If $X$ and $Y$ are zero mean independent Gaussian random variables with different variances, what is the density of $\sqrt{X^2+Y^2}$
According to your description, $x = Z \cdot \cos(W),\ y = Z \cdot \sin(W)$. Follow this answer for the derivation of joint PDF of $(Z, W)$ : Complex Gaussian Magnitude and Phase Joint PDF Derivation

You will reach the following expression after Methid of transformation:
$$f_{Z,W}(z,w) = |\mathbf J|.f_{X,Y}(z \cdot \cos(w), z \cdot \sin(w)), \ where \ \mathbf J \ is \ Jacobian$$
Since the computation of $\mathbf J$ will not change hence:
$$\mathbf J= z$$

Now, when you put the joint PDF of 2 independent Gaussian random varaibles $X, Y$ in the above expression you get the following :
$$f_{Z,W}(z,w) = |\mathbf J|\cdot f_{X,Y}(z\cdot \cos(w), z\cdot \sin(w)) $$ $$= z \cdot \frac{1}{\sqrt{2\pi \sigma_x^2}}\exp(-\frac{1}{2\sigma_x^2}(z\cdot \cos(w))^2)\cdot \frac{1}{\sqrt{2\pi \sigma_y^2}}\exp(-\frac{1}{2\sigma_y^2}(z\cdot \sin(w))^2)$$

Combine this and find the marginal PDFs of $Z$ and $W$ from the above expression. You will find that integrating from $-\pi$ to $\pi$ w.r.t. $w$ will not give you Rayleigh Distribution anymore. And, similarly, integrating from $0$ to $\infty$ w.r.t. $z$ will not give you $\frac{1}{2\pi}$, and hence not Uniform Distribution anymore.