= Solution
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>, $2\pi R^3/3$:
$$
\boxed{\int_{\partial V}\mathbf F\cdot d\mathbf S=3(\mathbf a\cdot\mathbf b)\frac{2\pi R^3}{3}
=2\pi R^3\mathbf a\cdot\mathbf b.}
$$
The constant field may have nonzero <flux> through the curved part alone, but its curved and base contributions cancel on the closed surface.
Back to article page
