Converting cartesian rectangular equation to it's corresponding polar equation I try to find a manner to convert cartesian rectangular equation like $y = f(x)$ to it's polar equation $r = r(\theta)$.
Given the function $f(x) = y = 5x^4$, lets try to get the polar equation:
$$x = r\cdot \cos(\theta), \quad y = r\cdot \sin(\theta)$$
So:
$$r\cdot \sin(\theta) = 5\cdot r^4 \cdot \cos^4(\theta)$$
$$ \Rightarrow r^3= \frac{\sin(\theta)}{5\cdot \cos^4(\theta)} $$
$$ \Rightarrow r(\theta) = \sqrt[3] {\frac {\sin(\theta)}{5\cdot \cos^4(\theta)}}$$
My question is, That's really the correct answer? if so, how we can convert it back to rectangular equation? I tried to convert the polar equation we received to a rectangular equation using the following equations
$$\theta = \arctan\Bigr(\frac yx\Bigr), \quad r = \sqrt{x^2+y^2}$$
The result I got was everything except $y = 5x^4$.
Can someone point me the right direction to get the correct solution?
Thanks for Help!!

Using the inverse relations, the Cartesian equation reads
$$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\sin\left(\arctan\dfrac yx\right)}{5\cos^4\left(\arctan\dfrac yx\right)}}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}}.$$
Now notice that $$\frac1{\cos^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\tan^2\theta+1$$ and
$$\frac1{\cos^3\theta}=(\tan^2\theta+1)^{3/2}.$$
Substituting, we have 
$$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}}
=\sqrt[3]{\dfrac  y{5x}\left(\tan^2\left(\arctan\dfrac yx\right)+1\right)^{3/2}}
=\sqrt[3]{\dfrac  y{5x}\left(\dfrac{y^2}{x^2}+1\right)^{3/2}}=\sqrt[3]{\dfrac  y{5x^4}}\sqrt{x^2+y^2}.$$
A final simplification yields
$$\frac y{5x^4}=1.$$
Easy as that :-)
