[This is not mean of ${ q - }$norm on ${ S ^{n} _{p} ,}$ but a related quantity in case it is helpful]

Let ${ p \in [1, \infty) }.$ By the computation of ${ f _Z (z) }$ above theorem ${ 2 }$ [here](https://mathoverflow.net/a/458639/124146), the density $${ \begin{align*} f _{Z} (z) &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \\ &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}}  \frac{1}{\Gamma(1 + \frac{1}{p}) ^{n+1}} \frac{1}{p} \Gamma\left(\frac{n+1}{p}\right) \end{align*} }$$

integrates over ${ z \in (S _{p} ^{n}) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} < 1 \rbrace }$ to give ${ 1 }.$   

> Here densities ${ f _1 (t) = \ldots = f _{n+1} (t) = \frac{1}{\Gamma(1 + \frac{1}{p})} e ^{- t ^p} }$ for ${ t \geq 0 }.$

So $${ \int _{ z \in S _{p} ^{n}} \left( \sum _{i=1} ^{n+1} \vert z _ i \vert ^{2(p-1)}\right) ^{-\frac{1}{2}} dA = p \frac{ 2 ^{n+1} \Gamma(1+\frac{1}{p}) ^{n+1}}{\Gamma\left(\frac{n+1}{p}\right)} .}$$ 

Ignoring the ${ p = 1 }$ case, $${ \int _{z \in S _p ^n} \left( \lVert z \rVert _{2(p-1)} \right) ^{-(p-1)} dA = p \frac{ 2 ^{n+1} \Gamma(1+\frac{1}{p}) ^{n+1}}{\Gamma\left(\frac{n+1}{p}\right)}. }$$

> Setting ${ p = 2 }$ gives area of ${ S _2 ^n }$ to be ${ \frac{2 \pi ^{(n+1)/2}}{\Gamma(\frac{n+1}{2}) } }.$