Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 2 / a

Solution

Let

p = K ρ^{1 + 1/ m}

\frac{d p}{d z} = - ρ g .

Substitution and integration with

ρ = 0

z = 0

gives

ρ = ρ_{0} (\frac{- z}{H})^{m}, z \leq 0,

where

H

fixes the normalization. Integrating the hydrostatic equation from

z

to the free surface gives the pressure directly:

p (z) = g \int_{z}^{0} ρ (z^{'}) d z^{'} = \frac{ρ _{0} g H}{m + 1} (\frac{- z}{H})^{m + 1} .

This indeed has

p \propto ρ^{1 + 1/ m}

and

p (0) = 0

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