= Solution
Let $a$ be any constant <vector> and take $u=\phi a$. The <product rule for divergence> gives $\nabla\cdot u=a\cdot\nabla\phi$. The <divergence theorem> on a bounded region with piecewise smooth boundary, oriented outward, therefore yields
$$
a\cdot\int_V\nabla\phi\,dV=a\cdot\int_S\phi\,d\mathbf S.
$$
Since $a$ is arbitrary, equality of all components proves
$$
\boxed{\int_V\nabla\phi\,dV=\int_S\phi\,d\mathbf S.}
$$
Here $\phi$ is <continuously differentiable> on a neighbourhood of the closed region.
For the side of this <right circular cone>, the parameter tangents are $x_r=(\cos\theta,\sin\theta,\sqrt3)$ and $x_\theta=(-r\sin\theta,r\cos\theta,0)$. Their <cross product> in the outward order is
$$
\boxed{d\mathbf S=(x_\theta\times x_r)\,dr\,d\theta
=(\sqrt3\cos\theta,\sqrt3\sin\theta,-1)r\,dr\,d\theta.}
$$
The sign is outward because the solid <right circular cone> lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the <oriented surface element>.
To check the integral identity, the closed boundary must include the top <Euclidean disk> $z=1$, radius $1/\sqrt3$, as well as the curved side. For $\phi=z^2$, horizontal components cancel on integrating $\theta$. The side contribution is
$$
\int_{\rm side}\phi\,d\mathbf S
=-2\pi\int_0^{1/\sqrt3}3r^3\,dr\,e_z
=-\frac\pi6e_z.
$$
On the top <Euclidean disk> $\phi=1$ and $d\mathbf S=e_z\,dA$, so its contribution is $(\pi/3)e_z$. The total is $(\pi/6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla\phi=2ze_z$, giving
$$
\int_V\nabla\phi\,dV
=\int_0^1 2z\frac{\pi z^2}{3}\,dz\,e_z
=\boxed{\frac\pi6e_z}.
$$
Thus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\varepsilon$ and let $\varepsilon\downarrow0$, with the extra boundary contribution vanishing.
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