Double Integrals
A double integral adds up something spread over a flat region. Chop the region into tiny rectangles, multiply each one's area by the value of there and add; when is a height, the total is the volume under the surface.
What the symbols mean
- Area of a small rectangle
Δx Δy.
- Height
The surface's height over that rectangle.
- Area element
The limiting tiny piece of area.
Explanation
A double integral is simply a way to add up something that's spread over a flat region, like rain falling on a field or charge sitting on a plate. The symbol looks intimidating, but it's just three small pieces: (the region you're covering), (a tiny patch of that region) and (how much stuff there is per unit area at that spot). Multiply each patch by its value of , then add up all the patches. That's the whole idea; the rest of this unit is about doing that adding efficiently.
Take a look at how the single integral worked: chop into thin strips, multiply each width by a height, add, and let the strips get thinner. This is the same Riemann sum one dimension up. The strips become small rectangles in the plane, each strip's area becomes a box's volume, and the sum of the boxes closes in on the volume under the surface.
What the total means depends on what is:
- With a height, is the volume of the solid between the surface and the region .
- With , every box has height , so the total is just the area of .
- Divided by the area of , it's the average value of over , exactly like in one variable.
A common mistake. It's tempting to read every double integral as a volume. That's only true when . Where is negative the boxes count as negative, just as area below the axis did for single integrals, so is really a net amount.
The midpoint rule estimates it by evaluating at the center of each small rectangle. It's surprisingly good: in the picture below, nine squares already land within of the exact volume, because over each square the surface rises on one side of the center about as much as it falls on the other. The next lesson shows how to get the exact value with two ordinary integrals.
See it
Worked examples
Estimate over with four squares and the midpoint rule.
- 1
Four equal squares of side cover , so each has area . Their centers are , , and .
Practice
Type your answer and check it. Solve them all to complete the lesson.
Use the four-square midpoint rule to estimate over .
Find over .
The integral of over is 8. What is the average value of there?
For , what does represent?
Find over the disk of radius 3.
A m by m plate carries electric charge with density μC/m², for , . Find the total charge.
The hardest one here. Get it, and you've got this lesson.
Find over .
Where it's used
Total rainfall over a county from a rainfall-rate map, total mass of a plate from its density, and the expected value of a function of two random variables are all double integrals.