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The divergence theorem states that for a continuously differentiable vector field on a bounded region with outward unit normal $\mathbf{n}$ and suitably regular, possibly piecewise smooth, boundary,For a differentiable homogeneous function, differentiate $f(k\mathbf{x})=k^{n}f(\mathbf{x})$ at $k=1$ to get Euler theorem for homogeneous functions:Only positive $k$ near one is needed for this derivative, so no interpretation of noninteger powers at negative $k$ is required. Applying the product rule gives $\nabla \cdot (f\mathbf{x})=3f+\mathbf{x}\cdot \nabla f=(n+3)f$. The volume integral of a homogeneous function is thereforeThe condition $n\ne -3$ is needed for division; at that degree the divergence theorem instead gives only zero net boundary flux. The field $f\mathbf{x}$ must also be regular on the region, or singularities must be treated by excision and limiting boundary terms. Neither issue arises for the polynomial in the final calculation.

Write $r=\sqrt{x^2+y^2}$. In the cone, $0\le z\le \alpha$ and $0\le r\le z/\alpha$. The polynomial $f=z^4+{\alpha }^4{r}^4$ has degree four. Its direct volume integral isOn the lateral cone the outward normal is proportional to $\alpha {\mathbf{e}}_r-{\mathbf{e}}_z$, so $\mathbf{x}\cdot \mathbf{n}$ is proportional to $\alpha r-z=0$. Thus the lateral contribution vanishes. The cap is $z=\alpha$, $0\le r\le 1$, with outward normal ${\mathbf{e}}_z$, soMultiplying by $1/(n+3)=1/7$ reproduces $4\pi {\alpha }^5/21$, verifying the formula. The cap edge and apex have zero surface measure, and the piecewise smooth version of the divergence theorem applies.