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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?

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I don't know, but you might find it useful that the integral of the Gaussian curvature over all the ellipsoid, according to Gauss Bonnet, is $4\pi$ – yohBS Mar 11 '12 at 17:45
@yohBS Isn't this kind of obvious for ellipsoids? And I still don't see the connection to the integral, which works out as $\frac{1}{\sqrt{a}}$ . – draks ... Apr 5 '12 at 22:56
I don't know about quick/clever, but the surface is given by $$z=a^{-1/2}\sqrt{1-x^2-y^2}$$ and the Gaussian curvature is given by $$K=\frac{z_{xx} z_{yy} -z_{xy}^2}{(1+z_x^2+z_y^2)^2}=\frac{a}{\left(a+r^2-a r^2\right)^2}$$ where $r=\sqrt{x^2+y^2}$. The total curvature is given by integrating over $r$ but you need to multiply by the surface element $\sqrt{1+z_x^2+z_y^2}$. This results in $$4\pi \sqrt{a}\underset{0}{\overset{1}{\int }}\frac{r}{ \left(a+(1-a)r^2\right)^{3/2}\sqrt{\left(r^2-1\right)}}dr=4\pi$$ whish is somewhat similar to the integral you proposed. Hope that helps. – yohBS Apr 8 '12 at 13:16

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 $K \rightarrow 0$.

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