Total curvature of an ovaloid.

Question

I have the following exercise that I have to solve without using Gauss-Bonnet theorem. We say that a compact surface $\Sigma \subset \mathbb{R}^3$ is an ovaloid if the Gaussian curvature $K(p)>0$ for all $p \in \Sigma$. I have to prove that the total curvature is exactly $4 \pi$. Is it correct my idea: we can calculate with bare hands the total curvature of the sphere of radius $R$ and we immeditly obtain: $$ \int_{S^2} \frac{1}{R^2} d\mathcal{A}_{S^2} = 4 \pi .$$ Then we can use the Gauss map $N: \Sigma \rightarrow S^2$ showing that if the Gauss map is a diffeoorphism then $$ \int_{\Sigma} Jac(N)(p) \,\, d \mathcal{A}_{\Sigma} = \int_{S^2} 1 \,\, d\mathcal{A}_{S^2} = 4 \pi .$$ Is it the correct way? Is there a way to solve the exercise using the covering spaces? Thank you.