How can I find $\theta$ when converting an equation to cylindrical coordinates?

Question

The equation is $x^2+y^2=4y$ and I need to convert it to cylindrical coordinates

Here is what I did:

$x = 2r\cos(\theta)$

$y = 2r\sin(\theta)$

$2r^2\cos^2(\theta)+2r^2\sin^2(\theta) = 4\sin(\theta)$

$2r^2(\cos^2(\theta)+\sin^2(\theta) = 4\sin(\theta)$

$2r^2=4\sin(\theta)$

$r^2=2\sin(\theta)$

$r=\sqrt{2\sin(\theta)}$

If the above is correct, how do I solve for $\theta$?

Answer 1

You have some mistakes in your derivation.

Cylindrical coordinates are given by: $$ x=r\cos \theta \qquad y=r\sin \theta \qquad z=0 $$ so your equation $x^2+y^2=4y$ becomes: $$ r^2\cos^2\theta +r^2\sin^2\theta=4r\sin \theta $$ $$ r=4\sin \theta \qquad or \qquad \theta=\arcsin (r/4) $$

Answer 2

The correct solution was given, but this is also solvable in Cartesian coordinates. Rewrite

$$x^2+y^2-4y+4=x^2+(y-2)^2=4.$$

This is the equation of a circle centered at $(0,2)$, with radius $2$. It passes through the origin.