Evaluate integral with gaussian curvature 
I thought evaluating it in the following way: $$\begin{align}
\int_0^{2\pi}\int_0^{\pi}K(x,y)\sqrt{\det(g_{ij})} \, dy\,dx &= \int_0^{2\pi}\int_0^\pi \sqrt{\det L_{ij}}\cdot \sqrt{{\frac{\det L_{ij}}{\det{g_{ij}}}}} \, dy\,dx\\
&= \int_0^{2\pi}\int_0^\pi \sqrt{\det{L_{ij}}}\cdot \sqrt{\det (g^{ij}L_{ij})} \,dy\,dx \\
&=\int_0^{2\pi}\int_0^\pi\sqrt{\det L_{ij}}\cdot \sqrt {\det{L^i_{\space\space j}}} \, dy\, dx
\end{align}$$
but from here I'm stuck. How can I continue evaluating the integral above?
 A: $
\renewcommand \t\theta
\renewcommand \f\phi
\renewcommand \p\phi
\renewcommand {\d}{\partial\,}
\newcommand \T \Theta
\newcommand \P \Phi
\newcommand \F \Phi
$
HINT: Recall that Gaussian curvature of a surface is equal to Jacobian determinant of the Gauss map for this surface.

One can compute integral $ 
\int_0^{2\pi}\!\int_0^{\pi}K(x,y)\sqrt{\left\vert g_{ij}\right\vert} \, dy\,dx  $ on surface $M$ by mapping $M$ to the unit sphere $S^2$, i.e. by making change of variables defined by a Gauss map $G$.
$$
\big(\, \t, \f\, \big) = G\big(\, x, y \, \big)
\iff
\begin{cases}
\t = \T(x,y) \\
\f = \P(x,y),
\end{cases} 
\quad 
\big(\, x, y \, \big) = G^{-1}\big(\, \t , \p \, \big)
\iff
\begin{cases}
x = X(\t, \p) \\
\f = Y(\t, \p).
\end{cases} 
$$
Changing variables in the integral requires multiplying the integrand to Jacobian determinant $J_{G^{-1}}$ of the map $G^{-1}:S^2\to M$.
$$
\int_0^{2\pi}\int_0^{\pi}K(x,y)\sqrt{\left\vert g_{ij}\right\vert}\, dy\,dx  
= \int_0^{2\pi}\int_0^{\pi}K J_{G}\sqrt{\left\vert g_{ij}\right\vert} \, d\t\,d\p,
$$
where $J_G$ is the determinant of Jacobian matrix for $G$:
$\quad 
J_G = 
\left\vert \dfrac{D \left(\, x , y \, \right)}{D\left(\, \t , \p \, \right)}\right\vert
=
\begin{vmatrix}
\frac{\d X}{\d \t} & \frac{\d X}{\d \p} \\
\frac{\d Y}{\d \t} & \frac{\d Y}{\d \p} 
\end{vmatrix} 
$
Since Jacobian of a map is reciprocal of Jacobian of the its inverse, 
$$
J_G = \left\vert\dfrac{D(x,y)}{D(\t,\p)}\right\vert 
= \left\vert\dfrac{D(\t,\p)}{D(x,y)}\right\vert^{-1} 
= \dfrac{1}{J_{G^{-1}}},
$$
and since Jacobian determinant of Gauss map is equal to Gaussian curvature, 
we get rid of $K(x,y)$ and compute the integral. 
I think you can pick it up from here and make the final step by plugging everything into the integral and computing it. 

To summarize, in order to compute this integral we use the following facts:


*

*Integration by substitution involves multiplying integral by Jacobian

*The Jacobian on an inverse transformation is reciprocal to the Jacobian of direct transformation

*Gaussian curvature is equal to the Jacobian of Gauss map



PS The integral you are computing looks a lot like total Gaussian curvature.
