Quarter-area as a sum of trapezoids on one radius
Divide the single radius into N parts: $$\Delta r=\frac{R}{N},\qquad x_k=k\,\Delta r\;(k=0,\dots,N).$$ Define the heights on the circle by Pythagoras: $$b_k=\sqrt{R^2-x_k^2},\qquad b_0=R,\; b_N=0.$$
Trapezoids
The k-th trapezoid (for k=1..N) uses the adjacent heights \(B=b_{k-1}\) and \(b=b_k\) and has height \(\Delta r\):
$$A_k=\frac{(b_{k-1}+b_k)}{2}\,\Delta r \;=\; \frac{(B+b)}{2}\,\frac{R}{N}.$$
- First trapezoid: $$A_1=\frac{(R+b_1)}{2}\,\frac{R}{N}.$$
- Last trapezoid (degenerates to a triangle since \(b_N=0\)): $$A_N=\frac{(b_{N-1}+0)}{2}\,\frac{R}{N}=\frac{b_{N-1}}{2}\,\frac{R}{N}.$$
Total quarter area
$$A^{(Q)}_{\text{trap}}(N)=\sum_{k=1}^{N}\frac{(b_{k-1}+b_k)}{2}\,\frac{R}{N} \;=\; \sum_{k=0}^{N-1}\frac{\Delta r}{2}\big(b_k+b_{k+1}\big).$$ By symmetry for the full disk: $$A_{\text{trap}}(N)=4\,A^{(Q)}_{\text{trap}}(N).$$
Optional recursive update (computational shortcut)
You can compute the next height without re-evaluating the square root from scratch:
$$b_{k+1}^2 \;=\; R^2-(x_k+\Delta r)^2 \;=\; b_k^2-2\,x_k\,\Delta r-(\Delta r)^2.$$
Summary. Yes: the area of the quarter is the sum of all these trapezoids. The last one is automatically a triangle because \(b_N=0\). This matches your picture and the loop a computer would run.
GRG — 3D extension: the sphere (surface area) + error behaviour
Work-in-progress. GRG keeps growing while we build the site. The “joined method” (angular sectors + one-radius trapezoids) lifts naturally from the circle to the sphere.
Setup (hemisphere)
- Choose two integers: Nz (vertical splits) and Nθ (angular sectors for circumferences).
- Vertical step: $$\Delta z=\frac{R}{N_z},\qquad z_k=k\,\Delta z\ (k=0,\dots,N_z).$$
- Latitude radius by Pythagoras: $$r_k=\sqrt{R^2-z_k^2}.$$
- GRG circumference at radius \(r\) with angular split \(N_\theta\): $$C_{GRG}(r;N_\theta)=8\,N_\theta\,r\,\sin\!\Big(\frac{90^\circ}{2N_\theta}\Big).$$
Surface area by “frusta” along a meridian
Between two adjacent latitudes \(k \to k+1\) we approximate the spherical band by a conical frustum:
- Meridian slant step (no π): $$s_k=\sqrt{(\Delta z)^2+\big(r_{k+1}-r_k\big)^2}.$$
- Band area (average circumference × slant step, π-free):
$$A_k \;=\; \frac{C_{GRG}(r_k;N_\theta)+C_{GRG}(r_{k+1};N_\theta)}{2}\;\cdot\; s_k.$$
Hemisphere and sphere:
$$A_{\text{hemi}}(N_z,N_\theta)=\sum_{k=0}^{N_z-1} A_k,\qquad A_{\text{sphere}}(N_z,N_\theta)=2\,A_{\text{hemi}}(N_z,N_\theta).$$
Circle area reminder (why accuracy improves)
For the quarter of a circle we already use trapezoids on the single radius:
$$A^{(Q)}_{\text{trap}}(N)=\sum_{k=0}^{N-1}\frac{\Delta r}{2}\big(b_k+b_{k+1}\big),\quad \Delta r=\frac{R}{N},\ b_k=\sqrt{R^2-(k\Delta r)^2}.$$
This has quadratic error in \(\Delta r\) (order \(N^{-2}\)), so the accuracy rises fast. The same idea—“average of neighbours” along one axis—drives the sphere construction above.
Error behaviour (what shrinks, and why)
- Angular sectors (per circumference): chord vs arc gives a relative error that scales like \( \mathcal{O}(N_\theta^{-2})\).
- Vertical frusta (along a meridian): the trapezoidal averaging in \(z\) (via the slant step \(s_k\)) yields error \( \mathcal{O}(N_z^{-2})\).
Combined effect: the sphere estimate satisfies $$A_{\text{sphere}}(N_z,N_\theta) \to 4\pi R^2 \quad\text{as}\quad N_z,N_\theta\to\infty,$$ with total error behaving like $$\mathcal{O}\!\big(N_z^{-2}\big)+\mathcal{O}\!\big(N_\theta^{-2}\big).$$ (We keep formulas π-free in the computation; π appears only when comparing to the classical continuous value.)
