Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Surface integration

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data
    This isn't actually homework, but just related to something I am working on.

    Basically I need to find the distance between two points along a sphere. Not the linear distance, but the distance along the surface.
    This would be akin to "arc length" except in three dimensions.

    2. Relevant equations

    2D arc length: [itex]s = r \alpha[/itex]
    [tex]ds = dr \hat{r} + r d\theta \hat{\theta} + r sin(\theta) d\phi \hat{\phi}[/tex]

    3. The attempt at a solution

    I'm sure I already have an equation and a method of solving this by changing the coordinates to Cartesian, and then taking the dot product to find the angle and then using the 2D arc length equation to get the distance along the surface of the sphere between the two points.

    However, I am writing this because I am interested in using the second equation I gave above to find the distance. I'm not sure if I'm doing it right but it looks like I could just integrate so I have "s" instead of "ds" and the unit vectors are constant so they come outside of the integrals:
    [tex]s = \hat{r}\int_{r}\int_{\theta}\int_{\phi}dr + \hat{\theta}\int_{r}\int_{\theta}\int_{\phi}r d\theta + \hat{\phi}\int_{r}\int_{\theta}\int_{\phi}r sin(\theta) d\phi[/tex]

    The first term would become {r-hat}*(r2-r1) right?
    But what I'm not sure of are the other terms because just looking at the second term by itself you can split it up in this way:
    [tex]\hat{\theta}\int_{r}r \int_{\theta}d\theta[/tex]

    The thing that concerns me is that there is no "dr", so how do I go about integrating this? Similarly with the third term.

    Can I not find the distance in this way? I don't see why I shouldn't be able to...
    perhaps the "r" is a constant in this case...

    I hope this question isn't too dumb, and if it is please just bare with me and try to help me understand. Any advice is appreciated.

  2. jcsd
  3. Jul 10, 2010 #2
    You're totally on the right path. r = constant, therefore
    dr = 0
    ds = r d\theta \hat{\theta} + r sin(\theta) d\phi \hat{\phi}
    but you also need to parametrize your theta a phi; i.e. a path is one dimensional, so you need to have only one free parameter (e.g. 't' for theta(t), phi(t) )
    plug that parametrization into 'ds' above, and integrate over 't' (e.g.).

    Because you're looking for the shortest distance (presumably), your solution will be a geodesic (or great circle in this case).
  4. Jul 10, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your 3D equation is the vector equation

    [tex]d\mathbf{s} = dr\,\mathbf{\hat{r}}+r\,d\theta\,\boldsymbol{\hat{\theta}}+r\sin \theta\,d\phi\,\boldsymbol{\hat{\phi}}[/tex]

    for the line element. If you integrate ds, you'll get the vector s, not a distance. What you want is its length ds where

    [tex]ds^2=d\mathbf{s} \cdot d\mathbf{s} = dr^2 + r^2 (d\theta^2+\sin^2\theta\, d\phi^2)[/tex]

    (On a side note, in spherical coordinates, the unit vectors are not constant. They're functions of r, θ, and ϕ, so you can't just yank them out of integrals like you did.)
  5. Jul 11, 2010 #4

    But um, how would I do:
    [tex]\int_{r} dr^{2}[/tex]

    is that even possible?
  6. Jul 11, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You wouldn't. It's probably easiest to see how you do this type of calculation by example.

    Say the path is on the surface of a unit sphere and satisfies ϕ=θ where the angles run from ϕ=θ=0 to ϕ=θ=π/2. (See the attached plot.) On the surface of a sphere, r=1 is constant, so dr=0. Since ϕ=θ, we get dϕ=dθ. Let's use θ as the integration variable, so we'll get

    ds &=\sqrt{ds^2}=\sqrt{dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)} \\
    & = \sqrt{0^2+1^2(d\theta^2+\sin^2\theta\,d\theta^2)} \\
    & = \sqrt{1+\sin^2\theta}\,d\theta

    The length would therefore be given by

    [tex]s=\int ds = \int_0^{\pi/2} \sqrt{1+\sin^2\theta}\,d\theta[/tex]

    Attached Files:

    Last edited: Jul 12, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook