How to calculate the area of a hyperplane Anyone knows how to calculate the area of a hyperplane defined by $\sum\limits_{i = 1}^n x_i  = a$ with restriction $0 \leqslant x_i \leqslant a$ by integration, or provide a reference? Thank you!

PS: I know how to compute the special case when $n=3$ using surface integral. I don't know how to do this in general.
In particular I want to solve the following problem.

Calculate the area $A$ defined by $\sum\limits_{i = 1}^k x_i\sqrt {np_i q_i} = 0$ with restriction $ - \sqrt {\frac{np_i}{q_i}}  \leqslant {x_i} \leqslant \sqrt {\frac{nq_i}{p_i}} $ where $n\in\Bbb{N^+}$ is a fixed positive integer, and $p_i, q_i\in[0,1]$ are also fixed constants for $i=1,2,\ldots,k$. The answer is $A = \frac{\sqrt {n^{k-1}} }{(k - 1)!} \sqrt {\frac{\sum\limits_{i=1}^k p_i q_i }{\prod\limits_{i = 1}^k p_i q_i }} $.

This is from a probability textbook. $n$ is the number of independent trials, $k$ is the number of possible outcome of each trial, with the $i$th outcome having probability $p_i$. $q_i$ is defined as $q_i=1-p_i$. All possible outcomes of the $n$-time independent trials are contained in the hyperplane $A$. The area is then used to prove the "integral limit theorem". The textbook states "it is easy to verify this area by integration", but I have no idea how to do this. Hope someone can help!
 A: 6 years late but as I stumbled upon the same question without finding an answer here I thought I'd share what I found in case anyone else finds this thread. Also I'm also only answering you original question, without going into the textbook exercise. Lastly, this does not require integration explicitly (if you're willing to accept some well known facts about simplices, which you could derive using calculus).

The section of the hyperplane in $\mathbb{R}^k$ fulfilling the requirements $0 \leq x_i \leq 1$ and $\sum_{i=1}^k x_i = 1$  is a standard (or unit) $(k-1)$-simplex. Your restriction is easily achieved by scaling a standard simplex by a factor $a$.
Note that the area of a $k$-dimensional standard simplex is equal to the volume of a $k-1$ dimensional standard simplex as they can be projected onto each other. Intuitively speaking, the volume of a $k-1$ dimensional simplex consist of all points inside the simplex which fulfill the restriction $\sum_{i=1}^{k-1} x_i \leq 1$. If you then define $x_k = 1 - \sum_{i=1}^{k-1} x_i$ it is actually dostributed like all other $x_i$ and then $\sum_{i=1}^k x_i = 1$, i.e., those points properly form the surface of a $k$-simplex. You can also reverse this process to project the surface of a $k$-dimensional simplex onto the volume of a $k-1$ dimensional simplex. This shift in perspective (from area to volume) makes it easier to reason about scaling.
Since a $(k-1)$-dimensional simplex has volume $1/(k-1)!$, and we are scaling by a factor of $a$ in each of the $k-1$ dimensions, the volume of the scaled simplex satisfying $0 \leq x_i \leq a$ and $\sum_{i=1}^k x_i = a$, and thus also by the $k$-dimensional sections hyperplane fulfilling these conditions, is $a^{k-1}/(k-1)!$.
