I calculated the Gauss curvature of a surface in $\mathbb{R}^3$ with two parameters, ($\rho$, $\phi$), from the first fundamental form $I = \text{sech}^2(\rho)(d\rho^2 +d\phi^2)$.
I got $K = 1$, which I believe indicates that the surface is a sphere. Is there a way I can check my result, or somehow recognise that it must be correct directly from the first fundamental form?
On
For a surface with metric conformal to the Euclidean one, so $F=0$ and $E=G=\lambda$, the Gauss curvature is given by (Δ being the usual Laplace operator): $$K=-\dfrac{1}{2\lambda}\Delta \ln\lambda=-\dfrac{1}{2\lambda}\{\dfrac{\partial^2\ln\lambda}{\partial\rho^2}+\dfrac{\partial^2\ln\lambda}{\partial\phi^2}\}$$
Then $\lambda=\operatorname{sech}^2\rho$
$$K=-\dfrac{1}{2\operatorname{sech}^2\rho}\{\dfrac{\partial^2\ln\operatorname{sech}^2\rho}{\partial\rho^2}+\dfrac{\partial^2\ln\operatorname{sech}^2\rho}{\partial\phi^2}\}$$ $$=-\dfrac{1}{2\operatorname{sech}^2\rho}\{\dfrac{\partial(-2\operatorname{tanh}\rho)}{\partial\rho}\}=-\dfrac{1}{2\operatorname{sech}^2\rho}\{-2\operatorname{sech}^2\rho\}=1$$
On
Your $K$ calculation is correct. However the result holds for many surfaces sharing the given first fundamental form. They are isometrically deformable among themselves.. sharing same K, geodesic curvature, area, Christoffel symbols etc. as well; they can all be "applied" one over the other touching in local isometry. With the second fundamental form defined and Gauss-Codazzi compatibility satisfied, a rigid surface such as a sphere can be uniquely defined upto arbitrary integration constants of spatial location.
Your calculation is correct. You do have $K=1$. However, without information about the second fundamental form, you cannot deduce that the surface is a sphere. There are lots of surfaces with $K=1$. [Fancy fact: We only have one coordinate patch here. But if we know the surface is compact and smooth, then Liebmann's Theorem tells us it must be a sphere.]