= Solution
The <divergence theorem> states that a continuously differentiable <vector field> on a bounded region $V$ satisfies
$$
\int_V\nabla\cdot\mathbf u\,dV=\int_{\partial V}\mathbf u\cdot\mathbf n\,dS,
$$
with 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 identity>
$$
\boxed{\int_V\nabla\Omega\,dV=\int_{\partial V}\Omega\,d\mathbf S.}
$$
For the cone and its cap, the volume has $0\le z\le1$ and $0\le r\le\sqrt3z$ in <cylindrical coordinates>. Its volume is $\int_0^1\pi(\sqrt3z)^2\,dz=\pi$. Since $\nabla\Omega=-\mathbf e_z$,
$$
\int_V\nabla\Omega\,dV=-\pi\mathbf e_z.
$$
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 $\mathbf R(r,\theta)=(r\cos\theta,r\sin\theta,r/\sqrt3)$, $0\le r\le\sqrt3$, $0\le\theta<2\pi$. The outward <vector area element> is
$$
d\mathbf S=(\mathbf R_\theta\times\mathbf R_r)\,dr\,d\theta
=\left(\frac r{\sqrt3}\mathbf e_r-r\mathbf e_z\right)dr\,d\theta,
$$
pointing 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 is
$$
-2\pi\int_0^{\sqrt3}\left(a-\frac r{\sqrt3}\right)r\,dr\,\mathbf e_z
=(-3\pi a+2\pi)\mathbf e_z.
$$
Adding cap and side gives
$$
\boxed{\int_{\partial V}(a-z)\,d\mathbf S
=\{3\pi(a-1)-3\pi a+2\pi\}\mathbf e_z
=-\pi\mathbf e_z
=\int_V\nabla(a-z)\,dV.}
$$
The 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.
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