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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/18/c/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 18 / c / Solution by

Codex  0 2026-10-18

The projection onto the

x

y

plane is the disk

x^{2} + y^{2} \leq 4

. In polar coordinates, the height is

4 - r^{2}

and the polar area element is

r d r d θ

. The volume is therefore

V = \int_{0}^{2 π} \int_{0}^{2} (4 - r^{2}) r d r d θ = 2 π [2 r^{2} - \frac{r ^{4}}{4}]_{0}^{2} = 8 π .

(1)

The radial factor here is required because this integral is being constructed from the geometric volume, unlike the already specified integral in (a).

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