How does the limiting circle-area argument derive the formula πR²?

Q: How does the limiting circle-area argument derive the formula πR²?

The argument places strips of height $2\pi r$ along the radius axis, forming a Riemann sum. As the maximum partition width tends to zero, these sums approach the area beneath the linear circumference function. The limiting region is a triangle with base $R$ and height $2\pi R$. The area of this triangle is $\frac{1}{2} \cdot R \cdot 2\pi R = \pi R^2$, which gives the area of the circle.

The argument places strips of height

2\pi r

along the radius axis, forming a Riemann sum. As the maximum partition width tends to zero, these sums approach the area beneath the linear circumference function. The limiting region is a triangle with base

R

and height

2\pi R

. The area of this triangle is

\frac{1}{2} \cdot R \cdot 2\pi R = \pi R^2

, which gives the area of the circle.

Conditions

The partition is refined such that the maximum width tends to zero.
The circumference function $2\pi r$ is treated as the height of the strips.

Reasoning, step by step

Place strips along the radius axis with heights $2\pi r$ .
Form a Riemann sum of their areas.
Take the limit as the maximum partition width tends to zero.
Identify the limiting region as a triangle with base $R$ and height $2\pi R$ .
Calculate the area of the triangle: $\frac{1}{2} \cdot R \cdot 2\pi R = \pi R^2$ .
Conclude that the circle's area is $\pi R^2$ .

Example

The script states: 'The limiting region is a triangle with base R and height 2πR, giving πR². On a uniform partition, the upper and lower sums differ by 2πR²/n. The error tends to zero...'

Common misconceptions

Believing that finite graphics constitute a complete proof.
Assuming the convergence rate is super-polynomial.

Watch the explanation

YouTubeThe essence of calculus

5:52 – 7:13Watch this moment ↗

Explore next

Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.