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.