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 $K \rightarrow 0$.