Let $T$ be a revolution Torus. My objective is to prove that $\chi(T)=0,$ but without computing $\iint Kd\sigma$. I already know that this integral is zero, but I can't use this.
I parametrized $T$ by
$$\mathbb{x}(u,v)=((a+r\cos u)\cos v,(a+r\cos u)\sin v,r\sin u),\; 0<u<2\pi, 0<v<2\pi. $$ I have two ideas to compute $\chi (T).$
1) Finding some homeomorphic surface $S$ to $T$ and computing $\chi(S)=0.$
2) Explicit computing $\chi(T)=0$
On 1), I know that $S$ is homeomorphic to $S^{1}\times S^{1},$ but I don't know if this could help.
On 2), I actually don't know how to take an easy triangulation for $T$ to compute $\chi(T)$. The easiest triangulations I taught for $T$ has $V=4$, $F=8$ and $E=12$, but write this triangulation explicit would be laborious. Can I do it in another way?