On a sphere of radius $R$ we randomly choose $N$ circles of radius $r$ , $r < R$ (this means that centers of circles uniformly distributed on on a sphere of smaller radius). Each of the circles uniquely determines the spherical cap. What is the expected area of the union of these spherical caps?
Because of caps intersection I think simple approach with repeated integral doesn't seem to be the way this problem should be solved.
Thank you.
I'm not 100% sure of the following line of reasoning, but it gives a reasonable-looking result...
Let the area of the sphere be $S$ (a function of $R$), and the area of a single spherical cap $C$ (a function of $R$ and $r$).
Then the probability that a randomly-chosen point on the sphere is not on the cap is $$1-\frac{C}{S}$$
For $N$ caps, the probability that a randomly-chosen point on the sphere does not lie on any of them is $$\left(1-\frac{C}{S}\right)^N$$
We could interpret this probability as the fraction of $S$ not expected to be covered by caps.
So you would expect the surface area to be $$S\left(1-\left(1-\frac{C}{S}\right)^N\right)$$
This is reasonable for $N=1$ (gives $C$) and $N\rightarrow \infty$ (gives S)