By a suitable choice of $\mathbf{u}$ in the divergence theorem

$$\int_{V} \nabla \cdot \mathbf{u} d V=\int_{S} \mathbf{u} \cdot d \mathbf{S}$$

show that

$$\int_{V} \nabla \phi d V=\int_{S} \phi d \mathbf{S}$$

for any continuously differentiable function $\phi$.

For the curved surface of the cone

$$\mathbf{x}=(r \cos \theta, r \sin \theta, \sqrt{3} r), \quad 0 \leqslant \sqrt{3} r \leqslant 1, \quad 0 \leqslant \theta \leqslant 2 \pi$$

show that $d \mathbf{S}=(\sqrt{3} \cos \theta, \sqrt{3} \sin \theta,-1) r d r d \theta$.

Verify that $(*)$ holds for this cone and $\phi(x, y, z)=z^{2}$.