= Solution
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,
$$
\int_V\nabla\cdot\mathbf u\,dV=\int_{\partial V}\mathbf u\cdot\mathbf n\,dS.
$$
For a differentiable <homogeneous function>, differentiate $f(k\mathbf x)=k^nf(\mathbf x)$ at $k=1$ to get <Euler theorem for homogeneous functions>:
$$
\boxed{\mathbf x\cdot\nabla f=nf.}
$$
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 therefore
$$
\boxed{\int_Vf\,dV=\frac1{n+3}\int_{\partial V}f\,\mathbf x\cdot\mathbf n\,dS\quad(n\ne-3).}
$$
The 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, $0\leq z\leq\alpha$ and $0\leq r\leq z/\alpha$. The polynomial $f=z^4+\alpha^4r^4$ has degree four. Its direct volume integral is
$$
\begin{aligned}
\int_V f\,dV
&=2\pi\int_0^\alpha\int_0^{z/\alpha}(z^4+\alpha^4r^4)r\,dr\,dz\\
&=2\pi\int_0^\alpha\left(\frac{z^6}{2\alpha^2}+\frac{z^6}{6\alpha^2}\right)dz
=\boxed{\frac{4\pi\alpha^5}{21}}.
\end{aligned}
$$
On 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 $z=\alpha$, $0\leq r\leq1$, with outward normal $\mathbf e_z$, so
$$
\int_{\partial V}f\,\mathbf x\cdot d\mathbf A
=2\pi\alpha^5\int_0^1(1+r^4)r\,dr
=\frac{4\pi\alpha^5}{3}.
$$
Multiplying 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.
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