= Flux of the position vector through a cone
{title2=$\int_C\mathbf x\cdot d\mathbf S=3V$}
Let a solid <cone> of <volume> $V$ have its planar base in a plane through the origin, and orient its curved surface $C$ outwards. The <vector field> $\mathbf F(\mathbf x)=\mathbf x$ has <divergence> three. Its <flux> through the base vanishes because the <position vector> lies in that plane and is <orthogonal> to its <normal vector>. The <divergence theorem> therefore gives the displayed curved-surface <flux>. For a circular base of radius $a$ and perpendicular height $b$, this is $\pi a^2b$.
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