 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/6/c/solution

Use the vector triple product expression from part (a). Put

s = Ω \cdot x

and

r^{2} = x \cdot x

. Since

u = Ω + β x

u \cdot u = Ω^{2} + 2 β s + β^{2} r^{2}, u \cdot Ω = Ω^{2} + β s .

Expanding

W = (u \cdot u) Ω - (u \cdot Ω) u

and canceling the zeroth-order terms gives

W (x) = β [(Ω \cdot x) Ω - Ω^{2} x] + β^{2} [r^{2} Ω - (Ω \cdot x) x] .

The first bracket is odd in

x

and integrates to zero over the centred ball. With

α = 4 π a^{5} /15

\int_{V} r^{2} d V = 3 α, \int_{V} (Ω \cdot x) x_{i} d V = Ω_{j} \int_{V} x_{j} x_{i} d V = α Ω_{i} .

Therefore

\int_{V} W_{i} d V = 2 α β^{2} Ω_{i} = \frac{8 π a ^{5} β ^{2}}{15} Ω_{i} .

The resulting vector is parallel to

Ω

, as the rotational symmetry of the centred domain requires.

 New to topics? Read the docs here!