 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/15/b/solution

For

h > 0

, parametrize the curved surface by

(r cos θ, r sin θ, r^{2} / a)

, with

0 \leq r \leq ah

and

0 \leq θ \leq 2 π

. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:

d S = r 1 + \frac{4 r ^{2}}{a ^{2}} d r d θ .

Integrating gives

S = 2 π \int_{0}^{ah} r 1 + 4 r^{2} / a^{2} d r = \frac{π a ^{2}}{6} [(1 + 4 r^{2} / a^{2})^{3/2}]_{0}^{ah} = \frac{π a ^{2}}{6} [(1 + \frac{4 h}{a})^{3/2} - 1] .

This is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at

h = 0

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