 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-resit/8/d/solution

The closed boundary includes the curved unit hemisphere and the disk at

y = 0

. On the disk,

F = (0, y, 0) = 0

, so that contribution vanishes. On the curved part the outward normal is

n = (x, y, z)

F \times n = (yz, 0, - x y)

Use spherical coordinates with the polar axis along positive

y

x = sin θ cos ϕ, y = cos θ, z = sin θ sin ϕ, 0 \leq θ \leq \frac{π}{2}, 0 \leq ϕ \leq 2 π .

The area element is

sin θ d θ d ϕ

. Consequently

\iint_{S} F \times d S = \int_{0}^{π /2} cos θ sin^{2} θ d θ \int_{0}^{2 π} (sin ϕ, 0, - cos ϕ) d ϕ = (0, 0, 0) .

This directly performs the surface integral and includes the planar cap, rather than applying the closed-surface theorem to only an open hemisphere.

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