How is any point on the Cartesian coordinates converted to cylindrical and spherical coordinates. Taking as an example, how would you convert the point (1,1,1)? Thanks in advance.


Cylindrical coordinates: $(r, \theta, z)$ are given by the relationship $x = r \cos\theta$, $y= r\sin \theta$ and $z=z$. Using these we can solve for $r$ and $\theta$ explicitly: $$ x^2 + y^2 = (r\cos\theta)^2+(r\sin\theta)^2= r^2(\cos^2\theta+\sin^2\theta)= r^2 $$ $$ \cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2+y^2}}$$

Spherical coordinates: $(\rho,\theta,\phi)$ can be given in a few different ways, I would check your textbook, but what I teach is $x = \rho\cos(\theta)\sin(\phi)$, $y= \rho\sin(\theta)\sin(\phi)$, $z = \rho \cos(\phi)$. You can use the same tricks as we used for cylindrical coordinates to isolate $\rho,\theta$ and $\phi$.

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    $\begingroup$ Are you sure about your spherical coordinates? If $z=\rho\sin\theta$ then $\theta$ is measured above the plane. Usually the vertical angle is taken from the $z$ axis. Also, if $z=\rho\sin\theta$, wouldn't $y=\rho\cos\theta\sin\phi$? $\endgroup$ – John Douma Jun 19 '13 at 20:01
  • $\begingroup$ Oh wow, the world of typos, thanks for picking those up. $\endgroup$ – James Jun 19 '13 at 21:18

cylindrical: $(r, \theta, z) = (\sqrt{2}, \pi/4, 1)$

spherical: $(\rho, \theta, \phi) = (\sqrt{3}, \pi/4, 0.955\text{ radians})$


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