 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/19/b/ii/solution

Using the same cylindrical-shell volume,

L = 2 πh \int_{0}^{a} r^{2} v (r) d r = 2 πh \int_{0}^{a} r^{3} ω (r) d r .

For unit density, this is the axial angular momentum. To bound it, apply the integral Cauchy-Schwarz inequality to

f (r) = r v (r)

and

g (r) = r^{3/2}

(\int_{0}^{a} r^{2} v (r) d r)^{2} \leq (\int_{0}^{a} r v (r)^{2} d r) (\int_{0}^{a} r^{3} d r) = \frac{K}{πh} \frac{a ^{4}}{4} .

(2 πh)^{2}

gives

L^{2} \leq πh a^{4} K

. Since the chamber's volume is

V = π a^{2} h

, this is precisely

K \geq \frac{L ^{2}}{V a ^{2}} .

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