Find the $P{X < Y}$ from this joint density function $f(x, y) = 6x^2y$ I have a joint density function given as: 
$$f(x, y) = 6x^2y \\
0\le x\le 1 \\
0\le y\le 1$$
Now I am asked to find the $P\{X < Y\}$, as well as the $P\{X < 2Y\}$
In order to solve this, I did the following integrals:
\begin{align}
P\{X < Y\} &= \int_0^1\int_0^Y6x^2y\,dx\,dy \\
&= 6\int_0^1y\,dy\int_0^Yx^2\,dx \\
&= 3\int_0^Yx^2\,dx \\
&= y^3
\end{align}
That seems correct to me, as plugging in a value of $1$ for $y$ results in a probability of $1$, which is intuitively correct.
And for $P\{X < 2Y\}$:
\begin{align}
P\{X < 2Y\} &= \int_0^1\int_0^{2Y} 6x^2y\,dx\,dy \\
&= 6\int_0^1y\,dy\int_0^{2Y}x^2\,dx
\end{align}
...same as above, but ending up with $\boxed{8y^3}$.
Now this intuitively seems correct as well, since plugging in a value of $0.5$ for $y$ is like asking what is the probability that $X$ is less than $2\cdot 0.5$, or 1, which should also be 1, and it is with $8\cdot(0.5)^3 = 1$
However what I'm confused about slightly now is the range. I assumed that the function range should be redefined as:
$$P\{X < 2Y\} =
\begin{cases}
8y^3,  & 0 \le Y \le 1/2 \\
1, & 1/2 \le Y \le 1  \\
\end{cases}$$
...since obviously the probability cannot be greater than $1$, and the function continues up to $8$ within the original range. Does this all seem correct?
 A: You aren't entirely off track, but instead, you should have $$\int_0^1\int_0^y6x^2y\,dx\,dy.$$ We can certainly pull the $y$ out of the first integrand, but note that our limits of integration still have a $y$ in them, so we can't simply turn this into a product of integrals. Instead, evaluating the inner integral first, we have $$\int_0^1y\left(\int_0^y6x^2\,dx\right)\,dy=\int_0^1y(2y^3)\,dy,$$ and I'm sure you can take it from there. The other problem has similar issues and fixes.
The thing to keep in mind, here, is that $Y$ is the name of a function, that can take on values $y$.
A: For the second integral, when you have $P(X<2Y)$, I'd suggest drawing the following region in $\mathbb{R}^2$: First draw the "unit box" with $0 \leq x \leq 1$ and $0 \leq y \leq 1$, that way you know that $(X,Y)$ will land in that box. Now, draw the line corresponding to $x=2y$ (which is, of course, is the line $y=1/2 x$ in the "standard form"). The line will enter the unit box at the origin and exit at the point $(1, 1/2)$. Since the region you are integrating over will be the inside of the box above the line (since this is where $x < 2y$) you need to set up your bounds accordingly. If you integrate $dxdy$ then the integral is
$$
P(X<2Y)=\int_0^{1/2} \int_0^{2y} 6x^2 y\,dxdy + \int_{1/2}^1 \int_0^1 6x^2y \,dxdy
$$
Or, if you want to integrate $dydx$
$$
P(X<2Y)=\int_0^1 \int_{\frac{1}{2}x}^1 6x^2y \,dydx
$$
