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The divergence theorem states that a continuously differentiable vector field on a bounded region $V$ satisfieswith outward unit normal $\mathbf{n}$ and a sufficiently regular boundary; piecewise smooth boundaries are also allowed. For $\mathbf{u}=\mathbf{c}\Omega$ with constant $\mathbf{c}$, the left side is $\mathbf{c}\cdot {\int }_V\nabla \Omega dV$ and the right side is $\mathbf{c}\cdot {\int }_{\partial V}\Omega \mathbf{n}dS$. Equality for every constant vector gives the gradient volume-to-boundary identityFor the cone and its cap, the volume has $0\le z\le 1$ and $0\le r\le \sqrt{3}z$ in cylindrical coordinates. Its volume is ${\int }_0^1\pi (\sqrt{3}z{)}^{2}dz=\pi$. Since $\nabla \Omega =-{\mathbf{e}}_z$,The cap at $z=1$ has area $3\pi$, outward normal ${\mathbf{e}}_z$ and $\Omega =a-1$, so its contribution to the surface integral is $3\pi (a-1){\mathbf{e}}_z$.

Parameterize the side by $\mathbf{R}(r,\theta )=(r\cos \theta ,r\sin \theta ,r/\sqrt{3})$, $0\le r\le \sqrt{3}$, $0\le \theta <2\pi$. The outward vector area element ispointing radially outward and downward from the region above the cone. The radial part integrates to zero around the circle. The vertical part of the side integral isAdding cap and side givesThe apex and circular rim have zero surface area. Alternatively, removing a tiny apex neighborhood and taking its radius to zero justifies applying the smooth-boundary theorem there; the additional flux vanishes because the scalar field is bounded.