 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/14b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

R = a ξ_{1}

M = 4 π \int_{0}^{R} ρ (r) r^{2} d r = 4 π a^{3} ρ_{c} \int_{0}^{ξ_{1}} θ^{n} ξ^{2} d ξ = 4 π a^{3} ρ_{c} [- ξ^{2} θ^{'} (ξ)]_{ξ = ξ_{1}} = 4 π a^{3} ρ_{c} Y_{n} .

Substituting the value of

a

M = \frac{1}{2 π} (\frac{( n + 1 ) K}{G})^{3/2} ρ_{c}^{(3 - n) / (2 n)} Y_{n}, Y_{n} = - ξ_{1}^{2} θ^{'} (ξ_{1}) .

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