How is the volume of a solid bounded above by a curved surface approximated using Riemann sums in the double integral definition?

Conditions

• The solid is bounded above by a smooth surface $z=f(x,y)$ and below by a planar region $D$.
• The base region $D$ is uniformly partitioned into tiny rectangular grids.
• $\Delta\sigma_i$ represents the area of the $i$-th sub-region.
• $(\xi_i, \eta_i)$ is a sample point chosen within the $i$-th sub-region.

Reasoning, step by step

1. Plot the smooth upward-bulging surface and its rectangular projection $D$ on the horizontal plane.
2. Partition the base region $D$ into numerous tiny rectangular grids.
3. Extrude upwards from each micro-grid to generate slender rectangular prisms.
4. Assign the altitude of each prism as the corresponding surface height $f(\xi_i, \eta_i)$.
5. Sum the volumes of all individual prisms to form an approximation of the total volume.
6. Let the grid partitions become increasingly dense (mesh size $\to 0$).
7. Observe that the stepped model converges infinitely close to the actual smooth surface volume.

Example

Isolating one prism: its base area is $\Delta\sigma_i$ and its height is $f(\xi_i, \eta_i)$. Adding their products gives the approximate volume. The script states: 'As the grid partitions become increasingly dense, this stepped approximation model converges infinitely close to the actual smooth surface.'

Common misconceptions

• Believing that increasing the number of prisms alone guarantees accuracy without refining the mesh size (maximum diameter of sub-regions must go to zero).
• Confusing the sample point height with the average height or maximum height of the surface over the cell.

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