# 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 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?

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Note to self: Example 3 on this page bit.ly/10QmumQ), shows why doing this algebraically is not foolproof. – Jeff Apr 9 '13 at 19:24

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:

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

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To paraphrase (and simplify): $0 \le y \le 3x\ ⇒\ 0 \le \frac y3 \le x$. We know $x \le 1$. We are solving for $x$ in terms of $y$, so we can discard the $0$ on the left and get $\frac y3 \le x \le 1$. – Jeff Apr 9 '13 at 15:22

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.

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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 $\frac{y}{3} \le x \le \frac{3}{3}=1$

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