 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/c/solution

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

Subtract the integrands before integrating:

G_{2} - G_{1} = (0, y^{3}, 0)

(G_{2} - G_{1}) \cdot d S = (x^{2} y + 2 y^{2} z) d x d y = (2 - 2 x^{2} - x^{2} y) d x d y .

This extends continuously to the lower edge, so

∣ I_{1} - I_{2} ∣ = \int_{0}^{1} \int_{0}^{2} (2 - 2 x^{2} - x^{2} y) d y d x = 2.

The individual flux integrals also converge:

G_{1} \cdot d S = x (3 + 2 y) d x d y

gives

I_{1} = 5

, and hence

I_{2} = 7

. Reversing the surface orientation changes both signs and leaves the requested magnitude unchanged.

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