 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/16c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\dot{M} = ρ \int_{0}^{2 π} \int_{0}^{R} w (r) r d r d θ = - \frac{ρ π R ^{4}}{8 μ} \frac{\partial p}{\partial z} .

Thus

\dot{M} = - \frac{ρ π R ^{4}}{8 μ} \frac{\partial p}{\partial z} .

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