 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/1/c/solution

J = det \frac{\partial ( x , y , z )}{\partial ( u , v , ϕ )} = (h_{u} \times h_{v}) \cdot h_{ϕ} .

Here

h_{u} \times h_{v} = (u^{2} + v^{2}) (- sin ϕ, cos ϕ, 0)

, so

J = uv (u^{2} + v^{2}), d V = uv (u^{2} + v^{2}) d u d v d ϕ .

The sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals

h_{u} h_{v} h_{ϕ}

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