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?
Best Answer
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:
$$\int_0^3\int_{y/3}^1 f(x,y)\,\mathrm dx\mathrm dy$$