Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 1 / d

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- Solution d

Solution

Since

F = \nablaΩ

, its divergence is the Laplacian of

Ω

:

\nabla \cdot F = \nabla^{2} Ω = 6 x + 4 z .

The divergence theorem changes the outward flux through the whole closed boundary into

I = \int_{V} (6 x + 4 z) d V

. Rotational symmetry makes the

x

term vanish. With

z = (u^{2} - v^{2}) /2

and the Jacobian determinant from part (c),

I = 2 \int_{- π}^{π} \int_{0}^{1} \int_{0}^{v} (u^{2} - v^{2}) uv (u^{2} + v^{2}) d u d v d ϕ = 4 π \int_{0}^{1} v [\frac{u ^{6}}{6} - \frac{v ^{4} u ^{2}}{2}]_{u = 0}^{u = v} d v = - \frac{4 π}{3} \int_{0}^{1} v^{7} d v = - \frac{π}{6} .

Both the curved face and the top disk are included in this surface integral through the divergence theorem.

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