find the region for CDF 
Let $X$ and $Y$ be two random variables with the following density function:$$f(x)=
\begin{cases}
6x(1-x),  & \text{if   } 0\le x\le 1 \\
0, & \text{otherwise}  \\
\end{cases} \\ g(y)=\begin{cases}
3y^2,  & \text{if   } 0\le y\le 1 \\
0, & \text{otherwise}  \\
\end{cases}$$ If $X$ and $Y$ are independent find the pdf of random variables $Z=\frac{X}{Y}\text{ and }U=XY$.

As $X$ and $Y$ are independent then the joint pdf is,
$$f_{X,Y}(x,y)=\begin{cases}
18xy^2(1-x),  & \text{if   } 0\le x\le 1,0\le y\le 1 \\
0, & \text{otherwise}  \\
\end{cases}$$
$(1)$
Let $F_Z(z)$ is the CDF of Z,
$$F_Z(z)=P(Z\le z)=P\left(\frac{x}{y}\le z\right)\stackrel{(a)}{=}P(x\le zy)$$
As $0\le y\le 1$, our inequality didn't change in $(a)$. Now it's time to draw the region,

$$F_Z(z)=\int_0^1\int_0^{zy}f_{XY}(x,y)\:dx\:dy$$ I can do the rest. Can anyone tell me Am I correctly done everything until here$?$
$(2)$
Let $F_U(u)$ is the CDF of U,
$$F_U(u)=P(U\le u)=P(xy\le u)$$
From here I can't image the region. Can Anyone help me to figure it out.

Any solution or hint will be appreciated. Thanks in advance.
 A: If you draw a square $0\leq x\leq 1$, $0\leq y\leq 1$ on the picture, the integration region will be a region inside the square above the line $x=zy$. If $z<1$, the line crosses the upper side of the square. For this case your integration bounds are valid:
for $z<1$ (and for $z=1$ too),
$$
\mathbb P(X\leq zY) = \int_0^1\int_0^{zy}f_{X,Y}(x,y)\:dx\:dy. 
$$ 
If $z>1$, the line crosses right side of the square in the point $x=1, y=\frac1z$.  The region above the line can be divided into two regions: 
1) $0\leq y\leq \frac1z$, $\ 0\leq x\leq zy$
and 
2) $\frac1z\leq y\leq 1$, $\ 0\leq x\leq 1$.

Therefore, for $z>1$
$$
\mathbb P(X\leq zY) = \int_0^{\frac1z}\int_0^{zy}f_{X,Y}(x,y)\:dx\:dy+
\int_{\frac1z}^1\int_0^1f_{X,Y}(x,y)\:dx\:dy. 
$$ 
The answer is 
$$
F_Z(z)=\begin{cases}\frac95z^2-z^3, & 0\leq z\leq 1, \cr
1-\frac{1}{5z^3}, & z>1, \cr
0, & z<0.\end{cases}
$$
For the second question: $xy<u$ for $0<x<1$, $0<y<1$, $0<u<1$ is the region $y<\frac{u}{x}$ inside the square $0<x<1$, $0<y<1$. This is the region below the hyperbola $y=u/x$. This hyperbola cross upper side in the point $y=1$, $x=u$. 
$$
F_U(u)=\int_0^u \int_0^1 f_{X,Y}(x,y) \, dy \, dx + \int_u^1 \int_0^{u/x} f_{X,Y}(x,y) \, dy \, dx.
$$
Since hyperbola is symmetric under the line $x=y$, one can also swap integrals
$$
F_U(u)=\int_0^u \int_0^1 f_{X,Y}(x,y) \, dx \, dy + \int_u^1 \int_0^{u/y} f_{X,Y}(x,y) \, dx \, dy.
$$

