 Solution

For the sphere, choose the inward normal so that the second fundamental form is $I/a$. Both principal curvatures are $1/a$, giving $K=1/a^2$; the outward choice reverses both signs and gives the same product.

The octant has area $4\pi a^2/8=\pi a^2/2$, so its curvature integral is $\pi /2$. Its sides are great-circle geodesics, hence have zero geodesic curvature. Each interior angle is $\pi /2$, so each exterior angle is also $\pi /2$. Thereforeverifying the polygon formula directly.