 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/12/a/i/solution

z

r (z) = R e^{- z / H}

and angle

Θ (z) = 2 π - π z / H

. Its area is

Θ (z) r (z)^{2} /2

, so slicing by horizontal cross-sections gives the volume integral

V = π R^{2} \int_{0}^{H} (1 - \frac{z}{2 H}) e^{- 2 z / H} d z .

t = z / H

. Then

V = π R^{2} H \int_{0}^{1} (1 - t /2) e^{- 2 t} d t .

(t /4 - 3/8) e^{- 2 t}

, whose endpoint difference is

(3 - e^{- 2}) /8

. Therefore

V = \frac{π R ^{2} H}{8} (3 - e^{- 2}) .

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