 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/12/c/solution

Centre cylindrical coordinates on the right-hand sphere, with the axial coordinate

s = z - α a

positive outwards. Using

R = β a

and

d = (α - γ) a

, the retained solid satisfies

- d \leq s \leq R, 0 \leq r \leq R^{2} - s^{2}, 0 \leq ϕ < 2 π .

In particular, it contains more than a hemisphere; it is the small cap on the inward side that has been removed. The volume integral for its mass, including the Jacobian determinant

r

, is

M_{e} = ρ \int_{- d}^{R} \int_{0}^{2 π} \int_{0}^{R^{2} - s^{2}} r d r d ϕ d s = π ρ \int_{- d}^{R} (R^{2} - s^{2}) d s = π ρ (\frac{2 R ^{3}}{3} + R^{2} d - \frac{d ^{3}}{3}) .

Thus each end piece has mass

M_{e} = π ρ a^{3} [\frac{2 β ^{3}}{3} + β^{2} (α - γ) - \frac{( α - γ ) ^{3}}{3}] .

R + d

, multiplied by the uniform mass density.

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