 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/12/b/solution

r d z d r d θ

. Integrating from the base to the curved surface gives

V = \int_{0}^{α} d θ \int_{0}^{R} r d r \int_{0}^{a (1 - r^{2} / b^{2})} d z = α a [\frac{r ^{2}}{2} - \frac{r ^{4}}{4 b ^{2}}]_{0}^{R} .

Hence

V = \frac{α a b ^{2}}{4} .

For a complete rotation this is half the volume of the enclosing cylinder.

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