Find Marginal CDF probability from PDF (2 random variables) Given the following PDF of continuous 2 random variables:
$$
f_{X,Y}(x,y)=\begin{cases}
       y^2 & 0\le y\le x\le 1;\newline
       0 & \text{otherwise}.
\end{cases}
$$
Graph showing the region of integration with x and y random variables:

Question is to find the marginal CDF $F_{Y}(y)$ and $P[Y > x]$
My attempt to find joint CDF: $F_{X,Y}(x,y)$
$$
\begin{alignedat}{0}
F_{X,Y}(x,y)&=\int_{-\infty}^{x}\int_{-\infty}^{x}y^2 dydx \newline
&=\int_{0}^{x}\int_{0}^{x}y^2 dy dx\newline
&=\int_{0}^{x}\frac{y^3}{3}\Bigg|_{0}^{x} dx=\int_{0}^{x}\frac{x^3}{3}dx\newline 
&=\frac{x^4}{12}\Bigg|_{0}^{x}=\frac{x^4}{12}
\end{alignedat}
$$
And then to find marginal CDF of: $F_{Y}(y)=F_{X,Y}(x,y)$
$$
F_{Y}(y)=F_{X,Y}(\infty,y)=
$$
But where do I go from there? What does setting x to $\infty$ even mean?
And the question to find $P[Y>X]$, shouldn't it be 0 since $0\leq y\leq x\leq 1$?
UPDATE 12/17/2022
Based on @Xi'an's tip,
$$
\int_{0}^{1}\int_{0}^{1}cy^2dydx=1\\
\text{ implies } c = 3\\
$$
The original PDF must have been:
$$f_{X,Y}(x,y)=\begin{cases}
  3y^2 & 0\le y\le x\le1 \newline
   0   & otherwise
\end{cases}
$$
Find PDF between $0\le y\le x\le1$
$$
\begin{alignedat}{0}
F_{X,Y}(x,y)&=\int_{0}^{x}\int_{0}^{y}3y^2dy_0dx_0\\
&= y^3x
\end{alignedat}
$$
Therefore, the joint CDF is
$$
F_{X,Y}(x,y)=
\begin{cases}
0 & y\le0,x\le0,\\
y^3x & 0\le y\le x\le1,\\
1 & y\ge1, x\ge1.
\end{cases}
$$
How do I compute for other conditions, for example $y>=1, 0<=x<=1$
$$
F_{X,Y}(x,y)=\int_0^x\int_1^\infty 3y^2 dydx\\
$$
y goes to infinity?
 A: Warning: The density $f_{X,Y}$ does not integrate to one (1). This error can be easily corrected and the proper multiplying constant $$c=1\Big/\iint y^2 \mathbb I_{0\le y\le x\le 1}\text dy\text dx$$ derived.
Hint: Writing
\begin{alignedat}{0}
F_{X,Y}(x,y)&=\int_{-\infty}^{x}\int_{-\infty}^{x} cy^2 \text dy\text dx
\end{alignedat}
is not correct, because the symbol $x$ takes two different meanings in this expression, namely upper bound value and integrand symbol. The upper bound $y$ is furthermore missing.
The correct entry is, for $x_0,y_0\ge 0$,
\begin{alignedat}{0}
F_{X,Y}(x_0,y_0)&=\int_{-\infty}^{x_0}\int_{-\infty}^{y_0} cy^2 \mathbb I_{0\le y\le x\le 1}\text dy\text dx\\
&= \int_{0}^{x_0\wedge 1}\left\{\int_{0}^{y_0\wedge x} cy^2 \text dy\right\}\text dx
\end{alignedat}
where $a\wedge b=\min\{a,b\}$. Once the computation of $F_{X,Y}(x,y)$ is correctly completed,
$$F_Y(y)=F_{X,Y}(1,y)$$
since $1$ is the maximal possible value of $X$.
Note: The same care must be taken when computing $P(Y>X)$ [if I assume there is a typo in the question about $P(Y>x)$].
