Let $P$ be the polygon with vertices $(0,0), (1,0), (\cos \alpha, \sin \alpha)$ where $\alpha \le \pi/2.$ I wish to find the surface area of the portion of the unit sphere that lies above $P$ or its interior. Unfortunately, every time I try to set up the integral, it becomes too messy.
1st attempt: The triangle has bounds $0 \le y \le \sin \alpha$ and $y \cot \alpha \le x \le 1+(\cos \alpha – 1)\frac{y}{\sin \alpha}.$ We can parameterize these bounds using $0 \le u,v \le 1,$ then add $z = \sqrt{1-x^2-y^2} = \sqrt{1-f(u,v)^2-g(u,v)^2}.$
2nd attempt: We should instead parameterize via $\sin \theta, \sin \phi$ and let $0 \le \theta, \psi \le \pi/2.$ This gives us $y = \sin \alpha \sin \theta, x = \cos \alpha \sin \theta + (1-\frac{\sin \theta}{\sin \alpha}) \sin \phi,$ and $z = \sqrt{1-x^2-y^2} = \sqrt{\cos^2 \theta – 2\cos \alpha \sin \theta – \sin^2 \phi \csc^2 \alpha (\sin \alpha – \sin \theta)^2}.$ In order to proceed, I need to compute $dS = ||\frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi}|| d\theta d\phi.$ Unfortunately, $\frac{dz}{\partial \theta}, \frac{dz}{\partial \phi}$ are garbage.
Is there a better parametrization? How would you come up with it and what's the motivation?
Best Answer
Instead of a different parameterization, I would subdivide the problem into two parts:
Part 1 can be determined by a simple ratio. Part 2 reduces to an easy integral after it is set up.