Let $U \sim \mathcal{E}(1)$ (exponential random variable with parameter $1$) and $V$ a uniform distributed random variable on $(0,1)$.
We assume that $U$ and $V$ are independent.
Define $X: = \sqrt{U}\cos(2\pi V)$ and $Y: = \sqrt{U} \sin(2\pi V)$.
I have to compute the distribution of $(X,Y)$.
Now what they do is the following:
Let $\phi: \mathbb{R}^2 \to \mathbb{R}_+$ be measurable.
$\mathbb{E}[\phi(X,Y)] = \int_0^{+\infty} \int_0^1 \phi(\sqrt{u} \cos(2\pi v), \sqrt{u} \sin(2\pi v)) e^{-u} du dv$
I don't how to derive this, the remaining steps are clear to me. I'm aware of the formula:
$\mathbb{E}[f(X)] = \int_Ef(X) d \mathbb{P}_X(X)$ but I don't how this was applied to my case.
I know that the denisty of $U$ is $e^{-x}$ and the one of $V$ is $1_{[0,1]}$ (indicator function). And since $U$ and $V$ are by assumption independent we have that the density of $(U,V)$ is $e^{-x}1_{[0,1]}$ (the product of the two densities).
Hint
Let $\varphi :(0,\infty )\times (0,1)\to \mathbb R^2$ defined by $\varphi (u,v)=(\sqrt u\cos(2\pi v),\sqrt u\sin(2\pi v))$, i.e. $(X,Y)=\varphi (U,V)$.
$$\mathbb P\{(X,Y)\in B\}=\mathbb P\{(U,V)\in \varphi ^{-1}(B)\}=\int_{\varphi ^{-1}(B)}f_{(U,V)}(u,v)dudv=\int_Bf_{(U,V)}(\varphi ^{-1}(x,y))|J_{\varphi ^{-1}(x,y)}|dxdy.$$
Therefore $$f_{(X,Y)}(x,y)=f_{(U,V)}(\varphi ^{-1}(x,y))|J_{\varphi ^{-1}(x,y)}|,$$
where $J_{g(s,t)}$ denote the jacobian of the function $g$ at $(s,t)$. Just compute $\varphi ^{-1}$ and you are done.