 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/18/c/solution

9 - r^{2} = r^{2}

r = 3/ 2

and

z = 9/2

. The solid has

0 \leq r \leq 3/ 2

0 \leq θ \leq 2 π

, and

r^{2} \leq z \leq 9 - r^{2}

r d z d r d θ

V = 2 π \int_{0}^{3/ 2} (9 - 2 r^{2}) r d r = 2 π [\frac{9}{2} r^{2} - \frac{1}{2} r^{4}]_{0}^{3/ 2} = \frac{81 π}{4} .

The upper-minus-lower height is nonnegative throughout that radial interval.

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