<s>[INST] 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.
 [/INST] 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$=W>F(1)$.
</s>