 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/19/b/solution

Choose the upward orientation. The boundary is the circle

x^{2} + y^{2} = 4

z = 1

, oriented counterclockwise from above. By Stokes theorem, the surface integral is

\oint_{\partial S} G \cdot d r

. On this boundary

d z = 0

and

z^{2} = 1

, so

\oint_{\partial S} G \cdot d r = \oint (d x - 3 x y d y) .

The first term integrates to zero around a closed curve. By Green's theorem, the second equals the disk integral

\int_{x^{2} + y^{2} \leq 4} (- 3 y) d x d y = 0

, by oddness in

y

. Thus

\int_{S} (\nabla \times G) \cdot n d S = 0.

Here reversing orientation leaves the zero value unchanged.

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