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It seems that it should be standard that the volume I am trying to compute should be known, but I was not successful in finding any reference. Anyway, I am trying to find the volume of the following:

Let $D,T > 0$ be parameters, and let $\textbf{u} = (u_1, \cdots, u_n), \textbf{v} =(v_1, \cdots, v_n) \in (\mathbb{R}^{+} \cup \{0\})^n$ be fixed vectors. Define the set $S(D,T,\textbf{u}, \textbf{v} )$ to be the intersection of the hyperplane section $\{(x_1, \cdots, x_n) \in \mathbb{R}^n : u_1 x_1 + \cdots + u_n x_n = D, x_i \geq 0, 1 \leq i \leq n\}$ and the set $\{(x_1, \cdots, x_n) \in \mathbb{R}^n : v_1 x_1 + \cdots + v_n x_n \leq T, x_i \geq 0, 1 \leq i \leq n\}$. Let $A(S) = A(S(D,T,\textbf{u}, \textbf{v}))$ denote the volume of $S$. My questions are:

1) Is there an explicit formula to compute $A(S)$? which depends only on $D,T, \textbf{u}, \textbf{v}$?

2) If such a general formula is not available, can one compute the value of $A(S)$ which is maximal for fixed $T, \textbf{u}, \textbf{v}$ and allowing $D$ to vary?

Thanks for any insights.

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I might be wrong, but it seems to me that there's no guarantee that the volume will not be infinite, for example if $u=v$ and $T\geq D$. –  tomasz Jun 24 '12 at 18:17
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It would also be easier to read if you wrote $\langle u,x\rangle=D$, $\langle v,x\rangle\leq T$. –  tomasz Jun 24 '12 at 18:20
    
If you allowed $T$ to vary instead of $D$, the maximum would be $\prod u_j/(\lVert u \rVert (n-1)!)$, I think. Or infinite if some $u_j=0$. –  tomasz Jun 24 '12 at 18:40
    
The volume is guaranteed to be finite since the set $\{\textbf{x} \in \mathbb{R}^n : \langle v, x \rangle \leq T, x_i \geq 0\}$ is compact, and since a hyperplane is closed their intersection must be compact, hence has finite volume. –  syxiao Jun 25 '12 at 2:40

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