Consider $x \in \mathbb{R}^n$, and $B = \{x: ||x|| < r\}$, the n-ball with radius $r$ centered at the origin. Let $V$ be its volume. Further, consider $1 \leq m \leq n$ linear inequality constraints of the form $A_ix \geq 0, \, 1 \leq i \leq m$, where $A_i \in \mathbb{R}^{1 \times n}$, and the $A_i$ are linearly independent. Let C be the set of points $x$ that fulfill all of these constraints, $$C = \bigcap_{i=1}^m\{x \in \mathbb{R}^n: A_ix \geq 0\}.$$
I am looking for the volume of the "piece" of the ball defined by the $m$ equality constraints, that is, the measure of the intersection $B \cap C$. In particular, I am wondering if the volume can be calculated without evaluating any integrals, for example through some kind of recursion. A few (informal) observations that make me think that this might be the case:
When there is only one constraint ($m = 1$), the volume of the intersection is $V_1 = V/2$. Then, $C$ consists of all the x "on one side" of the hyperplane $A_1x = 0$. As the hyperplane goes through the origin, $B$ is cut in half.
When we add another constraint $A_2$ to the first one, the hyperplane $A_2x = 0$ "cuts" the (already cut-in-half) ball at a certain angle. The volume of this smaller piece $V_2$ should be some function of $V_1$ and the angle between $A_1$ and $A_2$. For example, if $A_2$ is orthogonal to $A_1$, the ball section is cut in half again, and $V_2 = V_1/2$. If the angle is sharper than 90 degrees, $0 < V_2 < V_1/2$. If it is wider, $V_1/2 < V_2 < V_1$.
I am wondering if it is possible to generalize this to some recursive formula, where $V_{k+1}$ is a function of $V_k$ and the angles between $A_{k+1}$ and $A_1,...,A_k$.
If this is not possible, I be thankful for any suggestions on how to calculate/approximate this volume numerically.