Computing the distribution of a uniform r.v. with parameter being another uniform r.v. I have this:

Let $X\sim U(0,1)$, $Y\sim U(X,1)$.
  What is the distribution of variable $Y$?

My answer: I use a geometric approach since everything happens in the square $(0,1)\times (0,1)$, see the image.

so I found that 
$$F(y)= \mathbb{P}\{ Y \leq y \}=
\begin{cases}
0 & y<0 \\
y^2 & 0 \leq y\leq 1\\
1 & y>1
\end{cases}
$$
then
$$f(y)=
\begin{cases}2y & 0\leq y\leq 1\\
0 & \text{otherwise.}
\end{cases}
$$
Is this approach correct?
 A: A sanity check: verify you get the right expectation. (This is not sufficient, but at least it's a necessary condition for correctness).
Since $Y\sim U(X,1)$, we have
$$\mathbb{E}[Y\mid X] = \frac{1}{2}(1+X)$$
and therefore
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\mid X]] = \frac{1}{2}(1+\mathbb{E}[X])
= \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$$
since $X\sim U(0,1)$ (for the second-to-last equality).
Your solution, however, yields
$$\mathbb{E}[Y] = \int_{-\infty}^\infty dy f(y)y=\int_{0}^1 2y^2dy = \frac{2y^3}{3}\Big|_0^1=\frac{2}{3}
$$
so your answer cannot be correct.

A derivation: 
Fix any $t\in\mathbb{R}$. First of all, it is obvious that $F_Y(t) = 0$ if $t\leq 0$ and $F_Y(t) = 1$ if $t\geq 1$, so we can focus on $t\in(0,1)$. Then, we have
$$
F_Y(t) = \int_0^1 dx \int_x^t dy \frac{1}{1-x}\mathbb{1}_{\{t\geq x\}}
= \int_0^t dx \int_x^t dy \frac{1}{1-x}
= \int_0^t dx \frac{t-x}{1-x}
$$
and after computation, we get
$$
F_Y(t) = \begin{cases} 0 &t\leq 0\\
t+(1-t)\ln(1-t) & t\in(0,1)\\
1 & t\geq 1
\end{cases}
$$
(which after verification is consistent: $F_Y(0^+)=0, F_Y(1^-)=1$, and $F_Y$ is non-decreasing and continuous.)
Deriving this CDF, we obtain
$$
f_Y(t) = -\ln(1-t) \mathbb{1}_{(0,1)}(t),\qquad t\in\mathbb{R}
$$
which passes the (necessary) test
$$
\int_{-\infty}^\infty t f_Y(t) dt = -\int_0^1 t\ln(1-t) dt = \frac{3}{4}.
$$

Plot of the results obtained in the derivation:

PDF $f_Y$ of $Y$

CDF $F_Y$ of $Y$
