Let $(M,g)$ be a simply connected 2-dimensional Riemannian manifold without boundary, and let $K$ be the Gausssian curvature defined on $M$. If $M$ is compact, then by Gauss-Bonnet Theorem, we have $$\int_M K dA = 4\pi,$$where $dA$ is the area element of $M$ under the metric $g$.

If $M$ is not compact, then the above equality is no longer true. For example, let $M=\mathbb{R}^2$ and we define the conformal metric $g=e^{2u}\delta$, where $u=\ln(sech x)$ and $\delta$ is the Euclidean metric, then one can verify that $K \equiv 1$ on $M$, but the total area of $M$ is $\infty$, not $4\pi$.

Motivated by this example, my question is that, if we assume $(M,g)$ is a noncompact simply connected surface without boundary, and the total area of $M$ is finite, then is it true that

$$\int_M K dA =4\pi?$$
Or is it true at least for the conformal case?