The distribution is $R^2\sim \exp(\lambda)$ and I want to find the pdf(R). Here is what I did.

Let $$Y=R^2$$
Then $$P(R \le r)=P(\pm \sqrt{Y} \le r)=P(-r\le\sqrt{Y}\le r)$$

At this point, the probability would cancel out. I am not sure if I am on the right track. Any tips would be greatly appreciated!