 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/12/b/solution

x = r cos θ

y = r sin θ

d V = r d r d θ d z .

- d < z < 0

and

0 \leq r < a 1 + z / d

. Thus the volume is

V = \int_{- d}^{0} \int_{0}^{2 π} \int_{0}^{a 1 + z / d} r d r d θ d z = π a^{2} \int_{- d}^{0} (1 + \frac{z}{d}) d z = \frac{1}{2} π a^{2} d .

Equivalently, integrating vertically first gives

V = \int_{0}^{2 π} \int_{0}^{a} \int_{d (r^{2} / a^{2} - 1)}^{0} r d z d r d θ

. The region is bounded by an elliptic paraboloid and its top disc; its volume is half that of the cylinder with base radius

a

d

. Including or excluding the boundary surfaces does not change the volume.

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