Connection between the area of a n-sphere and the Riemann zeta function? The Riemann Xi-Function is defined as
$$
\xi(s) = \tfrac{1}{2} s(s-1) \pi^{-s/2} \Gamma\left(\tfrac{1}{2} s\right) \zeta(s)
$$
and it satisfies the reflection formula
$$
\xi(s) = \xi(1-s).
$$
But the area $A$ of a $s$-dimensional sphere is 
$$
A(s) = \frac{2 \pi^{s/2}}{\Gamma\left(\tfrac{1}{2} s\right)}
$$
so that we can write the Xi-Function like
$$
\xi(s) = s(s-1) \frac{\zeta(s)}{A(s)}.
$$
If we insert this into the reflection formula we get the following relation between the area of a n-sphere and the Riemann zeta function
$$
\frac{\zeta(s)}{\zeta(1-s)} = \frac{A(s)}{A(1-s)}
$$
Has this relation been noted in the literature? Why should there be such a connection between the area of a n-sphere and the zeta function? Can this relation be explained geometrically?
UPDATE: I agree with the comments that negative dimensional spheres are difficult to interpret geometrically. However if we insert the formula for the area $A$ into Eulers reflection formula
$$
\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin(\pi z)}
$$
we get
$$
A(1-s) = \frac{4\sin\left(\pi \frac{s+1}{2}\right)}{A(s+1)}
$$
Inserting this into our relation with the Riemann zeta function yields
$$
\frac{\zeta(s)}{\zeta(1-s)} = \frac{A(s)A(s+1)}{4\sin\left(\pi \frac{s+1}{2}\right)}
$$
which avoids the negative dimensions for $s>0$. Can this formula be interpreted geometrically?
 A: The relation $$\frac{\zeta(s)}{A(s)}=\frac{\zeta(1-s)}{A(1-s)}$$ stems from the fact that $$\frac{\zeta(s)}{A(s)}=\frac{1}{2}\mathcal{M}\left(\psi\right)\left(\frac{s}{2}\right),$$ where $\mathcal{M}$ denotes the Mellin transform and $$\psi(x)=\sum_{n=1}^\infty e^{-\pi n^2 x}.$$ This function is symmetric due to the functional equation for $\psi(x)$, and that yields the functional equation for the zeta function. The reason for the connection to the surface area appears in my previous answer https://math.stackexchange.com/a/1494471/6075, which provides a geometric explanation for integer dimensions - the case where we can actually interpret the surface area. In that answer, I showed that the factor of $A(s)$ appears due to the spherical symmetry of a naturally arising $s$-dimensional integral expression for $\zeta(s)$.
Remark 1: This gives an excellent way to remember the precise form of the functional equation!
Remark 2: Note that the surface area here, $A(s)$, corresponds to the $s-1$-dimensional sphere, that is the surface of the $s$-dimensional ball, whereas in the linked answer, $A_{k-1}$ is the surface area of the $k-1$-dimensional sphere, that is the surface of the $k$ dimensional ball.
