[This does not give the mean of ${ q - }$norm on ${ S ^{n} _{p} ,}$ but a related quantity] 

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

that is

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


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

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**A better form for above integral**:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ [here](https://www.jstor.org/stable/2695802). 

Consider the same expression $${ \begin{align*} f _{Z} (z) &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \\ &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( z _{n+1} u ) u ^{n} \, du \end{align*} }$$ for density of ${ Z \in (S _p ^n) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} < 1 \rbrace }.$ 

Set densities $${ f _1 (t) = \frac{p}{\Gamma(d _1 /p)} t ^{d _1 - 1} e ^{-t ^p}, \ldots , f _{n+1} (t) = \frac{p}{\Gamma(d _{n+1}/p)} t ^{d _{n+1} - 1} e ^{- t ^p }  }$$ for ${ t \geq 0 }$ (where parameters ${ d _1, \ldots, d _{n+1} > 0 }$).   
These are Generalised Gamma densities, as in section 3 of the draft [here](https://mathoverflow.net/a/456451/124146). 

Now we get $${ f _{Z} (z) = \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} z _1 ^{d _1 -1} \ldots z _{n+1} ^{d _{n+1} -1} p ^n \frac{\Gamma\left(\frac{\sum _{i=1} ^{n+1} d _i}{p}\right)}{\Gamma(\frac{d _1}{p}) \ldots \Gamma(\frac{d _{n+1}}{p})} }$$ for ${ z \in (S _p ^n ) ^{+} }.$

Hence $${\boxed{ \int _{z \in S _p ^n}  \left(\sum _{i=1} ^{n+1} \vert z _i \vert ^{2(p-1)} \right) ^{-\frac{1}{2}} \vert z _1 \vert  ^{d _1 -1} \ldots \vert z _{n+1} \vert  ^{d _{n+1} -1}  dA = 2 ^{n+1} \frac{1}{p ^n} \frac{\Gamma(\frac{d _1}{p}) \ldots \Gamma(\frac{d _{n+1}}{p})}{\Gamma\left(\frac{\sum _{i=1} ^{n+1} d _i}{p}\right)}   } }$$ for parameters ${ d _1, \ldots, d _{n+1} > 0 }.$ 

> Setting ${ p = 2 }$ gives the theorem in Folland's article.

I do not know how to proceed if the term ${ \left(\sum _{i=1} ^{n+1} \vert z _i \vert ^{2(p-1)} \right) ^{-\frac{1}{2}} }$ is removed from the integrand.