Why does approximating a thin ring as a rectangle introduce error, and how is it controlled in the derivation of circle area?

Conditions

• The ring has finite width $\Delta r$
• The partition is being refined ($\Delta r \to 0$)

Reasoning, step by step

1. Recognize that a ring of radius $r$ and width $\Delta r$ has inner circumference $2\pi(r-\Delta r/2)$ and outer circumference $2\pi(r+\Delta r/2)$.
2. Approximate the ring's area as a rectangle with length $2\pi r$ and height $\Delta r$.
3. Acknowledge that this ignores the difference between inner and outer perimeters (the 'curved' nature).
4. Understand that for a single ring, the error is proportional to $(\Delta r)^2$.
5. Summing over $n$ rings where $R = n \Delta r$, the total error scales roughly as $n (\Delta r)^2 = R \Delta r$.
6. As $\Delta r \to 0$, the total error $R \Delta r \to 0$, ensuring the limit yields the correct area.

Example

The script states: 'Finite thickness still introduces error; unrolling a curved ring does not make it exactly identical to a rectangle. The total error must be controlled as the partition is refined.'

Common misconceptions

• Believing that the rectangular approximation is exact for any non-zero width.
• Thinking that visualizing the unrolled strip proves the formula without taking a limit.

Watch the explanation

Explore next

• How does the upper and lower sum argument rigorously bound this error?
• What is the specific rate at which the error approaches zero?