Finding $P(X\leq 2Y)$ for joint pdf 

$$f_{x,y}(x,y) = \begin{cases}\frac{4}{3}(2-x)y, & 0\leq x\leq 1, 0\leq y\leq 1\\
0, & \text{otherwise.}\end{cases}$$
    Obtain $P(X\leq 2Y)$. Sketch the region of integration.


Is this the correct way to solve it?
$$\int_0^1\int_0^{2y}\frac{4}{3}(2-x)y,dy\,dx$$
Or do I need to make it so that it is: 
$$P(Y\leq \frac{x}{2})$$
And then solve form there? 
And how can I sketch the region of integration once I have solved it?
 A: So you need to integrate over this region.
This can be done by first freely choosing $x\in [0,1]$ then making $y$ vary :
$$
\int_0^1 dx \left( \int_{x/2}^1 dy\ f(x,y) \right)
$$
Or you can first choose $y$ and then $x$ but here we need to be careful since how $x$ will vary as function of $y$ must be split into two cases, when $y \in[0,1/2]$ or $y \in]1/2,1]$ :
$$
\int_{0}^{1/2} dy \left( \int_0^{2y} dx\ f(x,y) \right) + \int_{1/2}^{1} dy \left( \int_0^{1} dx\ f(x,y) \right) 
$$
A: Remember that in conjunction with the event $x<2y$, the support is bounded by $0<x<1$ and $0<y<1$, so in our integral, $x$ cannot exceed the minimum of $2y$ or $1$.
We thus need to split the integral into two parts.
$$\begin{align}\mathsf P(X<2Y)~&=~\int_0^1\int_0^{\min\{2y,1\}}\tfrac{4}{3}(2-x)y\operatorname d x\operatorname dy \\[1ex] &=~ \int_0^{1/2}\int_0^{2y} \tfrac 43(2-x)y\operatorname d x\operatorname d y+\int_{1/2}^1\int_0^1 \tfrac 43(2-x)y\operatorname d x\operatorname dy\end{align}$$
But, yes, you can take the alternate path
$$\begin{align}\mathsf P(Y>X/2)~&=~\int_0^1\int_{x/2}^1\tfrac{4}{3}(2-x)y\operatorname d y\operatorname dx\end{align}$$
Whichever you find easier to integrate.
