Converting from square to polar coordinants

[Bob19]

Hi

If I have the following equation:

x^2 - 2y = 0

I need to convert it into polar coordinants:

r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

Is this correct ?

/Bob

 

[arildno]

Yes, it is correct.

 

[HallsofIvy]

Hi

If I have the following equation:

x^2 - 2y = 0

I need to convert it into polar coordinants:

r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

Is this correct ?

/Bob

And, of course, as long as r is not 0, you can divide through by r to get
$$r cos^2(\theta)- 2sin(\theta)= 0$$
or
$$r= \frac{2sin(\theta)}{cos^2(\theta)}$$

(It is customary to use "theta" in polar coordinates rather than "phi")