Integrating $|\vec{a}\cdot \vec{u}|^h $ over unit sphere I have been studying analysis and the author mentions that the following is obvious: Let $S^n\subset \mathbb{R}^{n+1}$ be the unit sphere and $\vec{a}\in \mathbb{R}^{n+1}.$ Then
$$\int_{S^n} |\vec{a}\cdot \vec{u}|^h  \,du=C|\vec{a}|^h$$
for some constant $C$, where $h$ a real number. 
Could you please explain why is this formula true? What is $C$? 
I have been trying to understand why this is true. I would appreciate your help!
 A: Without lost of generality, we may assume $\vec{a} = (0, 0, 0, \ldots, |a|) \in \mathbb{R}^{n+1}$. Then it follows
\begin{align}
\int_{S^n}|\vec{a}\cdot \vec{u}|^h\ d\sigma(\vec{u}) = |a|^h\int_{S^n} |u_{n+1}|^h\ d\sigma(\vec{u}). 
\end{align}
Hence we see that
\begin{align}
C = \int_{S^n}|u_{n+1}|^h d\sigma(\vec{u}). 
\end{align}
Edit: Let us try to evaluate $C$. Observe
\begin{align}
C = 2\int_{S^n\cap \{u_{n+1}>0\}}u_{n+1}^h d\sigma(\vec{u})
\end{align}
and since $S^n\cap\{u_{n+1}>0\}$, the upper hemisphere, is a graph given by $u_{n+1} = \sqrt{1-u_1^2-u_2^2-\ldots-u_n^2}$.
Then we see that
\begin{align}
d\sigma(u) = \frac{du_1 du_2\ldots du_n}{\sqrt{1-u_1^2-\ldots -u_n^2}}
\end{align} 
which means
\begin{align}
2\int_{S^n\cap \{u_{n+1}>0\}}u_{n+1}^h d\sigma(u) = 2\int_D (1-u_1^2-\ldots-u_n^2)^{(h-1)/2}du_1\ldots du_n
\end{align}
where $D$ is the $n$-dimensional disk. Hence using polar coordinates, we get that
\begin{align}
2\int_D (1-|\vec{x}|^2)^{(h-1)/2} d\vec{x}=2\int^1_0 \int_{|\vec{x}|=r} (1-r^2)^{(h-1)/2}dS(x)dr= 2\omega_{n-1}\int^1_0 r^{n-1}(1-r^2)^{(h-1)/2}dr
\end{align}
where $\omega_{n-1}$ denotes the surface area of the $n-1$-dimensional unit sphere. The last integral is given by
\begin{align}
2\int^1_0 r^{n-1}(1-r^2)^{(h-1)/2} dr = \frac{\Gamma(\frac{h+1}{2})\Gamma(\frac{n}{2})}{\Gamma(\frac{1}{2}(h+n+1))}.
\end{align}
Thus, we have
\begin{align}
C= \omega_{n-1}\frac{\Gamma(\frac{h+1}{2})\Gamma(\frac{n}{2})}{\Gamma(\frac{1}{2}(h+n+1))}
\end{align}
and
\begin{align}
\int_{S^n}|\vec{a}\cdot \vec{u}|^h d\sigma(\vec{u}) = \omega_{n-1}\frac{\Gamma(\frac{h+1}{2})\Gamma(\frac{n}{2})}{\Gamma(\frac{h+n+1}{2})}|\vec{a}|^h.
\end{align}
In the case $S^2$ and $h=2$, we have that
\begin{align}
\int_{S^2} |\vec{a}\cdot \vec{u}|^2\ d\sigma(\vec{u}) = \frac{4\pi}{3}|\vec{a}|^2. 
\end{align}
A: An easy way to see this is to use the "rotation invariance" of  the surface area measure $du$ on the sphere $S.$ I.e., if $f\in C(S),$ then
$$\tag 1 \int_S f(T(u))\, du = \int_S f(u)\, du$$
for any orthogonal linear transformation $T$ on $\mathbb R^n.$ Recall that if $T$ is such a map, then $u\cdot v = (Tu)\cdot (Tv)$ for all $u,v\in \mathbb R^n.$ 
In our problem, for any nonzero $a$ we have
$$\int_S |a\cdot u|^h\, du = \int_S |(|a|(a/|a|)\cdot u|^h\ du= |a|^h\int_S |(a/|a|)\cdot u|^h\ du.$$
Let $T$ be orthogonal. Then the last integral equals
$$\tag 2 \int_S |T(a/|a|)\cdot Tu|^h\ du = \int_S |T(a/|a|)\cdot u|^h\ du.$$
We have used $(1)$ here. As $T$ runs through all orthogonal transformations, $T(a/|a|)$ runs through all unit vectors in $S.$ In other words, $v\to \int_S |v\cdot u|^h\ du$ is a constant function on $S.$ This gives the result.
