Can $A(S^-)>A(S^+)$?

Question

Let's be $S$ a surface which is homeomorphic to a $\mathbb{S}^2$ (the unit sphere).

Let's be $$S^+=\{p\in S : K(p) \geq 0\} \text{ and } S^-=\{p\in S : K(p) \leq 0\}$$ being $K$ Gaussian curvature.

Is it possible that $Area(S^-)>Area(S^+)$??

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I think I have to use the Gauss-Bonnet theorem.

Using Gauss-Bonnet theorem, I know that $Area(\mathbb{S}^2)=4\pi=\int\int_{\mathbb{S}^2}KdA$, because $K=1$.

I also know that Euler-Poincaré characteristic is invariant to homeomorphisms. So I know that $\chi(S)=\chi(\mathbb{S}^2)=2$ and $S=S^+\cup S^-$.

Can someone help me to say something about the relation of areas?