I wonder which way is the correct way to calculate a specific fraction of a solid angle.
I divided a hemisphere into a number of solid angles by using weights of gauss quadrature in the zenith direction ($W_j$) and in the azimuthal direction ($W_i$).
I denote the first zenith angle as $\theta_1$. If $W_j$ is the weight of $\theta_1$, the fraction of this solid angle as a part of the hemisphere is: $$\frac{W_i}{\sum_{i} W_i}*\frac{W_{j=1}}{\sum_{j} W_j}$$
Now, consider $\theta_0<\theta_1$.
I need to divide $\theta_1$ in the angular direction into two parts:$\theta_0$ (which is currently the first angle (near the z axis)) and $\theta_1- \theta_0$ which is currently the second angle.
My question is what is the correct way to calculate the solid angle near the z axis. I started with: $$\frac{W_i}{\sum_{i} W_i}*\frac{W_{j=1}}{\sum_{j} W_j}*\frac{\theta_0}{\theta_1}$$ But later I thought about two other options that may be more precise: $$\frac{W_i}{\sum_{W_i}}*\frac{W_{j=1}}{\sum_{j} W_j}*\frac{sin(\theta_0)}{sin(\theta_1)}$$ (which may be correct because: $\int_{2\pi} d\Omega=\int \int sin\theta d\theta d\phi={2 \pi \sum_{j} sin\theta_j}$).
($\phi$ is the azimuthal angle).
The third option is
$$\frac{W_i}{\sum_{i} W_i}*\frac{W_{j=1}}{\sum_{j} W_j}*\frac{1-cos(\theta_0)}{1-cos(\theta_1)}$$ (which may be correct because: $\int \Omega=\int_{0}^{\theta} \int_{0}^{\phi} sin\theta d\theta d\phi = \int_{0}^{\theta} sin\theta d\theta \int_{0}^{\phi} d\phi={\phi(1-cos\theta})$).
I confused with all this options. Can anyone help me? Thanks!