Approximating Integrals with Riemann Sums
Volumes using lower left corner for height
Example: Find 
, with A defined by the rectangle 
, 
, and 
with rectangles of length 
and of width 
Solution:
- Identify Domain of f (x, y) to be evaluated.
- Divide the domain into rectangles with length and width .
- Place the surface over the region.
- For each rectangle, find the height of the surface at the chosen point. In this case, find the value of f (x, y) in the lower left corner.
- For the rectangle,
, 
, the height f (0, 0) in the left hand corner is z = 4 so place the plane z = 4 above this region.
- For the rectangle, ,
, the height f (0, 2) in the left hand corner is z = 6 so place the plane z = 6 above this region.
- For the rectangle,
, , the height f (0, 2) in the left hand corner is z = 8 so place the plane z = 8 above this region.
- For the rectangle, , , the height f (2, 2) in the left hand corner is z = 4 so place the plane z = 8 above the region.
- For each rectangle, find the volume of the cuboid between the rectangle in the xy plane and the indicated plane above that region.
| Rectangle |
Length |
Width |
Height |
Volume |
| 1 |
2 |
2 |
4 |
2 * 2 * 4 = 16 |
| 2 |
2 |
2 |
6 |
2 * 2 * 6 = 24 |
| 3 |
2 |
2 |
8 |
2 * 2 * 8 = 32 |
| 4 |
2 |
2 |
14 |
2 * 2 * 14 = 56 |
- Approximate the volume beneath the surface and over the region as the sum of the volumes over the four rectangles
Volume = 16 + 24 + 32 + 56 = 128
