Gaussian Curvature Can anyone see the connection between the Gaussian curvature of the ellipsoid $x^2+y^2+az^2-1=0$ where $a>0$ and the integral $\int_0^1  {1\over (1+(a-1)w^2)^{3\over 2}}dw$?
I am guessing Gauss Bonnet, since that is what the related chapter is about. Is there a quick / clever way of finding the curvature of this surface?
 A: First of all I would like suggest a different parametrization ofr the ellipsoid.
\begin{equation}
{X}=\{\cos(u) \cos(v), \cos(u) \sin(v), a \sin(u)\}
\end{equation}
where $u \in \{-\pi/2,+\pi/2 \}$ and $v \in \{-\pi, +\pi \}$. The cartesian parametrization allows representing only half of the ellipsoid because of the sign of the square root. By using the above parametrization the gaussian curvature can be written as:
\begin{equation}
K = \frac{4 a^{2}}{(1+a^{2}+(a^{2}-1) \cos{(2 u)})^{2}}
\end{equation}
Than 
\begin{equation}
\sqrt{K} = \frac{2 a}{(1+a^{2}+(a^{2}-1) \cos{(2 u)})}
\end{equation}
where I assumed a>0. Now it is possible to prove that we can change the integration variable $w$ so that 
\begin{equation}
\int{\frac{1}{(1+(a-1) w^{2})^{3/2}} dw}={K}^{1/2}
\end{equation}
In order to prove the relation above use the substitution:
\begin{equation}
w^{2} = \frac{1}{1-a+\frac{1}{\sqrt{K}}}
\end{equation}
Then by differentiating:
\begin{equation}
2 w dw  = \frac{1}{2 K ^{3/2} \left(-a+\frac{1}{\sqrt{K}}+1\right)^2}dK
\end{equation}
And by replacing $w$ I get:
\begin{equation}
dw  = \frac{1}{4}\left(\frac{1}{K (1-a) +\sqrt{K }}\right)^{3/2} dK
\end{equation}
Then I have:
\begin{equation}
\int{\frac{1}{(1+(a-1) w^{2})^{3/2}} dw}= \frac{1}{4}\int{\frac{1}{K^{3/4}}} dK= \sqrt{K}
\end{equation}
Now consider the integration limits. $w$ ranges between 0 and 1. When $w=1$ we have $K=\frac{1}{\sqrt{a}}$, while when $w=0$ we have $K \rightarrow 0$.
