Regarding Transformation on Uniformly Distributed Unit Disk $(X,Y)$ is distributed uniformly on the unit disk.
The transformations are:

$$
Z = {X + Y \over \sqrt{2}}\,,\qquad
W = {X - Y \over \sqrt{2}}
$$

I solved these equations in terms of $X$ and $Y$ and got:

$X$ = $(Z+W)/\sqrt(2)$
$Y$ = $(Z-W)/\sqrt(2)$

which has Jacobian equal to 1. Joint distribution of $Z$ and $W$ should be
$f(z, w)$ = $1/2\pi$ but with range -$\sqrt(2)$ $<$ $z$ $<$ $\sqrt(2)$. I'm not sure how to restrict the range of W.
The questions of interest are:
(a) What is the distribution of Z? Find $E[Z]$ and $Var[Z]%|$.

It seems fairly easy to see that $E[Z] = 0$ since the expectation of $X$ and $Y$ are both zero (both random variables are centered around zero). However, how would we compute $E[X^2]$ in order to compute the variance?

(b) Are Z and W uncorrelated? And are they independent?

I think that they are uncorrelated, but that they are not independent. I think this part might get easier after figuring out the exact range of the joint distribution.

(c) What is the distribution of $X/Y$ (the ratio of X and Y)?

Making the transformation $U$ = $X/Y$ and $V$ = $Y$ we have Jacobian 1. I'm confused as to what the range of joint distribution should be.

(d) What is the distribution of $Z/W$?

I'd imagine this part would get easier after figuring out part (c)


The transformation you have chosen is a rotation of coordinates, and so the joint density of $W$ and $Z$ is the same as the joint density of $X$ and $Y$, that is,
having value $\pi^{-1}$ inside the unit disc and value $0$ outside.  While it is
true that the maximum values of $X$ and $Y$ are $1$, they cannot take on value
$1$ simultaneously and so your range $(-\sqrt{2}, \sqrt{2})$ is incorrect.
(a) To find $\displaystyle E[X^2] = \int_{-\infty}^\infty \int_{-\infty}^\infty 
x^2 f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$, change to polar coordinates. You will also need $\operatorname{cov}(X,Y)$ which can also be found by computing $E[XY]$
via the change to polar coordinates.
(b) Uncorrelated is easy to do since $E[ZW] = \frac{1}{2}E[X^2-Y^2] = 0$. Why?
They are not independent for the same reason that $X$ and $Y$ are not
independent: the support of the joint density is not a product set and
so the joint density does not factor into the product of the marginal
densities over the entire plane.
(c) Don't use Jacobians here, just ask yourself how you might compute
$P\{Y/X \leq \alpha\}$ directly. Drawing a picture will help tremendously
here. But if not, look at this answer.
(d) Since the joint density of $Z$ and $W$ is the same as the joint
density of $X$ and $Y$, ....
