I am trying to compute the total Gaussian curvature of $ z^2 = - (x^2 + y^2 -16)((x-2)^2 + y^2 -1) ((x+2)^2 + y^2 -1)$ in $\mathbb{R}^3$.
The first thing that comes in my mind is Gauss--Bonnet theorem. That is for a compact, simply connected space we have $\int_M K dA = 2 \pi \chi(M)$.
But I have no idea how to deal with this surface. For example, I can not image the shape of this surface.
My first trial is $X(u,v) = (u,v, \sqrt{-(u^2 + v^2 -16)((u-2)^2 + v^2 -1) ((u+2)^2 + v^2 -1)})$ and compute total Gaussian curvature. But I am not sure of the last coordinate. My domain was $(u,v) \in [-\infty, \infty] \times [-\infty, \infty]$, and after trial with mathematica the results was terrible.
It's a topology question. Let's consider the function: $$ f(x,y) = - (x^2 + y^2 -16)\times\Big[(x-2)^2 + y^2 -1\Big]\times\Big[(x+2)^2 + y^2 -1\Big] $$
The manifold $\partial\mathcal R: f(x,y) = 0$ consists of 3 non-intersecting circles. Moreover, $\mathcal R: f(x,y) > 0$ is a connected region:
When we consider, manifold $\Omega: z^2 = f(x,y)$ it basically inflates the region $\mathcal R$, keeping the boundary $\partial\mathcal R$ in place. (You can imagine the process of inflating the disk $x^2+y^2-1<0$ to the sphere $z^2=1-x^2-y^2$):
You don't need to plot this manifold in 3D to understand that it's homeomorhic to sphere with 2 handles. Can you use Gauss–Bonnet theorem now?