SUNY Geneseo Department of Mathematics

Some Applications of Multivariable Integrals

Tuesday, November 1

Math 223 01
Fall 2022
Prof. Doug Baldwin

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Some Applications of Multivariable Integration

Based on “Applications of Double Integrals” in section 4.1 of the textbook.

Questions

How do you find bounds for integrals?

Sometimes the problem gives them to you.

Other times you have to work them out from the context of the problem. For example, imagine a temporary garden building that stretches from y = 0 to y = 12, and has an arched cross section whose height (in feet) is given by h = 9 - x2:

Building shaped like a half cylinder except with parabolic cross section

If you wanted to find the volume of this building, you could use the given y bounds (0 ≤ y ≤ 12), but you’d have to work out the bounds on x based on the height formula and the understanding that buildings don’t have negative heights. So the x bounds must be x values that make height 0, which works out to ±3 (shown in the drawing, but because we calculated those numbers, not because they were there to begin with).

Key Ideas

Calculate the area of a region by integrating 1 over that region.

The average value of a function over a region is the integral of the function over that region divided by the area of the region.

The volume under a surface f(x,y) is the integral of f over the region of interest.

Examples

Find the volume under one peak of the “egg-carton” z = sin x cos y, 0 ≤ x ≤ π, -π/2 ≤ y ≤ π/2.

Here’s “one peak” plotted by Mathematica, to give you a sense of the volume in question:

Plot of a raised bump on a square base

Use the idea that volume is the integral of the function (sin x cos y in this case) over the region of interest (given as 0 ≤ x ≤ π, -π/2 ≤ y ≤ π/2):

Integrating sine X times cosine Y from with respect to X and then Y yields volume of 4

What’s the average height of that peak?

Use the idea that the average value of a function is the function’s integral (which we just calculated) divided by the area of the region (which is a square π units on a side):

1 over area times integral of F is 1 over Pi squared times 4 or 4 over Pi squared

Next

These volumes and average values, let alone areas, aren’t terribly interesting if they can only be calculated over rectangular regions. So let’s see if there’s a way to evaluate integrals over more general shapes.

Please read “General Regions of Integration,” “Double Integrals over Non-Rectangular Regions,” and “Changing the Order of Integration” in section 4.2 of the textbook.

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