Finding intersection of a sphere and a plane Suppose I have the sphere  $\ x^2 + y^2 + z^2 = 4  $ and the plane $\  x-y\sqrt{3} =0. $ 
How do I find the intersection curve and write it in polar terms?
In polar coordinates the sphere is just $\ r = 2 $ 
substituting $\ x = \sqrt3  y $ in ball equation:
$$ \ (\sqrt3 y)^2 + y^2 + z^2 = 4 \\ 3r^2\sin^2\theta\sin^2\phi + r^2 \sin^2\theta \sin^2 \phi + r^2 \cos^2\phi =4 \\ r^2 \sin^2\phi \cdot 3 + r^2\cos^2\phi = 4 \\ r^2(3\sin^2\phi+\cos^2\phi) = 4$$

Find orthogonal vectors each of length (radius) $2$ in the plane given:
$(1, \sqrt{3},0)$ and $(0,0,2)$
Now create a trigonometric (polar) sum of these vectors:
$$\cos \theta (1, \sqrt{3}, 0) + \sin \theta (0,0,2) = (\cos \theta, \sqrt{3} \cos \theta, 2 \sin \theta)$$

In traditional polar coordinates:
$$(r, \theta, \phi) \to (2, 60^\circ, \phi)\ {\rm for}\ 0 < \phi < 2 \pi $$
