I want to find the proportion of the surface area of those points on the surface of the unit $n$ sphere whose coordinates sum to a nonpositive number and for which the last coordinate is nonpositive. My attempt at doing this is to try and calculate $$\int_{Q} 1 \ dS$$ where $Q$ is the region in $\mathbb{R}^n$ given by the conditions $\sum_{i=1}^n x_i\leq 0$, $x_n\leq 0$ and $\sum_{i=1}^n x_i^2=1$ and then divide by the surface area of the $n$ sphere. For example, in $\mathbb{R}^2$, this is the arc length of the unit circle from $\pi$ to $7\pi/4$. The proportion is therefore $3\pi/4/ (2\ \pi)=3/8$. I am not sure if computing a surface integral is the best way to proceed but for now I cannot think of any other method of doing this.
I know that the $n$ sphere can be parameterized using spherical coordinates but my issue is that I cannot find the appropriate integral bounds to carry out this computation. In spherical coordinates this reduces to solving for $(\theta_1,\ldots,\theta_{n-1})\in [0,\pi]^{n-2}\times [0,2\pi)$ such that $$\cos(\theta_1)+\sin(\theta_1)\cos(\theta_2)+\ldots+\cos(\theta_{n-1})\Pi_{i=1}^{n-2}\sin(\theta_i)+\Pi_{i=1}^{n-1}\sin(\theta_i)\leq 0$$ and $$\Pi_{i=1}^{n-1}\sin(\theta_i)\leq 0.$$ Due to $\sin(x)\geq0$ for $x\in [0,\pi]$, the last condition gives $\theta_{n-1}\in [\pi,2\pi)$. I am trying to figure out how to work with the first condition.
I want to possibly compute "similar" surface areas on the unit $n$ sphere so I want to find a way of doing it that can be generalized.
$\DeclareMathOperator{\vol}{vol}$The region $Q$ is the intersection of the unit $(n - 1)$-sphere with the closed half-spaces having inward unit normals $\nu_{1} := -\frac{1}{\sqrt{n}}(1, 1, \dots, 1)$ and $\nu_{2} := (0, 0, \dots, 0, -1)$. The angle $\theta$ between these vectors is $\theta = \arccos (\nu_{1} \cdot \nu_{2}) = \arccos \frac{1}{\sqrt{n}}$. By rotating the sphere, we obtain a congruent region $Q'$ obtained by intersecting the sphere with the closed half-spaces having inward unit normals $$ \nu_{1}' = (0, \dots, 0, -\sin\theta, -\cos\theta) = \frac{1}{\sqrt{n}}(0, \dots, -\sqrt{n-1}, 1),\qquad \nu_{2}' = \nu_{2}. $$ We can now slice the sphere Cavalieri-style by planes parallel to the $(x_{n-1}, x_{n})$-plane, with the result that $$ \frac{\vol{Q}}{\vol S^{n-1}} = \frac{\vol{Q'}}{\vol S^{n-1}} = \frac{\pi - \theta}{2\pi}. $$ (If the question really was about the $n$-sphere in $\mathbf{R}^{n+1}$, replace $n$ by $n + 1$ in this formula.)