 Solution

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/b/solution

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

Choose

f = x^{2} (1 + y) + y^{2} z - 1

. Since

f_{z} = y^{2} > 0

on the stated open domain, its upward vector area element is

d S = (\frac{2 x ( 1 + y )}{y ^{2}}, \frac{x ^{2} + 2 yz}{y ^{2}}, 1) d x d y, z = \frac{1 - x ^{2} ( 1 + y )}{y ^{2}} .

The graph becomes unbounded near parts of

y = 0

; the flux integrals below are understood as limits from

y > 0

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