Let $(X,Y)$ be distributed over $[0,1]\times[0,1]$ according to $ f(x,y)=6xy^2$. Find $P(XY^3 \leq \frac{1}{2})$ . Let  $(X,Y)$  be distributed over  $[0,1]\times[0,1]$ according to $f(x,y)=6xy^2 $. Find  $P(XY^3 \leq \frac{1}{2})$.
I think this the double integration of f (X,Y) but i'm confused in limits.
 A: Guide:
This is the corresponding region.
It might be easier to describe the complement region. 
$$P(XY^3\le \frac12)=1-P(XY^3 > \frac12)$$

For the complement region, certainly we need $X > \frac12$, now fixing $x$, try to solve for $xy^3 > \frac12$, that is for this inequality, let $y$ be on the LHS and move $x$ to the RHS.
A: Let $(X,Y)$ be a random vector with density function $f_{(X,Y)}:\mathbb R^2 \to \Bbb R,$ $f_{(X,Y)}(x,y) = 6xy^2 \chi_{[0,1]^2}(x,y) $.
Let $\mu $ be distribution of $(X,Y)$, that is for $A \in \mathcal B(\mathbb R^2) $ (borel set), $ \mu(A) = \int_A f_{(X,Y)}(x,y) d\lambda_2(x,y) $
By that we have $\mathbb P(XY^3 \leq \frac{1}{2}) = \mathbb P(\{ \omega \in \Omega : (X,Y)(\omega) \in B\}) = \mu(B) $, where 
$B=\{(x,y) \in \mathbb R^2: x \leq \frac{1}{2y^3} \}$
So now we integrate:
$$\mu(B) = \int_B f_{(X,Y)}(x,y) d\lambda_2(x,y) = \int_0^\frac{1}{\sqrt[3]2}\int_0^1 6xy^2 dxdy + \int_\frac{1}{\sqrt[3]2}^1\int_0^\frac{1}{2y^3} 6xy^2 dxdy = $$
$$ = \frac{1}{2} + \frac{3}{4}\int_\frac{1}{\sqrt[3]2} ^1 \frac{dy}{y^4} = \frac{1}{2} - \frac{1}{4}(1 - 2) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$
So, $ \mathbb P(XY^3 \leq \frac{1}{2}) = \frac{3}{4} $
We had to find point $y= \frac{1}{\sqrt[3]{2}}$ (where $y^3 = \frac{1}{2}$), because $ x\leq \min\{\frac{1}{2y^3},1\} $ so for $y\in[0,\frac{1}{\sqrt[3]2}) $, $x \in [0,1)$, and for $y \in [\frac{1}{\sqrt[3]2},1] $, $x \in [0,\frac{1}{2y^3}]$
