 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/16/b/i/solution

Both patches project onto the unit disk

D

in the

x y

d S_{+} = (\frac{x}{ρ}, \frac{y}{ρ}, 1)^{T} d x d y, d S_{-} = (2 x, 2 y, - 1)^{T} d x d y .

The cone's apex has zero area and does not affect the surface integral. For

k = 0

(y, x, 1)^{T}

, so

Φ_{+} = \iint_{D} (\frac{2 x y}{ρ} + 1) d x d y = π, Φ_{-} = \iint_{D} (4 x y - 1) d x d y = - π .

The terms involving

x y

vanish by reflection symmetry of the disk. Thus the total outward flux is

Φ = 0 .

This calculation evaluates the two surface integrals directly.

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