Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 1 / c

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- Solution c

Solution

Close the hyperboloid segment with the two disks at

z = \pm 1

, and let

V

be its interior. The flux through either disk is zero because

A_{z} = 0

. Thus the desired surface integral is the total outward flux through the closed boundary. Since

\nabla \cdot A = 3 x^{2} + 3 y^{2},

the divergence theorem and cylindrical coordinates give

\int_{S} A \cdot \hat{n} d S = \int_{- 1}^{1} \int_{0}^{2 π} \int_{0}^{1 + z^{2}} 3 r^{2} r d r d θ d z = \frac{3 π}{2} \int_{- 1}^{1} (1 + z^{2})^{2} d z = \frac{3 π}{2} (2 + \frac{4}{3} + \frac{2}{5}) = \frac{28 π}{5} .

The extra factor

r

is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

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