 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/12/a/solution

The height is constant along each circular ring, whose area element after angular integration is

2 π r d r

V = 2 π \int_{0}^{R} r R e^{- r / R} d r = 2 π R^{3} \int_{0}^{1} u e^{- u} d u .

Here

u = r / R

\int_{0}^{1} u e^{- u} d u = [- (u + 1) e^{- u}]_{0}^{1} = 1 - 2/ e

. Hence

R = [\frac{V}{2 π ( 1 - 2/ e )}]^{1/3} .

The dimensionless coefficient is positive, and the cube root of a volume has the required physical dimension of length.

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