Converting cartesian equation to polar issue

Question

I have tried converting the equation $ y = x^3 $ to a polar equation using the following steps but the graph of the resulting polar equation also appears to include $ y=-x^3 $.

$ r^2 = x^2 + y^2 $

$ r^2 = x^2 + (x^3)^2$

$ r^2 = r^2\times cos(\theta)^2 + r^6cos(\theta)^6 $

$ 1 = cos(\theta)^2+r^4\times cos(\theta)^6$

$ \frac{sin(\theta)^2}{cos(\theta)^6}=r^4$

$ r = (\frac{sin(\theta)^2}{cos(\theta)^6})^{1/4}$

However, the graph of this is not $ y = x^3 $.

I am wondering why converting a cartesian equation to a polar equation does not appear to work this way. I understand that $ y=x^3$ can be easily converted to polar by substituting $ y=rsin(\theta)$ and $x=rcos(\theta)$ into $y=x^3$ but I don't understand why I do not get the same answer when substituting $y = x^3$ for $x$ in $r^2 = x^2+y^2$.

Thanks.

Answer 1

$y=x^3$

$r\sin\theta=r^3\cos^3\theta$

cancel $r$ and solve

$r^2=\frac{\sin\theta}{\cos^3\theta}$

to solve for $r$ we must have $\frac{\sin\theta}{\cos\theta}\ge 0\to \tan \theta \ge 0$ that is $0\le \theta<\pi/2\lor \pi\le \theta<3\pi/2$

$r=\sqrt{\frac{\sin\theta}{\cos^3\theta}};\;\theta\in [0,\pi/2)\cup[\pi,3\pi/2)$

Answer 2

$$r=\sqrt{x^2+x^6}$$ is fine, as the RHS is positive. Then

$$\tan\theta=-\frac{x^3}x=-x^2$$ and

$$r=\sqrt{-\tan^3\theta-\tan\theta}=\sqrt{-\dfrac{\sin\theta}{\cos^3\theta}}.$$