 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016/10/b/i/solution

Use the cone parametrization

r (x, θ) = (x, x cos θ, x sin θ)

for

2 \leq x \leq 3

0 \leq θ \leq 2 π

2 x d x d θ

. Since

x^{2} y^{2} = x^{4} cos^{2} θ

\iint_{S} x^{2} y^{2} d S = 2 \int_{2}^{3} x^{5} d x \int_{0}^{2 π} cos^{2} θ d θ = \frac{665 π 2}{6} .

This uses only the curved conical surface. There are no closing discs in the specified integration surface.

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