x^2+y^2-3y=0 solve for r?

Question

how do i solve this for r after converting to polar form?

Answer 1

x = rcosθ, y = rsinθ

 

x² + y² = r²cos²θ + r²sin²θ = r²(cos²θ + sin²θ) = r²(1) = r²

 

So, if x² + y² - 3y = 0, then r² - 3rsinθ = 0

 

                                         r(r - 3sinθ) = 0

 

                                         r = 0 or r = 3sinθ