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

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.

ii

For

k = 1

(x^{2} y, x y^{2}, z)^{T}

. On the cone,

F \cdot d S_{+} = (\frac{x ^{3} y + x y ^{3}}{ρ} + 1 - ρ) d x d y = (x y ρ + 1 - ρ) d x d y .

The first term has zero integral by symmetry. Using polar coordinates for the remaining double integral gives

Φ_{+} = 2 π \int_{0}^{1} (1 - ρ) ρ d ρ = \frac{π}{3} .

z = ρ^{2} - 1

F \cdot d S_{-} = (2 x^{3} y + 2 x y^{3} - (ρ^{2} - 1)) d x d y = (2 x y ρ^{2} + 1 - ρ^{2}) d x d y .

Again the

x y

term cancels by symmetry, giving

Φ_{-} = 2 π \int_{0}^{1} (1 - ρ^{2}) ρ d ρ = \frac{π}{2} .

Therefore direct surface integral yields

Φ = \frac{π}{3} + \frac{π}{2} = \frac{5 π}{6} .

No separate disk at

z = 0

is included: that rim joins the two surfaces and is not an additional face of the closed boundary.

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