Integral equality of 1st intrinsic volume of spheroid Computations suggest that
$$\int_{0}^{\infty}\int_{0}^{\infty} \sqrt{x+y^2} \cdot e^{-\frac{1}{2}(\frac{x}{s}+s^2y^2)}dxdy=\frac{2}{s}+\frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}.$$
The question is how to prove this equality.
Background: this is the mean width of an ellipsoid with semiaxes $s,s,1/s^2$ and this almost completes the proof of the uniqueness hypothesis of an ellipsoid with given intrinsic volumes.
 A: First of all make a replacement $x=t^2$ and go to polar coordinates
\begin{equation*}
    \int_{0}^{\infty}\int_{0}^{2\pi}
    \cos(\phi)r^3 \cdot e^{-\frac{\frac{r^2\cos^2(\phi)}{s}+r^2\sin^2(\phi)s^2}{2}}drd\phi.
\end{equation*}
After than notice that we can take integral by radius. Let $R=\frac{1}{2}(\cos^2(\phi)\frac{1}{s}+\sin^2(\phi)s^2)$ then
\begin{equation*}
       \left. \int_{0}^{\infty}
    r^3 \cdot e^{-r^2R}dr= -\frac{1}{2R}\int_{0}^{\infty}
    r^2\cdot de^{-r^2R}=-\frac{1}{2R}r^2e^{-r^2R}\right|_{0}^{\infty}+\frac{1}{R}\int_{0}^{\infty}
    re^{-r^2R}dr=
\end{equation*}
\begin{equation*}
=\frac{1}{R}\int_{0}^{\infty}re^{-r^2R}dr=\left.-\frac{e^{-r^2R}}{2R^2}\right|_0^{\infty}=\frac{1}{2R^2}
\end{equation*}
Back to the main integral
\begin{equation*}
    \int_0^{2\pi} \frac{\cos(\phi)}{R^2}d\phi=\int_0^{2\pi} \frac{\cos(\phi)}{(\frac{1}{2}\cos^2(\phi)\frac{1}{s}+\frac{1}{2}\sin^2(\phi)s^2)^2}d\phi \propto 
\end{equation*}
\begin{equation*}
\propto s^2\int_0^{2\pi} \frac{\cos(\phi)}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\phi=
\end{equation*}
\begin{equation*}
    =s^2\int_0^{2\pi} \frac{1}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\sin(\phi)=s^2\int_{-1}^1 \frac{1}{(1-t^2+s^3t^2)^2}dt
\end{equation*}
It is easy to chek that
\begin{equation*}
    \int\frac{1}{(1+x^2a)^2}dx=\frac{x}{2(ax^2+1)}+\frac{\arctan(\sqrt{a})}{2\sqrt{a}},
\end{equation*}
So we finaly obtain for $a=s^3-1$ when $F(s)$ - is the main width of spheroid with semiaxes $\sqrt{s},\sqrt{s},\frac{1}{s}$.
\begin{equation}
    F(s)=Const\cdot s^2\left(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right)=Const\cdot \left(\frac{1}{s}+\frac{s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right)
\end{equation}
And fact that $F(s)$ decrease when $s\in(0,1)$ and increase when $s\in(1,\infty)$ means that there are only two spheroids with volume$=1$ and mean width$=W>F(1)$.
A: we pose $x=t^2$
$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{t^2+y^2} e^{\frac{t^2}{s}+s^2 y^2} 2tdxdt $
by change of variable ,we get:
$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy =2\int^{\frac{\pi}{2}}_0\int^{\infty}_0 r^3 \cos (\theta) e^{\frac{-1}{2}(\frac{1}{s} \cos ^2\theta+s^2 sin^2 \theta )r^2} drd\theta $
we pose 
$A= \frac{1}{2}(\frac{1}{s} \cos^2 (\theta)+s^2 \sin^2(\theta) )$
$\int^{+\infty}_0 r^3 e^{-Ar^2} dr=\frac{-r^2}{2A} e^{-Ar^2}\Bigg |^{+\infty}_0+\int^{+\infty}_0 \frac{r}{A}e^{-Ar^2}dr
= \frac{-1}{2A^2} e^{-Ar^2} \Bigg |^{+\infty}_0=\frac{1}{2A^2}$
$$\int^{\frac{\pi}{2}}_0\frac{\cos \theta}{2 A}d\theta =2 s^2 \int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta$$
$\int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta =\int^1_0 \frac{1}{(1+(s^3-1)t^2)^2}dt$
we pose
$a=s^3-1$
$\int^1_0 \frac{1}{(1+at^2)^2}dt =\Bigg [ \frac{t}{2(1+at^2)} + \frac{\arctan(\sqrt{a}t)}{2\sqrt{a}}\Bigg ]^1_0
=\frac{1}{2}(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}})$
So
$$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\frac{2}{s}+ \frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}$$
