1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Line integral in polar/spherical system?

  1. Dec 21, 2009 #1
    Hello, sorry for my English;D
    1. The problem statement, all variables and given/known data
    Can a vector field exist in polar/spherical system? is it possible to define line integral in these systems? does it make any sense a vector field defined in polar system, ex. [tex]\vec A\left(r,\varphi\right)=r^3[/tex]? and a line integral from [tex]r_1,\varphi_1[/tex] to [tex]r_2,\varphi_2[/tex] like this [tex]\int\limits_L\vec A\left(r,\varphi\right)\mbox{d}r+r\vec A\left(r,\varphi\right)\mbox{d}\varphi[/tex], where L is a line defined by [tex]r\left(\phi\right)[/tex] equation or [tex]r=r(t),\quad\phi=\phi(t)[/tex]
    and for spherical system [tex]\int\limits_L\vec A\left(r,\varphi,\phi\right)\mbox{d}r+r\vec A\left(r,\varphi,\phi\right)\mbox{d}\varphi+r\vec A\left(r,\varphi,\phi\right)\mbox{d}\phi[/tex]?
  2. jcsd
  3. Dec 22, 2009 #2


    User Avatar
    Science Advisor

    In polar coordinates, [itex]x= r cos(\phi)[/itex] so [itex]dx= cos(\phi)dr- r sin(\phi)d\phi[/itex]. [itex]y= r sin(\phi)[/itex] so [itex]dy= sin(\phi)dr+ r cos(\phi)d\phi[/itex].

    [itex]dx^2= cos^2(\phi)dr^2- 2r cos(\phi)sin(\phi)drd\phi+ r^2 sin^2(\phi)d\phi^2[/itex]
    [itex]dy^2= sin^2(\theat)dr^2+ 2r cos(\phi)sin(\phi)drd\phi+ r^2cos^2(\phi)d\phi^2[/itex]

    so [itex]dx^2+ dy^2= dr^2+ r^2 d\phi^2[/itex] so the "differential of arc length", ds, is given by [itex]ds= \sqrt{dr^2+ r^2d\phi^2}[/itex].

    If r and [itex]\phi[/itex] are given in terms of a parameter, t, then
    [itex]ds= \sqrt{\left(\frac{dr}{dt}\right)^2+ r^2\left(\frac{d\phi}{dt}\right)^2}dt[/itex]

    In spherical coordinates:

    Since [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]dx= cos(\theta)sin(\phi)d\rho- \rho sin(\theta)sin(\phi)d\phi+ \rho cos(\theta)cos(\phi)d\phi[/itex].

    Since [itex]y= \rho sin(\theta)sin(\phi)[/itex], [itex]dx= sin(\theta)sin(\phi)d\rho+ \rho cos(\theta)sin(\phi)d\phi+ \rho sin(\theta)cos(\phi)d\phi[/itex].

    Since [itex]z= \rho cos(\phi)[/itex], [itex]dz= cos(\phi)d\rho- \rho sin(\phi)d\phi[/itex].

    The "differential of areclength", ds, is given by [itex]ds= \sqrt{dx^2+ dy^2+ dz^2}[/itex]. Use the above equations to write that in terms of [itex]d\rho, d\theta[/itex], and [itex]d\phi[/itex]. it starts out messy but there is a lot of cancelling at the end.
    Last edited by a moderator: Dec 22, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook