Discussion
The right cylinder over the circle \(x^{2} + y^{2} = 1\) has the parametrization \(\bar{x} : U \to \mathbf{R}^{3}\), where :
\[ \begin{aligned} U = \{&(u, v) \in \mathbf{R}^{2} \mid 0 \lt u \lt 2\pi, \\ &-\infty \lt v \lt \infty\}, \end{aligned} \]
\(\bar{x}(u, v) = (\cos u, \sin u, v)\)
Then the coefficients E, F, G in the first fundamental form of the cylinder are given by :
Answer: B
The coefficients are E = 1, F = 0, G = 1.
For \(\bar{x}(u, v) = (\cos u, \sin u, v)\): \(\bar{x}_{u} = (-\sin u, \cos u, 0)\) and \(\bar{x}_{v} = (0, 0, 1)\). So
\[ \begin{aligned} E &= \bar{x}_{u} \cdot \bar{x}_{u} = \sin^{2}u + \cos^{2}u = 1 \\ F &= \bar{x}_{u} \cdot \bar{x}_{v} = 0 \\ G &= \bar{x}_{v} \cdot \bar{x}_{v} = 1 \end{aligned} \]
This is why the cylinder is locally isometric to the plane: its first fundamental form is \(du^{2} + dv^{2}\). \(\sin^{2}u\) and \(\cos u\) are single components of \(\bar{x}_{u}\), not its squared length, and F = 1 would mean the coordinate curves are not perpendicular.
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