Relating an integral over the n-ball with the surface area of the (n-1)-sphere I'm asking for help in order to prove the following formula:
$$\int_{|\mathbf{x}|<1}x_1^2 d\mathbf{x}= \frac{\omega_{n}}{n\left(n+2\right)} $$
Where $\mathbf{x}=(x_1,\dots,x_n)\in \mathbb{R}^n $, and $\omega_n$ is the area of the $(n-1)$-sphere. \ 
I already proved the formula in the cases $n=2$ and $n=3$, using respectively polar coordinates and Fubini's theorem. However I don't have knowledge of any theorem that could help me in the general case aside from Fubini's theorem, which allows me to proceed in the following way:
$$\int_{|\mathbf{x}|<1}x_1^2 d\mathbf{x}= \int_{-1}^{1}\left(\int_{D_{\bar{x}_1}}x_1^2\mathrm{d}x_2\dots\mathrm{d}x_n\right)\mathrm{d}x_1= \int_{-1}^{1}x_1^2 \mathcal{L}^{n-1}(D_{\bar{x}_1})\mathrm{d}x_1 $$
Where $D_{\bar{x}_1}= \left\{|\mathbf{x}|<1\right\}\cap \left\{x_1=\bar{x}_1\right\}$, which is an $(n-1)$-ball of radius $\sqrt{1-x_1^2}$, and $\mathcal{L}^{n-1}$ denotes the Lebesgue measure on $\mathbb{R}^{n-1}$.
However, this doesn't help as I'm supposed to relate the integral to the measure of the $(n-1)$-sphere, not the $(n-1)$-ball. Even then I could only go as far as saying $$\mathcal{L}^{n-1}(D_{\bar{x}_1})= \left(1-x_1^2\right)^{\frac{n-1}{2}}\mathcal{L}^{n-1}(D) $$
Where $D$ is the unit ball in $\mathbb{R}^{n-1}$, but I still can't compute the following integral when $n$ is even:
$$\int_{-1}^{1}x^2(1-x^2)^{\frac{n-1}{2}}\mathrm{d}x $$
I guess that I need some form of Stokes's theorem, but I only studied the classical results in $\mathbb{R}^3$, and the generalized version that I've found is expressed through differential forms. Is it possible to solve this in a more elementary way?
Edit: the number of dimensions was incorrect in some instances
 A: In the closed ball $x_1^2+\ldots+x_n^2\leq 1$, for any $k\in[-1,1]$ the section $x_1=k$ is given by a $(n-1)$-dimensional closed ball with radius $\sqrt{1-k^2}$. The volume of the $n$-dimensional closed ball of radius $\rho$ is given by $\frac{\pi^{n/2}}{\Gamma(1+n/2)}\rho^n$, hence by Cavalieri's principle
$$\int_{x_1^2+\ldots+x_n^2\leq 1}x_1^2 \,d\mu = \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}\int_{-1}^{1}k^2\sqrt{1-k^2}^{n-1}\,dk=\color{red}{\frac{\pi^{\frac{n}{2}}}{2\,\Gamma\left(2+\frac{n}{2}\right)}}$$
where the last identity follows from Euler's Beta function.
 The same can be proved by computing $\int_{x_1^2+\ldots+x_n^2\leq 1}\left(x_1^2+\ldots +x_n^2\right) \,d\mu$ and exploiting symmetry.
A: (some of) other answers compute the quantities of intersect explicitly in order to relate them. But actually the motivation is to find such a relation and inductively compute beginning with circle of dim 1.
To connect $V_n$ the volume of n-ball, to the surface area $S_n$ of its boundary, which is the $n-1$-sphere, all we need to do is an integration in polar coordinates (a.k.a) the coarea formula:
$$
V_n = \int_B 1 \  dV = \int_0^1 \left(\int_{S(r)} 1 \ dS \right) \ dr,
$$
where $S(r)$ stands for sphere of radius $r$ centered at origin. But this equals, by observing that surface area scales by power $n-1$, the following
$$
V_n = \int_0^1 S_n r^{n-1} dr = \frac{S_n}{n} \ .
$$
A: I revisited this old question of mine today. Looking at it now, I'm surprised no one gave the right answer, but here is what I was looking for.
Let
$$c:=\int_{|x|<1} x_1^2 $$
Then by symmetry
$$c=\int_{|x|<1}x_i^2,\qquad \forall i=1,\dots, n $$
So
$$nc=\sum_{i=1}^{n}\int_{|x|<1} x_i^2=\int_0^1\int_{|x|=\rho}\sum_{i=1}^{n}x_i^2\,\mathrm{d}\rho=\int_0^1\rho^2 |\left\{|x|=\rho\right\}|\mathrm{d}\rho=\omega_n\int_0^1 \rho^{n+1}\,\mathrm{d}\rho=\frac{\omega_n}{n+2}.$$
Hence, $$c=\frac{\omega_n}{n(n+2)}.$$
