Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 19

The divergence is $\nabla \cdot F=-1+0+1=0$. Take the hemisphere's outward orientation, and close it with the unit disk $D$ at $z=0$, whose outward normal is $-e_z$. The divergence theorem then gives ${\int }_{S}F\cdot dS+{\int }_{D}F\cdot dS=0$. On the disk, $F\cdot (-e_z)=-x^2$, and polar coordinates giveConsequently the curved hemisphere has fluxThe reversed hemisphere orientation would reverse the sign.

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.