A Holy Moment

For simplicity, rotate the problem so that the solid is defined by the inequalities $x^2 + y^2 + z^2 \le R^2 \hspace{2cm} x^2 + y^2 \ge \tfrac14R^2$ and let us find the moment of inertia about the $x$ -axis. Using cylindrical polar coordinates $(r,\theta,z)$ , the solid is defined by the inequalities $\tfrac12R \le r \le R \hspace{1cm} 0 \le \theta \le 2\pi \hspace{1cm} |z| \le \sqrt{R^2-r^2}$ and so, assuming constant density $\rho$ , the mass of the solid is $M \; = \; \int_0^{2\pi} \,d\theta \int_{\frac12R}^R\,dr \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}r\rho \,dz \; = \; 4\pi\rho\int_{-\frac12R}^R r\sqrt{R^2-r^2}\,dr \; = \; 4\pi\rho\Big[-\tfrac13(R^2-r^2)^{\frac32}\Big]_{\frac12R}^R \; = \; \tfrac12\pi\rho\sqrt{3} R^3$ On the other hand, the moment of inertia about the $x$ -axis is $\begin{aligned} I & = \; \int_0^{2\pi} \,d\theta \int_{\frac12R}^R\,dr \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}\rho r(y^2 + z^2)\,dz \; = \; \rho\int_0^{2\pi} \,d\theta \int_{\frac12R}^R\,dr \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}(r^3\sin^2\theta + rz^2)\,dz \\ & = \; \pi\rho \int_{\frac12R}^R\,dr \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}(r^3 + 2rz^2)\,dz \; = \; \tfrac23\pi\rho \int_{\frac12R}^R r(r^2+2R^2)\sqrt{R^2-r^2}\,dr \\ & = \; \tfrac23\pi\rho R^5 \int_{\frac12}^1 u(u^2+2)\sqrt{1-u^2}\,du \; = \; \tfrac{17}{80}\pi\rho\sqrt{3}R^5 \end{aligned}$ which makes $\alpha = \tfrac{17}{40}$ and hence $\lfloor 1000\alpha \rfloor = 1000\alpha = \boxed{425}$ .