]]>

but it is otherwise probly ok, even if not closed form —

it is easy to show that for

|t| < 1,

we have

n > pi / (2*pi – 1) (~0.6),

and we can show that

n^2 / (2*n-1) < (n-1)/n) using a special case to avoid the weirdness in the integrals

But never mind, it’s not a bad idea to eliminate

dφ² + sin²φ dθ² = Nπ sin²φ dφ²

dθ² = (Nπ − csc²φ) dφ²

. . so we integrate `Sqrt[n*Pi-csc[x]^2]`

and get —

— not even an attempt. Wow.

Well, the point about square roots suggests that my blunder was earlier in the process.

]]>also, the square roots are problematic, since some terms are negative at t=±1: for example,

n*pi*(1-t^2)-1

is -1 when t=±1, and is positive for the midrange.