 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/12/b/solution

The PDF diagram includes a solid base of thickness

d

, occupying

- d \leq z \leq 0

and

0 \leq r \leq R

. Above it,

0 \leq z \leq h

, the metal lies between the two specified curved surfaces. In cylindrical coordinates, the Jacobian determinant contributes a factor

r

, so the mass is

M = ρ \int_{- d}^{0} \int_{0}^{2 π} \int_{0}^{R} r d r d ϕ d z + ρ \int_{0}^{h} \int_{0}^{2 π} \int_{(9/10) R (1 + z^{2} / h^{2})}^{R (1 + z^{2} / h^{2})} r d r d ϕ d z .

The base contributes

π ρ R^{2} d

. The wall contributes

π ρ R^{2} (1 - \frac{81}{100}) \int_{0}^{h} (1 + \frac{z ^{2}}{h ^{2}})^{2} d z = π ρ R^{2} \frac{19}{100} h (1 + \frac{2}{3} + \frac{1}{5}) = \frac{133}{375} π ρ R^{2} h .

Consequently

M = π ρ R^{2} (d + \frac{133 h}{375}) .

The base thickness

d

comes from the diagram; omitting it would omit part of the cup's volume.

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