 Flux of the position vector through a cone

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 $\pi a^{2}b$.