 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-resit/question-3/a/ii/solution

Choose the outward unit normal vector on the upper hemisphere. The curl of

F = (4 y, x, 2 z)

\nabla \times F = (0, 0, 1 - 4) = (0, 0, - 3) .

s (φ, θ) = a (sin φ cos θ, sin φ sin θ, cos φ), 0 \leq φ \leq \frac{π}{2}, 0 \leq θ \leq 2 π .

Here

n_{z} = cos φ

and

d S = a^{2} sin φ d φ d θ

. The surface integral is therefore

\iint_{Σ} (\nabla \times F) \cdot n d S = - 3 a^{2} \int_{0}^{2 π} \int_{0}^{π /2} cos φ sin φ d φ d θ = - 3 π a^{2} .

The compatible boundary orientation follows increasing

t

r (t) = (a cos t, a sin t, 0)

0 \leq t \leq 2 π

. On this circle,

F (r (t)) = (4 a sin t, a cos t, 0), r^{'} (t) = (- a sin t, a cos t, 0) .

\oint_{\partial Σ} F \cdot d r = a^{2} \int_{0}^{2 π} (- 4 sin^{2} t + cos^{2} t) d t = - 3 π a^{2},

which agrees with the surface integral as required by the Stokes theorem.

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