 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/ii/solution

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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