Gaussian integral over a surface I have to solve the following integral:
$$
I(s)=\int_{S(s)} \frac{1}{2\pi^{3/2}}e^{-\frac{x^2+y^2+2z^2}{2}}dxdydz
$$
where $S(s)$ is the surface defined by $s=\sqrt{(x-y)^2+4z^2}$.
I parametrised $S(s$) using:
\begin{cases}
x = \frac{\xi+s\cos(\theta)}{2} \\ 
y = \frac{\xi-s\cos(\theta)}{2} \\ 
z = \frac{s\sin(\theta)}{2}
\end{cases}
with $-\infty<\xi<+\infty$ and $ 0 \leq \theta\leq2\pi$, for which $S(s)$ the element of area is equal to:
$$
dS=\frac{s}{2\sqrt{2}}\sqrt{2-\cos^2(\theta)}
$$
So the integral became:
$$
I(s)=\frac{1}{2\pi^{3/2}}\frac{s}{2\sqrt{2}}\int_{-\infty}^{+\infty}e^{-\frac{\xi^2+s^2}{4}}d\xi\int_{0}^{+2\pi}\sqrt{2-\cos^2(\theta)}d\theta
$$
but the later integral is not elementary. 
On the other hand, using delta function seems to resolve all. Writing 
$$
I(s)=\int_{-\infty}^{+\infty}\frac{1}{2\pi^{3/2}}e^{-\frac{x^2+y^2+2z^2}{2}}\delta\left(s-\sqrt{(x-y)^2+4z^2}\right)dxdydz
$$
and using the change of variables
\begin{cases}
x = \frac{\xi+r\cos(\theta)}{2} \\ 
y = \frac{\xi-r\cos(\theta)}{2} \\ 
z = \frac{r\sin(\theta)}{2}
\end{cases}
(the Jacobian is equal to $\frac{-r}{4}$) the integral becomes 
$$
I(s)=\frac{1}{8\pi^{3/2}}\int_{0}^{+\infty}r\delta(s-r)dr\int_{0}^{+2\pi}d\theta\int_{-\infty}^{+\infty}e^{-\frac{\xi^2+r^2}{4}}d\xi=\frac{s}{2}e^{-\frac{s^2}{4}}
$$
So, what is wrong with my first attempt?
Thanks
 A: Your first attempt is entirely correct and the value of the integral is indeed
\begin{align}
I(s) &= \frac{1}{2\pi^{3/2}}\frac{s}{2\sqrt{2}}\int \limits_{-\infty}^\infty \mathrm{e}^{-\frac{\xi^2+s^2}{4}} \mathrm{d} \xi \int \limits_0^{2\pi}\sqrt{2-\cos^2(\theta)} \, \mathrm{d} \theta = \frac{s \mathrm{e}^{-s^2 /4}}{2\pi} \int \limits_0^{2\pi} \sqrt{1 - \frac{1}{2} \cos^2(\theta)} \, \mathrm{d} \theta \\
&= \frac{2}{\pi} \operatorname{E}\left(\frac{1}{\sqrt{2}}\right)s \mathrm{e}^{-s^2 /4}
\end{align}
with the complete elliptic integral of the second kind $\operatorname{E}$.
The mistake lies in your second approach. As discussed here, an additional factor must be taken into account when expressing a surface integral in terms of the delta function. Let $g(x,y,z) = \sqrt{(x-y)^2 + 4 z^2}$. Then the correct expression for your integral is
$$ I(s) = \int \limits_{\mathbb{R}^3}  \frac{1}{2\pi^{3/2}} \mathrm{e}^{-\frac{x^2+y^2+2z^2}{2}} \delta(s - g(x,y,z)) \left|\nabla g(x,y,z)\right|\, \mathrm{d} x \, \mathrm{d}y \, \mathrm{d}z $$
and the gradient term turns into the missing square root after the change of variables.
