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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]?
    thanks!
     
  2. jcsd
  3. Dec 22, 2009 #2

    HallsofIvy

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    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: Dec 22, 2009
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