Q: Integral over a surface area $\int_{S^{n-1}}x_{1}^{2}dS$ I want to evaluate the following integral:
$$\int_{S^{n-1}}x_{1}^{2}dS$$
And I think i'm supposed to use the fact that
$$
\int_{S^{n-1}}dS=\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}
$$
Attempt:
As $S^{n-1}=\left\{ \left(x_{1},\ldots,x_{n}\right)\in\mathbb{R}^{n}\mid x_{1}^{2}+\ldots+x_{n}^{2}=1\right\} \subset\mathbb{R}^{n}$
we can parametrize the hemisphere using $r\colon B^{n-1}\subset\mathbb{R}^{n-1}\to\mathbb{R}^{n}$
defined by 
$$r\left(x\right)=\left(f\left(x_{2},\ldots,x_{n}\right),x_{2},\ldots,x_{n}\right)
$$ where $f\colon\mathbb{R}^{n-1}\to\mathbb{R}$ is defined by 
\begin{align*}
 & f\left(x\right)=\sqrt{1-\left(x_{2}^{2}+\ldots+x_{n}^{2}\right)}=\sqrt{1-\left|x\right|^{2}}\\
\Rightarrow\quad & \left|\nabla f\right|^{2}=\frac{\left|x\right|^{2}}{1-\left|x\right|^{2}}
\end{align*}
Now for $g\colon\mathbb{R}^{n}\to\mathbb{R}$ defined by $g\left(x_{1},\ldots,x_{n}\right)=x_{1}^{2}$
we can get
\begin{align*}
\int_{S^{n-1}}g\left(x\right)dS & =\int_{B^{n-1}}\left(g\circ r\right)\left(x\right)\sqrt{1+\left|\nabla f\right|^{2}}dx_{2}\ldots dx_{n}=\\
 & =\int_{B^{n-1}}f\left(x_{2},\ldots,x_{n}\right)^{2}\sqrt{1+\left|\nabla f\right|^{2}}dx_{2}\ldots dx_{n}=\\
 & =\int_{B^{n-1}}\left(1-\left|x\right|^{2}\right)\sqrt{\frac{1}{1-\left|x\right|^{2}}}dx_{2}\ldots dx_{n}=\\
 & =\int_{B^{n-1}}\sqrt{1-\left|x\right|^{2}}dx_{2}\ldots dx_{n}
\end{align*}
Is this correct? I don't even know if it helps. How can I evaluate
the initial integral?

A: Hint: By symmetry, $$\int_{S^{n-1}} x_1^2 dS = \int_{S^{n-1}} x_2^2 dS = \dots = \int_{S^{n-1}} x_n^2 dS$$
What's $\int_{S^{n-1}} x_1^2 + x_2^2 + \dots + x_n^2 dS$?
