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Problem 6 The First Fundamental Form of a Surface Let & : D CR' - D(D) = S be a parametrization of a surface S; in coordinates, we express as usual D(u, v) = (x(u, v), y(u, v), z(u, v)). Let adll E( u, V ) = F(u, v) = - du av' G(u, v) = av The matrix I ( u, v ) = (E(u, v) F(u, v) F(u, v) G(u, v) is known as the first fundamental form of the surface S. Show that | |Tu x Tull = Vdet I(u, v), where as usual, Tu = OQ/Ou, To = 06/Ov. That is, show that the "Jacobian" of the surface area element is equal to the square root of the determinant of the first fundamental form. Hint: Use Lagrange's identity for computing the magnitude of a cross product, lla x bll = V llall2115/12 - (a . 5)2, choosing a to be Tu and b to be To. Remark. This problem shows that the first fundamental form encodes the "Jacobian" of the surface area element dS and hence, encodes information about the area of the surface S. In fact, the first fundamental form encodes all of the intrinsic geometric properties of the surface S, such as length, area, and curvature. In the field of differential geometry, this is known as the "metric tensor" of the surface