 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/12/b/ii/solution

z \geq 0

and becomes

0 \leq ϑ \leq α

. The unit ball gives

0 \leq ρ \leq 1

. Hence

V = \int_{0}^{2 π} \int_{0}^{α} \int_{0}^{1} ρ^{2} sin ϑ d ρ d ϑ d φ = \frac{2 π}{3} (1 - cos α) .

For an independent calculation in cylindrical coordinates, the lower boundary is

z = r cot α

and the upper boundary is

z = 1 - r^{2}

. They meet at

r = sin α

, since

r^{2} cot^{2} α = 1 - r^{2}

. Thus

V = 2 π \int_{0}^{s i n α} r (1 - r^{2} - r cot α) d r = \frac{2 π}{3} [1 - cos^{3} α - sin^{3} α cot α] = \frac{2 π}{3} (1 - cos α) .

The equality uses

sin^{3} α cot α = sin^{2} α cos α

. The positive cosine on the stated range removes any square-root sign ambiguity.

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