Double integrals measure volume, and are defined as limits of double Riemann Sums. We can estimate them by forgetting about the limit, and just looking at a Riemann sum; essentially this means we're adding up the volume of boxes that fit "under" the surface z=f(x,y).
The word "under" is in quotes because the boxes aren't always under the surface. We start with a rectangle in the xy-plane, and then draw a box over it. (Or below, if f(x,y) is negative there!) To decide the height of the box, we evaluate f(x,y) somewhere in the rectangle. Common choices include the lower-left corner of the rectangle, or the midpoint of the rectangle. In the examples below you can compare these choices for a certain function.
The pictures below show visually how you estimate the double integral of
on the rectangle R = [0,1] x [0, π]. The three estimates shown are:
You can click on the pictures and rotate them; "shift"+(click and drag up/down) will zoom in and out. Press the "Home" key to reset the images.
16 boxes, with the height evaluated at a corner of each rectangle
Estimated Value: 2.16431Actual Value: about 2.42743
16 boxes, with the height evaluated at the midpoint of each rectangle
Estimated Value: 2.41994Actual Value: about 2.42743
200 boxes, with the height evaluated at the midpoint of each rectangle
Estimated Value: 2.42713Actual Value: about 2.42743
Created using Live Graphics 3D.
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