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[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (The term is ${ 1 }$ when ${ p = 2 }$, so it is known how to integrate polynomials on ${ S _2 ^n }$).

[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (The term is ${ 1 }$ when ${ p = 2 }$, so we know how to integrate polynomials on ${ S _2 ^n }$).

[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (The term is ${ 1 }$ when ${ p = 2 }$).

[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (The term is ${ 1 }$ when ${ p = 2 }$, so it is known how to integrate polynomials on ${ S _2 ^n }$).

[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (Thankfully the term is ${ 1 }$ when ${ p = 2 }$).

[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, 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

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

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

A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

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.

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 don't 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. (The term is ${ 1 }$ when ${ p = 2 }$).