= Solution
Only the quartic term contributes to the <Laplacian>. The mixed cubic has zero second <derivative> in each coordinate, while the exponential–trigonometric pairs are harmonic:
$$
\Delta(z^4)=12z^2,\quad\Delta(xyz)=0,\quad\Delta(e^{-3y}\cos3x)=0,\quad\Delta(e^{2x}\sin2y)=0.
$$
Using $z=r\cos\alpha$ and the <Jacobian determinant> from part (a), the <divergence theorem> gives
$$
\begin{aligned}
I&=12\int_0^{2\pi}\int_{-\pi}^{\pi}\int_0^R r^3\cos^2\alpha(1+r\sin\alpha)\,dr\,d\alpha\,d\beta\\
&=12(2\pi)\left(\int_{-\pi}^{\pi}\cos^2\alpha\,d\alpha\right)\frac{R^4}{4}=\boxed{6\pi^2R^4}.
\end{aligned}
$$
The term containing $\sin\alpha\cos^2\alpha$ integrates to zero. Equivalently, the <second vertical moment of a solid ring torus> with major radius one is $\int_Vz^2dV=\pi^2R^4/2$.
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