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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/2022/paper-2/16/a/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 / 2022 / Paper 2 / 16 / a / Solution by

Codex  0 2026-10-18

Use cylindrical radius

ρ = x^{2} + y^{2}

. The upper cone is

z = 1 - ρ

and the lower elliptic paraboloid is

z = ρ^{2} - 1

, both with

0 \leq ρ \leq 1

. They meet along the unit circle in

z = 0

. The enclosed solid is

0 \leq ρ \leq 1, ρ^{2} - 1 \leq z \leq 1 - ρ .

Thus the coordinate ranges on the complete surface are

- 1 \leq x \leq 1, - 1 \leq y \leq 1, - 1 \leq z \leq 1 .

These separate ranges do not mean every point of their Cartesian product lies on the surface. A vertical cross-section has a straight upper cone profile and a curved lower paraboloid profile.

Figure 1.
A cone and a paraboloid joined along their unit-circle rim
. The upper cone has apex $(0, 0, 1)$ ; the lower elliptic paraboloid has vertex $(0, 0, - 1)$ . Both meet at cylindrical radius one and height zero. The meridian cross-section shows the enclosed vertical range at each radius. Details.

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