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Double Integrals as Volume

UniversityMathematicsSigned midpoint Riemann columns with analytic polynomial integrals over rectangles and circular disks

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About this experiment

Build a signed volume from small columns over a chosen domain. Inspect individual cell contributions and separate quadrature error from the extra boundary-area error of a circular midpoint mask.

Learning goals

  • Interpret a double integral as a signed accumulation.
  • Compare midpoint sums with an analytic integral.
  • Recognise that circular boundary sampling can give nonmonotonic grid error.

How to use it

  1. 1Select a column and verify height times full cell area.
  2. 2Refine a rectangular grid and compare the signed sum with the analytic value.
  3. 3Allow negative heights and explain why negative columns subtract from the integral.
  4. 4Use a circular domain and compare represented cell area with the exact disk area.

The model

A circular domain retains cells whose midpoint lies inside the disk.
This exact formula applies to the displayed polynomial.
The disk formula includes the mean squared radial distance.

Assumptions

  • The integrand is h + ax + by + c times x squared plus y squared. Domains are axis-aligned rectangles or disks with a movable centre.
  • The circular boundary uses a midpoint-inclusion mask. It does not use exact partial-cell areas, so convergence can oscillate.
  • Display gaps separate columns, but sums use full cell areas. Height is compressed by a reported scale and the continuous surface extends beyond the integration domain.

Try this in class

  1. Q1Why can a signed integral vanish even when many columns have nonzero height?
  2. Q2Why can a finer circular grid temporarily have a larger error?

Frequently asked questions

Is a negative column a negative geometric volume?

It is a negative contribution to the signed integral. The physical magnitude of the drawn column is positive and its colour and placement indicate the sign.

Does the analytic value use the same midpoint mask as the approximation?

No. It integrates the exact polynomial over the exact rectangle or disk, giving an independent reference for both quadrature and boundary errors.