Double integration over two random variables order Suppose that $f(x,y) = 1$ with $0 \leq x \leq 1$ and $0 \leq y \leq 1$ for simplicity. You want to find $P(X \geq 3Y)$ where $X$ and $Y$ are some random variable.
I don't understand why changing the order of integrals gives me a different answer (I am likely missing something fundamental). I proceed the way that supposedly is the right answer:
$$\int_{0}^{1} \int_{0}^{x/3} 1\,dydx = \int_{0}^{1} \frac{x}{3} = \frac{x^2}{6} \biggr\rvert_{0}^{1} = \frac{1}{6}$$
Then I try and see what happens if I do it the other way where instead of isolating the value to the right of the inequality, I just go ahead with the integration and solve for $P(X \geq 3Y)$ as follows:
$$\int_{0}^{1} \int_{0}^{3y} 1\, dxdy = \int_{0}^{1} 3y\,dy = \frac{3y^2}{2} \biggr\rvert_{0}^{1} = \frac{3}{2}$$
I suspect I'm doing some invalid order changing and don't understand why. It has been a long time since I've had to do integrals.

The first thing to do it so draw a picture. So draw the square on which the joint density function "lives." Draw the line $x=3y$, which may look more familiar as $y=x/3$. 
We want the probability that $Y\le X/3$, so we want the probability that the  pair $(X,Y)$ lands in the part $K$ of our square that is below (or on) the line $y=x/3$.  
Since the joint density function is the constant $1$, the required probability is just the area of $K$. But since you are integrating, and since we want to deal also with non-constant density functions, let us solve the problem by integrating.
We express the double integral $\iint_K (1)\,dx\,dy$ as an iterated integral.
Let us integrate first with respect to $y$. Then (look at the diagram) for any fixed $x$ between $0$ and $1$, the variable $y$ travels from $0$ to $x/3$, and then $x$ travels from $0$ to $1$. We get your first (correct) integral.
Now let us instead integrate first with respect to $x$. For fixed $y$, the region $K$ starts at $y=x/3$, or equivalently at $x=3y$. So $x$ travels from $3y$ to $1$.
After that, $y$ travels from the bottom of triangle $K$ to the top. The top is reached when $y=(1/3)(1)$. So the correct integral is
$$\int_{y=0}^{1/3}\left(\int_{x=3y}^1 (1)\,dx\right)\,dy.$$
