 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/b/ii/solution

The horizontal cross-section is a triangle, so its area is half its width times its depth:

A (z) = L D (1 - \frac{z}{H}) 1 - \frac{z}{2 H} .

Thus the volume integral, with

t = z / H

, is

V = L DH \int_{0}^{1} (1 - t) 1 - t /2 d t .

u = 1 - t /2

. The endpoint values are

1

and

1/2

, and

1 - t = 2 u - 1

\frac{V}{L DH} = 2 \int_{1/2}^{1} (2 u - 1) u^{1/2} d u = 2 [\frac{4}{5} u^{5/2} - \frac{2}{3} u^{3/2}]_{1/2}^{1} = \frac{4 + 2 2}{15} .

Therefore

V = \frac{( 4 + 2 2 ) L DH}{15}

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