Orientation reversal of surface curvature Created 2026-09-24 Updated 2026-09-24
Replacing the chosen unit normal by replaces the shape operator by . On a surface, Gaussian curvature is unchanged, while mean curvature changes sign.
Past exam of the mathematics course of the University of Cambridge 2005 ib Paper 4 12A Solution Created 2026-09-24 Updated 2026-09-24
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At a regular point, choose the unit normal . The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, , independent of the normal orientation.
For the given graph, and , so with ,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 3 2E Solution Created 2026-09-24 Updated 2026-09-25
For a geodesic triangle with interior angles , the local Gauss-Bonnet theorem isbecause its geodesic sides have zero geodesic curvature.
Triangulate a closed oriented surface into geodesic triangles, with edges and vertices. Summing the local formula, the angles around each vertex total , soSince every triangular face has three edges and every edge belongs to two faces, . Hencewhich is the global Gauss-Bonnet theorem.
For the sphere , the unit normal is . Its shape operator is, up to the conventional sign, on each tangent plane. Both principal curvatures therefore have magnitude , and the Gaussian curvature is
An octant has one eighth of the sphere's area:Thus . Its three great-circle sides meet at three right angles, soThe two sides of the local Gauss-Bonnet formula agree directly.
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 26F c Solution Created 2026-09-24 Updated 2026-09-25
For a chosen unit normal field , the second fundamental form isIn a local parametrization , its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.