Systems of inequalities

In order to set up multiple integrals, it is necessary to solve systems of inequalities, a topic that isn't given much attention in Algebra courses. It's also necessary to present the solutions in a specific form. For purposes of multiple integration, a system of inequalities is solved if it takes a form such as this: where x, y, and z are the variables in the system, a and b are constants with a ≤ b, f and g are functions with the property that f(x) ≤ g(x) whenever a ≤ x ≤ b, and h and k are functions of two variables such that h(x, y) ≤ k(x, y) whenever a ≤ x ≤ b and f(x) ≤ y ≤ g(x). (In other words, if you get through none or some of the inequalities in the list with values of the variables so far that make them all true so far, then the next inequality is true for at least some value of the next variable.)

This solution corresponds to an iterated integral of the form

∫ab ∫f(x)g(x) ∫h(x,y)k(x,y) ⋯ dz dy dx.
Of course, there could be more or fewer than 3 variables, and they don't have to come in alphabetical order. Also, we will consider the system solved if it's broken into cases, each of which takes the form above. (This corresponds to when you must write a sum of iterated integrals.) In principle, some or all of the inequalities in the solution could be strict, although the ones that we need will always be weak (so that the domain of integration will be closed). Similarly, one side or the other of some or all of the compound inequalities could be left out, but ours will never do this (so that the domain of integration will be bounded).

One way to solve inequalities is to turn them into equations first, then test potential solutions on each side of the solutions to the equations. (This requires the expressions involved to be continuous, so that the Intermediate Value Theorem applies.) For compound inqualities such as we have here, the direction of the inequality is usually straightforward. Besides that, often a domain of integration is given as bounded by certain equations rather than by inequalities, and then you have no choice but to start with the equations.

Sometimes the relevant equations will have only one solution (or even none), and you'll find the other bound (or even both) by setting the two bounds on the next line equal. For example, in the solution template above, you might find a and/or b as the solutions to f(x) = g(x) rather than directly from given equations or inequalities. Similarly, you might find f and/or g by solving h(x, y) = k(x, y) for y.

For example, let's solve this system of inequalities:

I'll start at the bottom and work my way up. I could start with any variable, but to match the pattern at the top of the page, I'll start with z. If I start with the equations z = 0 and x + y + z = 1 (by turning the inequalities that involve z into equations), then the solutions for z are 0 and 1 − x − y. Setting these equal and solving for y, I get y = 1 − x; turning the only remaining inequality involving y into an equation, I also get y = 0. Setting 1 − x and 0 equal, I get x = 1; turning the last remaining inequality, involving only x, into an equation, I get x = 0. At this point, my results look like this: I still need to turn these into compound inequalities. Obviously, 1 > 0. Choosing a number in between, such as 1/2, for x, I see that 1 − x > 0, because 1 − x = 1/2 when x = 1/2. Keeping x = 1/2 and choosing a number between 0 and 1/2, such as 1/4, for y, I see that 0 < 1 − x − y = 1/4. Therefore, the final solution is So to integate over this region, I'd set up an integral of the form
∫01 ∫01−x ∫01−x−y ⋯ dz dy dx.
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This web page was written in 2019 by Toby Bartels, last edited on 2019 February 13. Toby reserves no legal rights to it.

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