How do you convert #x^2+y^2=z# into spherical and cylindrical form?

Question

35599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-08T11:52:20

Spherical form+ #r=cos phi csc^2 theta#.
Cylindrical form: #r=z csc^2theta#

The conversion formulas,

Cartesian #to# spherical::

#(x, y, z)=r(sin phi cos theta, sin phi sin theta, cos phi), r=sqrt(x^2+y^2+z^2)#

Cartesian #to# cylindrical:

#(x, y, z)=(rho cos theta, rho sin theta, z), rho=sqrt(x^2+y^2)#

Substitutions in #x^2+y^2=z# lead to the forms in the answer.

Note the nuances at the origin:

r = 0 is Cartesian (x, y, z) = (0, 0, 0). This is given by

#(r, theta, phi) = (0, theta, phi)#, in spherical form, and

#(rho, theta, z)=(0, theta, 0)#, in cylindrical form...
.