Evaluate integral with gaussian curvature

Question

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?

Answer 1

$ \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.

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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.

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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

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