# Joint density of dependent random variables

I'm dealing with this problem

Suppose that $X, Y$ are iid $N(0,1)$ random variables and let $R= \sqrt{X^2+Y^2}$. Find the joint pdf of $R$ and $R^2$.

I'm not sure how to solve the joint pdf of dependent random variables. I just tried to find cumulative distribution and differentiate it.

By change of variables, I got $f_R(r) = \frac{1}{2\pi}r\exp{-\frac12r}$. $$f_{R,R^2}(x,y) = \frac{\partial^2}{\partial x \partial y} P(R \le x, R^2 \le y)= \frac{\partial^2}{\partial x \partial y} P(R \le x , -\sqrt{y} \le R \le \sqrt{y} )$$ so this is nonzero only if $-\sqrt{y} \le x \le \sqrt{y}$, and this is equal to $\frac{\partial^2}{\partial x \partial y} \int^x_{-\sqrt{y}} f_R(r) dr =0$, but this doesn't make sense. How should I deal with this?

You are right that there is an issue here. In general if $X$ is an RV, $X$ and $X^2$ will not have a joint density (though they will certainly have a joint distribution). The issue is that the 2D distribution will be singular. This is because $X^2$ can only take one value conditional on $X,$ so the density will be concentrated along a $1$-dimensional curve in $X-X^2$ space.
This isn't true for all dependent variables. For instance if $X$ and $Y$ are jointly Gaussian with correlation $.5,$ they are dependent but they have a joint density that is a function. However if they have correlation $1,$ they will be singular since one is just a multiple of the other.
However, we can still write down a density in terms of delta functions. Let $Y=X^2$ to clean up the notation a little bit. Then if $f_X(x)$ is the PDF for $X$ we will have $$f_{X,Y}(x,y) = f_X(x)\delta(y-x^2)$$ where $\delta$ is the Dirac delta function.