Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 4

For a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, oriented by the outward unit normal vector, and a continuously differentiable vector field on a neighborhood of $\overline{V}$, the divergence theorem statesFor an oriented piecewise smooth surface $S$ with piecewise smooth boundary $C=\partial S$, and a continuously differentiable vector field on a neighborhood of $S$, Stokes theorem statesThe boundary orientation follows the right-hand rule relative to the chosen normal vector. Reversing that orientation reverses both sides of the latter identity.

The divergence is $\nabla \cdot F=2xyz+2xyz+2xyz=6xyz$. With outward orientation, the divergence theorem converts the surface flux into a volume integral:Directly, the three faces lying in coordinate planes contribute zero, and each opposite face contributes $1/4$, confirming the sign and normalization.

Take the outward orientation on the upper hemisphere; its boundary circle is then counterclockwise viewed from above. Parametrize that circle by $r(\varphi )=(r\cos \varphi ,r\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Then $G\cdot dr=r^2(3{\cos }^2\varphi -4{\sin }^2\varphi )d\varphi$, so Stokes theorem givesEquivalently $\nabla \times G=(0,0,-1)$, and the upward projected area of the hemisphere is $\pi r^2$. The opposite orientation gives the negative of this answer.