Split circle into horizontal lines

Question

Can you create a function that splits circle into $n$ lines that their sum gives the area of circle. For some x I want the function or formula to return the length of a horizontal line in the circle

Answer 1

One way to parameterize the first quadrant of a circle is $p(t) =(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2})$ for $0 \le t \le 1$. This works because $(2t)^2+(1-t^2)^2 =4t^2+1-2t^2+t^4 =1+2t^2+t^4 =(1+t^2)^2 $.

I'm going to use this to try to get the area of the circle.

Consider the lines from $p(-t)$ to $p(t)$ for $t = 0, 1/n, 2/n, ... n/n$. These will slice the top half of the circle into $n$ parts.

The length of the $k$-th one is $2\frac{2k/n}{1+(k/n)^2} =\frac{4kn}{n^2+k^2} $.

The distance between the lines is

$\begin{array}\\ \frac{1-((k-1)/n)^2}{1+((k-1)/n)^2}-\frac{1-(k/n)^2}{1+(k/n)^2} &=\frac{n^2-(k-1)^2}{n^2+(k-1)^2}-\frac{n^2-k^2}{n^2+k^2}\\ &=\frac{(n^2+k^2)(n^2-(k-1)^2)-(n^2-k^2)(n^2+(k-1)^2)}{(n^2+(k-1)^2)(n^2+k^2)}\\ &=\frac{(4 k - 2) n^2}{(n^2+(k-1)^2)(n^2+k^2)}\\ \end{array} $

The area of the $k$-th slice is thus about

$\begin{array}\\ \frac{4kn}{n^2+k^2}\frac{(4 k - 2) n^2}{(n^2+(k-1)^2)(n^2+k^2)} &=\frac{4k/n}{1+(k/n)^2}\frac{(4k /n - 2/n)}{(1+(k/n-1/n)^2)(1+(k/n)^2)}\frac1{n}\\ &\approx\frac{(4k/n)^2}{(1+(k/n)^2)^3}\frac1{n}\\ &=16\frac{(k/n)^2}{(1+(k/n)^2)^3}\frac1{n}\\ \end{array} $

Using these as a Riemann sum, the area of the top half of the circle is, letting $n \to \infty$,

$\begin{array}\\ \lim_{n\to \infty} \sum_{k=1}^n16\frac{(k/n)^2}{(1+(k/n)^2)^3}\frac1{n} &=16\int_0^1 \frac{x^2dx}{(1+x^2)^3}\\ &=\dfrac{\pi}{2}\\ \end{array} $

since, according to Wolfy,

$\int \dfrac{x^2dx}{(1+x^2)^3} = \frac18 \left(\frac{x (x^2 - 1)}{(x^2 + 1)^2} + \tan^{-1}(x)\right) $