Algorithm (summary)
- For k = 0..Nz: compute \(z_k, r_k, C_k=C_{GRG}(r_k;N_\theta)\).
- For k = 0..Nz−1: compute \(s_k=\sqrt{(\Delta z)^2+(r_{k+1}-r_k)^2}\) and $$A_k=\tfrac{C_k+C_{k+1}}{2}\,s_k.$$
- Sum \(A_k\) for the hemisphere; double for the whole sphere.
Notes
- The polar band automatically degenerates (like the “last trapezoid → triangle” in 2D), so the loop stays uniform.
- To balance errors, pick \(N_z\approx N_\theta\) (e.g., 24–48 each gives already small relative errors).
- Volume of the sphere can be done analogously using cylindrical shells with \(C_{GRG}(r;N_\theta)\) and a height term; we can add that next.
GRG — Joined method for the circle: perimeter by angular sectors, area by single-radius trapezoids
Work-in-progress. Generative Radial Geometry (GRG) is evolving as we build the website. The idea below emerged today (23 August 2025) and we document it in real time. GRG is a young, living framework: definitions, figures, and formulas grow together with the site.
Setup (quarter of a circle)
Let the radius be R and consider the first-quadrant quarter of the circle. Fix an integer N ≥ 1. We couple two distinct discretisations:
- Perimeter (angular division): split the right angle into N equal parts with step $$ \Delta\theta=\frac{\pi}{2N} $$
- Area (single-radius heights): split one radius into N equal parts with $$ \Delta r=\frac{R}{N},\qquad x_k=k\,\Delta r,\qquad y_k=\sqrt{R^2-x_k^2} $$
- Adjacent heights \(y_k,y_{k+1}\) form a trapezoid that tiles the quarter.
Formulas (π-free perimeter and trapezoidal area)
Quarter arc length (GRG, from angular sectors):
$$ C_N^{(Q)} \;=\; 2\,N\,R\,\sin\!\Big(\frac{\pi}{4N}\Big) $$(full circle: \(C_N=8\,N\,R\,\sin(\tfrac{\pi}{4N})\)).
Quarter area (trapezoids on one radius):
$$ A_{\text{trap}}^{(Q)}(N)\;=\;\sum_{k=0}^{N-1}\frac{\Delta r}{2}\,\big(y_k+y_{k+1}\big),\qquad \Delta r=\frac{R}{N},\quad y_k=\sqrt{R^2-x_k^2},\; x_k=k\,\Delta r $$Equivalently, with \(B_k:=y_k\) and \(b_k:=y_{k+1}\), each trapezoid is
$$ A_k=\frac{(B_k+b_k)}{2}\,\frac{R}{N},\qquad A_{N-1}=\frac{(B_{N-1}+0)}{2}\,\frac{R}{N} $$Full disk by symmetry:
$$ A_{\text{trap}}(N)=4\,A_{\text{trap}}^{(Q)}(N) $$Figure (joined method)
Blue = angular sectors for the perimeter. Red = heights on one radius and the adjacent trapezoids for the area. Dashed red rays start from the intersection points and do not, in general, coincide with the blue rays.
- angular sectors (perimeter)
- trapezoids on single-radius heights (area)
Formulas (π-free perimeter and trapezoidal area)
Quarter arc length (GRG, from angular sectors):
$$ C_N^{(Q)} \;=\; 2\,N\,R\,\sin\!\Big(\tfrac{\pi}{4N}\Big) $$Full circle:
$$ C_N \;=\; 8\,N\,R\,\sin\!\Big(\tfrac{\pi}{4N}\Big) $$Quarter area (trapezoids on one radius)
Using heights at \(x_k=k\,\Delta r\) with \(y_k=\sqrt{R^2-x_k^2}\), the quarter-area is:
$$ A_{\text{trap}}^{(Q)}(N)\;=\;\sum_{k=0}^{N-1}\frac{\Delta r}{2}\,\big(y_k+y_{k+1}\big) $$ $$ \Delta r\;=\;\tfrac{R}{N} $$ $$ x_k\;=\;k\,\Delta r $$ $$ y_k\;=\;\sqrt{R^2-x_k^2} $$Equivalently, with \(B_k:=y_k\) and \(b_k:=y_{k+1}\), each trapezoid is:
$$ A_k\;=\;\frac{(B_k+b_k)}{2}\,\frac{R}{N} $$ $$ A_{N-1}\;=\;\frac{(B_{N-1}+0)}{2}\,\frac{R}{N} $$Full disk by symmetry:
$$ A_{\text{trap}}(N)\;=\;4\,A_{\text{trap}}^{(Q)}(N) $$