrical coordinates

    [itex]x = \rho cos \theta[/itex]
    [itex]y = \rho sin \theta[/itex]
    [itex]z = z[/itex]

    the length element ds is given by

    [itex]ds^{2} = dx^{2} + dy^{2} + dz^{2} = d \rho^{2} + \rho^{2} d \theta ^{2} + dz^{2}[/itex]

    2. Relevant equations

    --

    3. The attempt at a solution

    Notice [itex]\rho = \sqrt{x^{2} + y^{2}}[/itex]

    I have found expressions for ∂x/∂θ, ∂x/∂ρ, ∂y/∂θ, ∂y/∂ρ, ∂ρ/∂x, ∂ρ/∂y, ∂θ/∂x, ∂θ/∂y, and of course it is trivial that dz = dz. Given all of these, how do i start using chain rule and total derivatives to find ds2? I guess because i have 8 partials but end up only ρ, θ, and z, i don't really know which ones to start with.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 27, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Start from
    [tex]
    dx = \frac{\partial x}{\partial \rho}d\rho + \frac{\partial x}{\partial \theta}d\theta
    [/tex]
    Squaring both sides gives [itex]dx^2[/itex].
     
  4. Sep 27, 2013 #3
    yeah i got it. thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: The length element in cylindrical coordinates
  1. Cylindrical coordinates (Replies: 25)

Loading...