What are the cylindrical coordinates of the point whose spherical coordinates are (4, 2, 1π/6) ?

1 Answer
Jun 16, 2018

Assuming the usual spherical coordinate system, #(r,theta,phi)=(4,2,pi/6)# equates to #(R,psi,Z)=(2,2,2sqrt(3))#.

Explanation:

There are several different conventions as to the order of the angles in spherical coordinates. See http://mathworld.wolfram.com/SphericalCoordinates.html for more info.

I will assume that the order of coordinates wanted here is radial, azimuthal, polar, which matches that link, also has the advantage of being a right-handed coordinate system, and is the system most often seen.

Relation of spherical to Cartesian coordinates:

Spherical radius: #r=sqrt(x^2+y^2+z^2)#
Spherical azimuth: #theta=arctan(y/x)#
Spherical polar angle: #phi=arccos(z/sqrt(x^2+y^2+z^2))#

Cylindrical coordinates are less confused, and the standard definition is shown here: http://mathworld.wolfram.com/SphericalCoordinates.html.

Cylindrical radius: #R=sqrt(x^2+y^2)#
Cylindrical azimuth: #psi=arctan(y/x)#
Height: #Z=z#

From these, we may relate the cylindrical coordinates to the spherical ones:

#R=rsinphi# (via a little trig simplification using the below #Z#)
#psi=theta#; the two azimuths are identical
#Z=rcosphi#

So in this case, #(r,theta,phi)=(4,2,pi/6)# equates to #(R,psi,Z)=(2,2,2sqrt(3))#. Assuming I chose the same spherical system as you did - the details are easily enough altered if some other spherical system is wanted.