How to calculate volume given by inequalities? I need to find the volume of the 3d space that is given by the following conditions:
\begin{array}{c}
 0 < x_1 < 1\\
 0 < x_2 < 1\\
 0 < x_3 < 1\\
x_1 + x_2  + x_3 < a.
\end{array}
I also need to solve this problem for the $n$-dimensional space. Could anybody, please, tell me if this problem is solvable analytically and how one can find the solution?
ADDED


*

*I need results only for a between 0 and 1.

*I know how to solve the 2D case. It is trivial. I think that I could even manage to solve the 3D case, but I need a general solution scalable to higher dimensions.
 A: If $X_1,\ldots,X_n$ are inedependent uniform$(0,1)$ random variables, then 
$$
{\rm P}(X_1+ \cdots +X_n \leq a) = {\rm volume}(A),
$$
where 
$$
A = \{ (x_1 , \ldots ,x_n ) \in (0,1)^n :x_1  +  \cdots  + x_n  < a\} .
$$
For the probability density function of the sum $X_1+ \cdots +X_n$, see this answer.
EDIT: When $0 < a \leq 1$, it holds
$$
{\rm volume}(A) = \frac{{a^n }}{{n!}}.
$$
EDIT 2: Probabilistic proof for the case $0 < a \leq 1$. 
Let $X_1,X_2,\ldots$ be independent uniform$(0,1)$ variables. We want to show that, for any $0 < a \leq 1$,
$$
{\rm P}(X_1+ \cdots +X_n \leq a) = \frac{{a^n }}{{n!}}.
$$
This can be easily done by induction, as follows. The case $n=1$ is trivial: ${\rm P}(X_1 \leq a) = a$. Assume that the result is true for $n$, and let $m = n+1$.
By the law of total probability,
$$
{\rm P}(X_1+ \cdots +  X_m \leq a) = \int_0^1 {{\rm P}(X_1  +  \cdots  + X_m  \le a|X_m  = u)\,du} 
$$
$$
= \int_0^a {{\rm P}(X_1  +  \cdots  + X_m  \le a|X_m  = u)\,du} = \int_0^a {{\rm P}(X_1  +  \cdots + X_n  \le a - u)\,du}.
$$
Hence, by the induction hypothesis,
$$
{\rm P}(X_1+ \cdots +  X_m \leq a) = \int_0^a {\frac{{(a - u)^n }}{{n!}}\,du} =  - \frac{{(a - u)^{n + 1} }}{{(n + 1)!}}\bigg|_0^a  = \frac{{a^{n + 1} }}{{(n + 1)!}}.
$$
The result is thus proved.
A: Here's a more geometric take.  You're asking about the volume of the convex hull of the $n+1$ points ${\bf 0}, {\bf a}_1, {\bf a}_2, \ldots {\bf a}_n$, where ${\bf a}_i$ is the point with an $a$ in coordinate $i$ and $0$'s elsewhere.
The convex hull of $n+1$ affinely independent points is called an $n$-simplex.  The volume of an $n$-simplex is known to equal the volume of its corresponding $n$-parallelotope, divided by $n!$.
Since in this case the corresponding $n$-parallelotope is the $n$-cube $\{ (x_1 , \ldots ,x_n ) \in (0,a)^n\}$, which obviously has volume $a^n$, the volume you want is $a^n/n!$.
