How to convert $x^2 - (y^3) = x$ to polar coordinates

Question

I understand that $x = r \cos(\theta)$ and $y = r \sin(\theta)$, and $x^2 + y^2 = r^2$, but I cannot figure out the algebraic manipulation needed for this problem. I need to isolate $r$ onto one side.

Thanks for you help!

Answer 1

Substitute your equations into the cartesian one:

$$x^2-y^3=x$$

$$\implies(r\cos\theta)^2 -(r\sin\theta)^3=r\cos\theta$$

Now the equation is in polar coordinates, and you can manipulate it into the form you desire.

Edit after comment:

Then we have $$r(r\cos^2\theta-r^2\sin^2\theta-\cos\theta)=0$$

$$r\sin^2\theta(r^2-\frac{\cos^2\theta}{\sin^2\theta}r+\frac{\cos\theta}{\sin\theta})=0,\quad \sin\theta\neq0$$

$$(r-\frac{1}{2}\cot^2\theta)^2+\cot\theta-\frac{1}{4}\cot^4\theta=0,\quad\sin\theta\neq0,\quad r\neq0$$

$$\implies r=\frac{1}{2}\cot^2\theta\pm\sqrt{\frac{\cot^4\theta-4\cot\theta}{4}} =\frac{\cot^2\theta\pm\sqrt{\cot^4\theta-4\cot\theta}}{2},\quad r\neq0,\sin\theta\neq0$$