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 $x^2+y^2=4$ at $z=1$, oriented counterclockwise from above. By Stokes theorem, the surface integral is ${\oint }_{\partial S}G\cdot dr$. On this boundary $dz=0$ and $z^2=1$, soThe first term integrates to zero around a closed curve. By Green's theorem, the second equals the disk integral ${\int }_{x^2+y^2\le 4}(-3y)dxdy=0$, by oddness in $y$. ThusHere reversing orientation leaves the zero value unchanged.