Algebraic way to change limits of integration of a double integral I know how to graphically change the limits of integration of a double integral. That is, by graphing the region and eyeballing (a.ka.a "looking at") it to determine the new limits. But an answer to a question hints that there is an algebraic method - I hope I'm using that term correctly - to finding the limits, and I want to know what that is.
The problem is to change the order of integration of
$$\int_0^1 \int_0^{3x} f(x,y)\ dy\ dx.$$
The answer is
$$\int_0^3 \int_{\frac y3}^1 f(x,y)\ dx\ dy.$$
The book's solution reads:

The region of integration is $0 \le x \le 1,\ 0 \le y \le 3x$. Writing
   $y=3x$ as $x=\frac y3$, we see that the inequalities translate into $0
 \le y \le 3,\ \frac y3 \le x \le 1$.

That suggests that there is an algebraic way to find the new limits.
So, I can get one of the intervals by just following the solution's directions (i.e. substitute $x=\frac y3$ into the first inequality)
$$ 0 \le x \le 1 \\
0 \le \frac y3 \le 1 \\
0 \le y \le 3.$$
So far so good.
But the same process does not work for the second inequality. I just get $0 \le x \le x$.
What's the procedure to change the limits of integration for this (or any) problem algebraically?
 A: The best way I figure these things out, despite your desire for a "pure" algebraic way, is to sketch the region of integration.  In this case, you are integrating first from $y \in [0,3x]$, over $x \in [0,1]$; that is, the region below the line.
In switching the order of integration, you are still below the line $x=y/3$ (which is the same line as $y=3 x$.  But now, to go horizontal first, you must begin at the line and end at $x=1$.  This translates to
$$\int_0^1 dx \: \int_0^{3 x} dy \: f(x,y) = \int_0^3 dy \: \int_{y/3}^{1} dx \: f(x,y)$$
This procedure works very generally for bijections between $x$ and $y$. When in one direction you go from the axis to the curve, in the other direction you go from the curve outward.
A: Being a lazy sort, I typically use Iverson brackets for the purpose. Recall that $[p]$ is $1$ if condition $p$ is true, and $0$ if condition $p$ is false.
With this, we can write your integral as
$$\iint [0\leq x\leq 1][0\leq y\leq 3x]f(x,y)\,\mathrm dy\mathrm dx=\iint [0\leq x\leq 1]\left[0\leq \frac{y}{3}\leq x\right]f(x,y)\,\mathrm dy\mathrm dx$$
This can be treated as an integral with doubly infinite limits; the Iverson brackets zero things out outside the domain of validity.
Now, Iverson brackets have the property that $[p\text{ and }q]=[p][q]$; we can use this property to give the alternate representation
$$\iint \left[0\leq \frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
where I have already taken the liberty to swap out the differentials.
Now, we can factor the Iverson bracket as
$$\iint \left[0\leq \frac{y}{3}\leq 1\right]\left[\frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
or
$$\iint \left[0\leq y\leq 3\right]\left[\frac{y}{3}\leq x\leq 1\right]f(x,y)\,\mathrm dx\mathrm dy$$
You can then translate this back into the usual notation:
$$\int_0^3\int_{y/3}^1 f(x,y)\,\mathrm dx\mathrm dy$$
A: In the first integral the bounds are as follows
$$0 \le x \le 1 \ \text{and} \ 0 \le y \le 3x$$
In the second one
$$0 \le y \le 3 \ \text{and} \ \frac{y}{3} \le x \le 1$$
Try to make a picture of your situation and see how the bounds change. Since $0 \le x \le 1$ and $0 \le y \le 3x$ we know the maximum value for $x$ is $1$. Hence the maximum value for $y$ is 3. Now we know, by changing order of the integral; $0 \le y \le 3$. Additionally, $y \le 3x \leftrightarrow \frac{y}{3} \le x$ and the maximum value for $y$ was $3$, because the maximum of $x$ was $1$. Thus $\frac{y}{3} \le x \le \frac{3}{3}=1$
