# Spherical coordinates length from differential length

1. Mar 25, 2014

### user3

is it logical to ask this question in Spherical coordinates:

Using the differential length dl , find the length where r=1 0<Θ<∏/4 ∏/2< θ <∏/4 where Θ is the azimuthal angle.

What I mean by ∏/2< θ <∏/4 is that the line is a "diagonal" line which has an ascention of ∏/4 from the xy plane. I dont know how else to write it.

Is the differential length dl = sqrt( (rdΘ)^2 + (dθ)^2 ) ?

Last edited: Mar 25, 2014
2. Mar 25, 2014

### Simon Bridge

Suspect not - but I'm not sure what you are asking.

if $\vec r(t)=(r(t), \theta(t),\phi(t))^t$ parameterzes a curve in spherical-polar coordinates, and $dl$ is the length element along that curve in the direction of increasing parameter $t$, then $dl$ can be expressed in terms of the curve $\vec r(t)$ and the spherical-polar unit vectors.

See:
http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf [Broken]
section 1.5 onwards.

Last edited by a moderator: May 6, 2017