SUNY Geneseo Department of Mathematics

Integration over General Regions

Tuesday, April 11

Math 223
Spring 2023
Prof. Doug Baldwin

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Integrals over General Regions

Based on “General Regions of Integration” and “Double Integrals over Non-rectangular Regions” in section 4.2.

Key Ideas

There are two basic types of region: ones bounded by constant x values, and ones bounded by constant y values.

Fubini’s Theorem (strong form): double integrals over these types of regions can be evaluated as iterated integrals in which the outer integral has constant bounds, and the inner integral has one or both bounds being functions of the variable from the outer integral. (This rule extends to more than 2 variables, too.)

This idea of bounds being functions of variable(s) from outer integrals is the big new idea today.

You can split a region into subregions to make it easier to integrate over.

Examples

What’s the volume between the surface z = 1 + x2 - y and the triangle with vertices at (0, 0), (1, 0), and (0, 1)?

The first thing we needed to do was come up with precise bounds for an integral. To do this, we noticed that x coordinates in the triangle range from 0 to 1. Then, for each x coordinate, y coordinates range from 0 to 1 - x:

Triangular region with 0 less or equal to X less or equal to 1 and 0 less or equal to Y less or equal to 1 minus X

Now we can use 0 to 1, and 0 to 1 - x, as bounds for integrals, and integrate. The integration happens just like it did before, except that when you plug in the upper bound to evaluate the inner integral, you replace y with an expression and not a number. After you do that, you can usually simplify somewhat:

Integrating 1 plus X squared minus Y from 0 to 1 and from 0 to 1 minus X gives 5 over 12

As an example where it’s helpful to split the region, suppose we wanted to integrate 1 + x2 - y again, but this time over a region that’s one side of a parabola for negative values of x, and the triangle from above for positive values:

Region with triangle on right and left bounded by curve 1 plus X all squared

Because the upper bound function has to be different on the left and right sides of this region, it’s convenient to split it into 2 regions. Then, from one of the properties of integrals, the integral over the whole region is the sum of the integrals over the individual parts. And we even already know one of those integrals, from the previous example.

Integral over entire region is integral over left part plus integral over right part

Now we evaluated the first integral. This proceeded just like before, except that halfway through we found ourselves working with high powers of x + 1, and thought they’d be a pain to calculate. We dealt with them by doing a simple u substitution so that we wouldn’t have to expand them in terms of x. As part of this substitution, we changed the bounds of the integral to be values of u instead of values of x:

Let U be X plus 1 in integral of X plus 1 squared plus X squared times X plus 1 squared minus 1 half X plus 1 squared

Now we could finish the integral:

Integral from 0 to 1 of U squared plus U minus 1 squared times U squared minus 1 half U to the fourth is 4 over 15

Finally, remember that the full integral we want is the sum of the one we just evaluated and the first example. So the total volume over this combined region is 4/15 + 5/12 or 41/60. 

Next

More sophisticated integrations over non-rectangular regions.

Please read “Changing the Order of Integration” in section 4.2 of the textbook, and “Triple Integrals over a General Bounded Region” in 4.4.

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