Covariance Matrix Polar to Cartesian
The covariance matrix is given by

$$C_{X,Y}=\begin{bmatrix}E(XX)&amp; E(XY) \\ E(YX )&amp; E(YY) \end{bmatrix}$$

This can be written as below:

$$C_{X,Y}=\begin{bmatrix}E(R^2cos^2(\Theta) )&amp; E(Rcos(\Theta)Rsin(\Theta)) \\ E(Rsin(\Theta)Rcos(\Theta) )&amp; E(R^2sin^2(\Theta)) \end{bmatrix}$$

Since $R$ and $\Theta$ are independent the expectation will distribute. Now the task is to calculate expectation of $Cos^2(\Theta)$
, $Sin^2(\Theta)$ and $Cos(\Theta)Sin(\Theta)$ or equivalently $Sin(2\Theta)$. These can be found by by calculating the expectations of each of these functions for ex: for $Cos^2(\Theta)$ this will be $$E(cos^2(\Theta)) = \int_{ - \infty}^{\infty}\, Cos^2(\theta) f_\Theta(\theta) \, d\theta$$

Where $f_{\Theta}(\theta)$ is the PDF of random varible $\Theta$

Hint: you can also utilize symmetries while calculating these integrals, $Sin(2\Theta)$ is an odd function and it multiples a Gaussian with zero mean which is an even function. Similarly $Cos^2(\Theta)$ and $Sin^2(\Theta)$ are even functions

EDIT, Solving the integral:

$$\cos^2\theta=0.5+0.5\cos 2\theta=0.5+0.25(e^{2i\theta}+e^{-2i\theta})$$

$$I=\int_{-\infty}^{\infty} \cos^{2}(\theta) f_{\Theta}(\theta) d\theta=0.5+0.25\int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-\theta^2\over 2}(e^{2i\theta}+e^{-2i\theta})d\theta$$

$$\int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}d\theta {= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}e^{-{(ki)^2\over 2}}e^{{(ki)^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-(\theta-ki)^2\over 2}e^{-{k^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{-{k^2\over 2}}d\theta \\= \sqrt{2\pi}e^{-{k^2\over 2}} }$$ hence $$
I=0.5+0.5e^{-2}$$

Original answer to the integral  https://math.stackexchange.com/questions/3650453/integral-of-int-infty-infty-cos2-theta-f-theta-theta-d/3650494#3650494