How do I find the centroid of a lune?

Question

"A lune is the concave-convex area bounded by two circular arcs, while a convex-convex area is termed a lens." The larger circle is centered at the origin and the smaller circle is centered at (d,0). The center of the smaller circle is within the larger circle: top left image. I know the centroid will be between (rad_large, 0) and (d + rad_small,0) but I haven't been able to find an equation of where the centroid would be. I will also need to find the area of the lune.

Answer 1

The reasoning presented here will be similar to the one already posted in my previous answer :

The centroid is the center of mass of the entire figure. Obviously, the $($horizontal$)$ line uniting the centers of the two circles splits both the lens and lune into two equal halves, so it clearly contains the centroid. Now we must find another line, perpendicular on the first, which also cuts the lune into two parts of equal area. Let's say that the larger circle is centered at the origin, and has radius R, while the smaller one is centered at $(a,0)$, and has radius r. First, we must compare $$\int_X^R\Big(\sqrt{r^2-(x-a)^2}-\sqrt{R^2-x^2}~\Big)~dx~\lessgtr~\int_R^{a+r}\sqrt{r^2-(x-a)^2}~dx,$$ where $X=\dfrac a2+\dfrac{R^2-r^2}{2a}$ is the horizontal coordinate of the two points of intersection of the

two circles. If they are equal, then the centroid lies exactly at $(X,0).~$ If $LHS<RHS$, then

we must find u so that $~LHS~+~\displaystyle\int_R^u\sqrt{r^2-(x-a)^2}~dx~=~\int_u^{a+r}\sqrt{r^2-(x-a)^2}~dx.~$

Otherwise, $$\int_X^u\Big(\sqrt{r^2-(x-a)^2}-\sqrt{R^2-x^2}~\Big)~dx~=~\int_u^R\Big(\sqrt{r^2-(x-a)^2}-\sqrt{R^2-x^2}~\Big)~dx~+$$ $+~RHS.~$ Note: A similar casuistry with regard to the position of u relative to X should have been included in my other answer as well.