Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 16 / b

Use the orientation induced by the boundary order $O\to A\to B\to O$. The triangle's vector area element integrates to $\frac{1}{2}\mathbf{a}\times \mathbf{b}$. Every point $\mathbf{r}$ in its plane is a linear combination of $\mathbf{a},\mathbf{b}$, so $\mathbf{r}\cdot (\mathbf{a}\times \mathbf{b})=0$. The radial term of the field therefore contributes no flux. The constant term givesReversing the orientation reverses this sign. If $\mathbf{a}$ and $\mathbf{b}$ are collinear, the triangle has zero area and the displayed value is zero by the degenerate-area convention.

Take the outward orientation of the complete closed surface, including its flat base. The constant part of the vector field has zero divergence, and the radial part has divergence $3\mathbf{a}\cdot \mathbf{b}$. The divergence theorem gives its flux as that constant times the solid hemisphere's volume, $2\pi R^3/3$:The constant field may have nonzero flux through the curved part alone, but its curved and base contributions cancel on the closed surface.