Are there different ways to embed surface with nonvanishing curvature in a higher-dimensional Euclidean space? The two-dimensional Euclidean space $E^2$ can be embedded isometrically in many ways in $E^3$. Its First Fundamental form will always be $\delta_{\mu\nu}$, but the Second Fundamental form $II$ will depend on the embedding. If it is embedded as a plane, then $II$ vanishes, but if it's a wave or a cylinder (never mind the global topology), then $II$ and at least one principal curvature are non-zero.
I find it difficult to imagine the same thing for spaces with non-vanishing intrinsic curvature. For example the sphere $S^2$, I can't imagine it being embedded in different ways in $\mathbb{R}^3$. Of course you can always move the sphere around, but I suspect any other operation (scaling it, skew it etc.) would affect the First Fundamental form.
Is that true? Or is it possible to alter the Second Fundamental form of embedded surfaces without changing its First Fundamental form? If it's possible, are there any examples?
EDIT: Intuitively spoken, a sheet of paper can be bent, because its curvature is zero. The bending will change $II$, but not $I$. But if you bend a sphere made of paper, I guess you will also bend its First Fundamental form. Right?
 A: Your intuition is correct for the sphere: Liebmann's theorem shows that any compact surface (without boundary) in $\mathbb R^3$ with constant positive Gaussian curvature is a standard round sphere. But for noncompact surfaces, there can be non-rigid deformations. For example, there's a family of surfaces of revolution with constant Gaussian curvature $+1$, so they are all locally isometric to the unit sphere; but they cannot be obtained from it (even locally) by a rigid motion because they have different mean curvatures. 
Here's a nice picture of two such surfaces (borrowed from these notes by Ilya Kapovich):

You can realize these physically by cutting a piece out of a rubber ball and pinching it.
A: First of all, there are lots of ways to embed the plane non-isometrically in $\Bbb R^3$. For example, take the graph of a general smooth function $f\colon\Bbb R^2\to\Bbb R$.
However, if you are embedding isometrically (so that the first fundamental form is the identity), then it follows that the curvature of the resulting surface must be $0$. This follows from Gauss's Theorema Egregium, which says that curvature can be computed just knowing the first fundamental form.
