Line integral in polar/spherical system?

[player1_1_1]

Hello, sorry for my English;D

Homework Statement

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. \vec A\left(r,\varphi\right)=r^3? and a line integral from r_1,\varphi_1 to r_2,\varphi_2 like this \int\limits_L\vec A\left(r,\varphi\right)\mbox{d}r+r\vec A\left(r,\varphi\right)\mbox{d}\varphi, where L is a line defined by r\left(\phi\right) equation or r=r(t),\quad\phi=\phi(t)
and for spherical system \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?
thanks!

 

[HallsofIvy]

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

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

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

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

In spherical coordinates:

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

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

Since z= \rho cos(\phi), dz= cos(\phi)d\rho- \rho sin(\phi)d\phi.

The "differential of areclength", ds, is given by ds= \sqrt{dx^2+ dy^2+ dz^2}. Use the above equations to write that in terms of d\rho, d\theta, and d\phi. it starts out messy but there is a lot of cancelling at the end.