Volume of Region in 5D Space I need to find the volume of the region defined by
$$\begin{align*}
a^2+b^2+c^2+d^2&\leq1,\\ 
a^2+b^2+c^2+e^2&\leq1,\\ 
a^2+b^2+d^2+e^2&\leq1,\\ 
a^2+c^2+d^2+e^2&\leq1 &\text{ and }\\
b^2+c^2+d^2+e^2&\leq1.
\end{align*}$$
I don't necessarily need a full solution but any starting points would be very useful.
 A: Here's how I'd tackle the problem (i.e. a starting point). For some intuition, reduce the problem to 3D. Now, you have 3 right-angled cylinders intersecting each other:



This article gives a solution to compute the volume of the intersection in 3D ; do you think it can be extended to 5D?
A: It is clear that the valid region lies between the hyperspheres of radius $r_1=1$ and $r_2=\sqrt{5/4}$. This already gives some trivial bounds, but the upper bound is not tight. 
Lets try to compute the excess volume, beyond the unitary sphere. 
Let fix some $1<r<r_2$. The valid region over that surface corresponds to the points with $|x_i|>\sqrt{r^2-1}$ ($\forall i$). We wish to compute $p(r)$, the proportion of the surface that lies in that region, i.e. the probability that if we throw a random point uniformily over a hypershpere surface, all its coordinates are greater than some value.  This happens to be closely related to other recent question. The approximation I proposed there, using independent  gaussians, can also be applied here, and we can expect to perform better (because we are interested in a more "probable" region). Applying the same reasoning, ($n$ gaussians with variance $r/n$ here, and $\epsilon = \sqrt{r^2-1}$), we get the approximation:
$$p(r) \approx \left[ 1 - erf\left(\sqrt{ n \, \frac{r^2-1}{2 r}}\right)\right]^n$$
and from this we can compute the total volume integrating:
$$ V(n) = V_1(n) + V_e(n) = \frac{\pi^{n/2}}{\Gamma{(\frac{n}{2}+1})} + n \frac{\pi^{n/2}}{\Gamma{(\frac{n}{2}+1})} \int_1^{r_2} p(r) \; r^{n-1} dr$$
where $r_2 = \sqrt{\frac{n}{n-1}}$. Pluggin the above approximation, I get these values:
 n       A       J        Ae      Je
---------------------------------------
 2    3.7664   4.0000   0.6248   0.8584
 3    4.6359   4.6863   0.4471   0.4975
 4    5.2619   5.2746   0.3271   0.3398
 5    5.5004   5.5036   0.2366   0.2398
 6    5.3353   5.3361   0.1676   0.1684
 7    4.8406   4.8408   0.1158   0.1160
 8    4.1366   4.1367   0.0779   0.0780

where column A is my approximation, J are the values from Joriki's answer (Montecarlo integration for  $n>5$). The columns Ae,Je show the respective excess volumes over the unitary hypersphere, they are actually the relevant quantities to be compare to judge the goodness of the approximation. (I must say I'm surprised that gives so good results for small $n$, and the speed of convergence)
Here goes the Octave/Matlab code:
clear
global N=5
r2=sqrt(N/(N-1));

function pp=pp(r)
 global N;
 pp = (1-erf( sqrt( N* (r.^2 - 1)./(2*r) ) )).^N .* r.^(N-1);
endfunction

(pi^(N/2)/gamma(N/2+1)) * ( 1 + quad ("pp",1,r2) * N)

A: There's reflection symmetry in each of the coordinates, so the volume is $2^5$ times the volume for positive coordinates. There's also permutation symmetry among the coordinates, so the volume is $5!$ times the volume with the additional constraint $a\le b\le c\le d\le e$. Then it remains to find the integration boundaries and solve the integrals.
The lower bound for $a$ is $0$. The upper bound for $a$, given the above constraints, is attained when $a=b=c=d=e$, and is thus $\sqrt{1/4}=1/2$. The lower bound for $b$ is $a$, and the upper bound for $b$ is again $1/2$. Then it gets slightly more complicated. The lower bound for $c$ is $b$, but for the upper bound for $c$ we have to take $c=d=e$ with $b$ given, which yields $\sqrt{(1-b^2)/3}$. Likewise, the lower bound for $d$ is $c$, and the upper bound for $d$ is attained for $d=e$ with $b$ and $c$ given, which yields $\sqrt{(1-b^2-c^2)/2}$. Finally, the lower bound for $e$ is $d$ and the upper bound for $e$ is $\sqrt{1-b^2-c^2-d^2}$. Putting it all together, the desired volume is
$$V_5=2^55!\int_0^{1/2}\int_a^{1/2}\int_b^{\sqrt{(1-b^2)/3}}\int_c^{\sqrt{(1-b^2-c^2)/2}}\int_d^{\sqrt{1-b^2-c^2-d^2}}\mathrm de\mathrm dd\mathrm dc\mathrm db\mathrm da\;.$$
That's a bit of a nightmare to work out; Wolfram Alpha gives up on even small parts of it, so let's do the corresponding thing in $3$ and $4$ dimensions first. In $3$ dimensions, we have
$$
\begin{eqnarray}
V_3
&=&
2^33!\int_0^{\sqrt{1/2}}\int_a^{\sqrt{1/2}}\int_b^{\sqrt{1-b^2}}\mathrm dc\mathrm db\mathrm da
\\
&=&
2^33!\int_0^{\sqrt{1/2}}\int_a^{\sqrt{1/2}}\left(\sqrt{1-b^2}-b\right)\mathrm db\mathrm da
\\
&=&
2^33!\int_0^{\sqrt{1/2}}\frac12\left(\arcsin\sqrt{\frac12}-\arcsin a-a\sqrt{1-a^2}+a^2\right)\mathrm da
\\
&=&
2^33!\frac16\left(2-\sqrt2\right)
\\
&=&
8\left(2-\sqrt2\right)\;.
\end{eqnarray}$$
I've worked out part of the answer for $4$ dimensions. There are some miraculous cancellations that make me think that a) there must be a better way to do this (perhaps anon's answer, if it can be fixed) and b) this might be workable for $5$ dimensions, too. I have other things to do now, but I'll check back and if there's no correct solution yet I'll try to finish the solution for $4$ dimensions.
