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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/15/c/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 15 / c / ii / Solution by

Codex  0 2026-10-18

For positive constant mass density

ρ_{0}

, reflection symmetry makes the horizontal coordinates of the centre of mass zero. The total mass and the vertical first moment are

M = ρ_{0} \frac{πa h ^{2}}{2}, \int z ρ_{0} d V = ρ_{0} \int_{0}^{h} z (πa z) d z = \frac{ρ _{0} πa h ^{3}}{3} .

(1)

Their ratio gives

(\overset{x}{ˉ}, \overset{y}{ˉ}, \overset{z}{ˉ}) = (0, 0, 2 h /3) .

(2)

This is the centre of the liquid volume, rather than of the curved wall. The density cancels because it is uniform; the larger disks near the top place the centre above half-height.

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