= Solution
Use the orientation induced by the boundary order $O\to A\to B\to O$. The triangle's <vector area element> integrates to $\tfrac12\mathbf a\times\mathbf b$. 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 gives
$$
\boxed{\int_{OAB}\mathbf F\cdot d\mathbf S
=(\mathbf a\times\mathbf b)\cdot\frac{\mathbf a\times\mathbf b}{2}
=\frac12|\mathbf a\times\mathbf b|^2.}
$$
Reversing 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.
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