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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/4/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 4 / a / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Where

f_{z} \neq = 0

, the implicit function theorem makes the surface a local graph

z = h (x, y)

with

h_{x} = - f_{x} / f_{z}

and

h_{y} = - f_{y} / f_{z}

. Its upward vector area element is

(- h_{x}, - h_{y}, 1) d x d y

. Therefore the vector area element of an implicit surface is

d S = \frac{\nabla f}{f _{z}} d x d y, F = \frac{\nabla f}{f _{z}}, F \cdot k = 1.

Dividing by

f_{z}

fixes the upward orientation even if the initially chosen gradient points downward. The restriction excludes a locally vertical graph, where this particular projected-area representation cannot be used.

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