 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/18/c/solution

x = r cos θ

y = r sin θ

. The upper plane gives

z = 4 - r (cos θ + sin θ) > 0

throughout the quarter disk, and the Jacobian determinant contributes

r

. Therefore

V = \int_{0}^{π /2} \int_{0}^{1} [4 - r (cos θ + sin θ)] r d r d θ = \int_{0}^{π /2} [2 - \frac{cos θ + sin θ}{3}] d θ,

V = π - \frac{2}{3} .

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