Given $U_1, U_2, \ldots, U_n$ i.i.d $\sim \text{Unif}(-1, 1)$, what's the probability that $U_1^2 + U_2^2 + \ldots + U_n^2 \le 1$? 
Given $U_1, U_2, \ldots, U_n$ i.i.d $\sim \text{Unif}(-1, 1)$, what's the probability that $U_1^2 + U_2^2 + \ldots + U_n^2 \le 1$? 

I tried to reduce it to $U_1^2 + U_2^2 \le 1$ and interpret the result geometrically but I'm not able to reason about it properly. 
$U_1^2 + U_2^2 \le 1$ represents the circle with radius $1$ centred in $0$, but how can you interpret $|U_1| + |U_2|$ ? 
If $-1 \le U_1 \le 1$ and $-1 \le U_2 \le 1$ then $0 \le |U_1| + |U_2| \le 2$. 
If my reasoning is correct: $$P(U_1^2 + U_2^2 \le 1) = \frac{Area of circle}{Area of |U_1| + |U_2|} = \frac{\pi}{Area of |U_1| + |U_2|}$$
After that the same principle can be applied in a higher dimension, in 3 dimensions could be the volume of sphere divided the volume of the shape defined by $|U_1| + |U_2| + |U_3|$, etc.
I'd appreciate some guidance on this.  
 A: The $n$-dimensional volume of the $n$-hypercube $[-1,+1]^n$ is $2^n$.  The volume of an $n$-hypersphere of radius $1$ is $\dfrac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}$.  Since $U_1,U_2,\ldots,U_n$ are independent random variables uniformly distributed on $[-1,+1]$, the joint probability measure $\mathbb{P}$ is the normalized volume measure on $[-1,+1]^n$.  (The volume of $[-1,+1]^n$ is normalized in the sense that $\mathbb{P}\big([-1,+1]^n\big)=1$; in other words, $\mathbb{P}(E)=\dfrac{\text{vol}_n(E)}{2^n}$ for all measurable subsets $E$ of $[-1,+1]^n$.  Here, $\text{vol}_n$ is the standard $n$-dimensional Legesgue measure.)  Since the event $U_1^2+U_2^2+\ldots+U_n^2\leq 1$ forms a unit $n$-hypersphere, the probability of this event is the normalized volume of the unit $n$-hyphersphere, which is
$$\frac{1}{2^n}\,\left(\dfrac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}\right)=\frac{1}{\Gamma\left(\frac{n}{2}+1\right)}\,\left(\frac{\sqrt{\pi}}{2}\right)^n=\left\{
\begin{array}{ll}
\frac{\pi^{\frac{n-1}2}}{2^{\frac{n-1}{2}}\,n!!}&\text{if $n$ is odd}\,,\\
\frac{\pi^{\frac{n}2}}{2^{\frac{n}{2}}\,n!!}&\text{if $n$ is even}\,.
\end{array}
\right.\,.$$
The definition of the double factorial $n!!$ of each integer $n\geq 0$ can be found here.
A: Let $C_n$ be the cube $[-1,1]^n$ and $\lambda_n$ be its (unique!) invariant measure w.r.t Lebesgue, defined by $\lambda_n(B) = vol(B \cap C_n)/ vol(C_n)$ for every Borell $B \subseteq \mathbb R^n$. The random vector $U=(U_1,\ldots,U_n)$ that you've specified is distributed according to $\lambda_n$. Therefore, for every Borell $B$, we have $P(X \in B) = \lambda_n(B)$. In particular, if $B$ is the unit sphere $S_n$, you get $P(X \in S_n) = vol(S_n \cap C_n) / vol(C_n)$. The rest is mechanical computations...
