A question about integral transformation from the Unit sphere to Euclidean space. I want to show the following identity about integral on spherical coordinate system:
Let $\mathbb{S}^{d-1}$ denote the unit sphere of the Euclidean space $(\mathbb{R}^d,||\cdot||)$ equipped with
the surface Lebesgue measure $\sigma$ normalized by $\sigma(\mathbb{\mathbb{S}}^{d-1}) = 1$.  Prove that for $1\leq q<\infty$:
$$\Big(\int_{\mathbb{s}^{d-1}}|x\cdot y|^qd\sigma(y)\Big)^{1/q} = ||x||_2\Big(\frac{2\Gamma(d/2)}{\Gamma(1/2)\Gamma((d-1)/2)}  \int_{0}^1 x_1^q(1-x_1^2)^{\frac{d-3}2} dx_1\Big)^{1/q},$$
where $||x||_2=(\sum_{i=1}^d|x_i|^2)^{1/2}$.
My understanding is that this identity transfers the integral about sphere to integral about one variable $x_1$. I feel that this  maybe using the rotation invariance of the measure $d\sigma$ and doing the spherical coordinate transformation. But I have no idea to prove this identity.
Any suggestions would be welcome and thank you!
 A: Here's a way to construct a proof. Your intuition is correct, but doing a rotation in spherical coordinates can be a bit problematic, albeit probably doable. A solution to this is to rewrite the expression as an integral over $\mathbb{R}^d$ and do the rotation there, where it's pretty easy to express explicitly.
Rewrite the original integral as
$$I=\int_{S^{d-1}}|x\cdot y|^qd\sigma_y=\mu^{-1}(S_{d-1})\int_{\mathbb{R}^d}d^d y~\delta(||y||_2-1)|x\cdot y|^q$$
Now perform an arbitrary rotation in $y$ expressed as a change of variables $y'=Ry~,~ R^TR=I~,~d^dy'=d^dy$. The integral now reads
$$I=\frac{1}{\mu(S_{d-1})}\int_{\mathbb{R}^d}d^d y'~\delta(||y'||_2-1)|(Rx)^T y'|^q$$
Keeping in mind that we want to introduce multidimensional spherical coordinates eventually, we pick $Rx=||x||_2\hat{e_1}$. This is always possible. With this choice $(Rx)^T y'=||x||_2||y'||_2\cos\varphi_1$ and finally reintroducing spherical coordinates we obtain
$$I=\frac{||x||^q}{\mu(S_{d-1})}\int_0^{\infty} dr  ~r^n\delta(r-1)\int d\Omega_{d-1}|\cos\varphi_1|^q$$
The radial integral is, as expected, trivial. The angular integral can be expressed in terms of the spherical surface element as
$$\int d\Omega_{d-1}|\cos\varphi_1|^q=\int d\Omega_{d-2}\int_{0}^{\pi}d\varphi_1|\cos\varphi_1|^q\sin^{d-2}\varphi_1=2\mu(S_{d-2})\int_0^{\pi/2}d\varphi_1\cos^{q}\varphi_1\sin^{d-2}\varphi_1=2\mu(S_{d-2})\int_{0}^1dx_1 ~x_1^q(1-x_1^2)^{\frac{d-3}{2}}$$
where the last step above has been achieved by setting $x_1=\cos\varphi_1$. Substituting in the known values for the surface area of the spheres we get the desired result
$$I=||x||^q\frac{2\Gamma\left(\frac{d}{2}\right)}{\sqrt{\pi}~\Gamma\left(\frac{d-1}{2}\right)}\int_{0}^1 dx_1~ x_1^q(1-x_1^2)^{\frac{d-3}{2}}$$
PS the last integral can be done in terms of Gamma functions as well!
