Calculating Gaussian curvature given first fundamental form only Given a first fundamental form, i.e. $$\frac{(du)^2 + (dv)^2}{u^2 + v^2}.$$ How can I calculate the Gaussian curvature $K$? I do not really know how to approach the problem, since the formulas for the Gaussian curvature involve the second fundamental form. Is there a way to calculate the second fundamental form out of the first one? 
Edit. Here is the theorema egregium as stated in Riemannian Manifolds by John M. Lee:
Let $M \subseteq \mathbb{R}^3$ be a $2$-dimensional submanifold and $g$ the induced metric on $M$. For any $p\in M$ and any basis $(X,Y)$ for $T_pM$, the Gaussian curvature of $M$ at $p$ is given by $$K= \frac{Rm(X,Y,Y,X)}{|X|^2|Y|^2-\langle X,Y\rangle^2}$$ Therefore the Gaussian curvature is an isometry invariant of $(M,g)$.
Where $Rm$ denotes the Riemann curvature tensor and $$Rm(X,Y,Z,W) = \langle R(X,Y)Z,W\rangle$$ where $R$ is the Riemann curvature endomorphism.
 A: You have $g(u, v)$ as a matrix with entries $g_{ij}(u, v)$ that are functions of $u, v$. You can compute the inverse of this matrix (it's diagonal, after all!), and call its entries $g^{ij}(u, v)$. 
Define
$$
\Gamma^\ell_{ij} (u, v) = \frac{1}{2} \sum_{k=1}^2 g^{kl} 
\left ( 
  \frac{\partial g_{ik}}{\partial u^j}
- \frac{\partial g_{ij}}{\partial u^k}
+ \frac{\partial g_{kj}}{\partial u^i}
\right )
$$
where $u^1$ means $u$, and $u^2$ means $v$, and everything on the right hand side is actually a function of $u$ and $v$. 
Now define
$$
R^\ell_{ijk} =
  \frac{\partial \Gamma_{ik}^\ell}{\partial u^j}
- \frac{\partial \Gamma_{ij}^\ell}{\partial u^k}
+ \sum_{p = 1,2} \left (\Gamma_{ik}^p \Gamma_{pj}^\ell - \Gamma_{ij}^p \Gamma_{pk}^\ell \right)
$$
(I'm not making this up!) Also: everything above is a function of $u$ and $v$, of course. 
Then the theorem egregium says that the curvature $K$ is given by 
$$
K(u, v) = \frac{1}{g(u, v)} \sum_{\ell = 1,2} R^\ell_{121} g_{\ell 2}
$$
where $g(u, v)$ in this formula denotes the determinant of the matrix of $g_{ij}$s. 
Now I'm not man enough to do all those calculations and derivatives for you, but the vast majority of the terms you compute will simply be zero (yay!). I copied these formulas from Millman and Parker's Elements of Differential Geometry; I mention that so that you can double-check that I got the indices correct. :( 
