= Lens-shaped intersection of two disks
{title2=$A=2r^2\arccos[d/(2r)]-(d/2)\sqrt{4r^2-d^2}$}
For two <disks> of equal radius $r$ whose centres are separated by $0\leq d\leq2r$, their intersection consists of two circular segments. Each segment is a sector of angle $2\theta$ minus an isosceles triangle, where $\cos\theta=d/(2r)$. Hence its total <area> is
$$
A=2r^2\theta-r^2\sin(2\theta)
=2r^2\arccos\frac{d}{2r}-\frac d2\sqrt{4r^2-d^2}.
$$
The limiting cases are $A=\pi r^2$ at $d=0$ and $A=0$ at $d=2r$. This geometric region is distinct from an <optical lens>.
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