Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 1 / c

First compute the curl directly:

\nabla \times F = (0, 1, 2 x) .

x

makes the last component integrate to zero. The middle component gives the hemisphere's volume, so

\int_{V} \nabla \times F d V = (0, \frac{2 π R ^{3}}{3}, 0) .

For an independent surface integral, the closed boundary consists of the curved hemisphere and its flat base. The right-hand side in part (b) is

\int_{S} n \times F d S

On the curved part use spherical coordinates with

0 \leq θ \leq π /2

and

0 \leq ϕ < 2 π

. There,

n F n \times F = (sin θ cos ϕ, sin θ sin ϕ, cos θ), = (R cos θ, R^{2} sin^{2} θ, 0), d S = R^{2} sin θ d θ d ϕ, = (- R^{2} cos θ sin^{2} θ, R cos^{2} θ, R^{2} sin^{3} θ cos ϕ - R sin θ cos θ sin ϕ) .

The last component has zero angular integral. The first two components give

\int_{S_{curved}} n \times F d S = (- 2 π R^{4} \int_{0}^{π /2} cos θ sin^{3} θ d θ, 2 π R^{3} \int_{0}^{π /2} cos^{2} θ sin θ d θ, 0) = (- \frac{π R ^{4}}{2}, \frac{2 π R ^{3}}{3}, 0) .

On the base,

z = 0

and the outward normal vector is

- e_{z}

. Thus

n \times F = (x^{2} + y^{2}, 0, 0)

\int_{S_{base}} n \times F d S = (2 π \int_{0}^{R} ρ^{3} d ρ, 0, 0) = (\frac{π R ^{4}}{2}, 0, 0) .

The first components cancel, and therefore both required sides equal

(0, \frac{2 π R ^{3}}{3}, 0), α = \frac{2 π R ^{3}}{3} .

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