Partial Sphere Moment

The mass of the shell is $M \; = \; \int_0^\alpha 2\pi(R\sin\phi)\rho \times R\,d\phi \; = \; 2\pi R^2\rho \int_0^\alpha \sin\phi\,d\phi \; = \; 2\pi R^2\rho(1 - \cos\alpha)$ and the moment of inertia is $\begin{aligned} I & = \; \int_0^\alpha 2\pi(R\sin\phi)\rho \times (R\sin\phi)^2\times R\,d\phi\; = \; 2\pi R^4\rho \int_0^\alpha \sin^3\phi\,d\phi \\ & = \; 2\pi R^4\rho \int_0^\alpha \sin\phi(1 - \cos^2\phi)\,d\phi \; = \; 2\pi R^4\rho \Big[-\cos\phi + \tfrac13\cos^3\phi\Big]_0^\alpha \\ & = \; 2\pi R^4 \rho \big(\tfrac23 + \cos\alpha - \tfrac13\cos^3\alpha\big) \; = \; \tfrac23\pi R^4\rho (2 + 3\cos\alpha - \cos^3\alpha) \\ &= \; \tfrac23\pi R^4 \rho(1 - \cos\alpha)^2(2 + \cos\alpha) \; = \; \tfrac83\pi R^4\rho (2 +\cos\alpha)\sin^2\tfrac12\alpha \end{aligned}$ where $\rho$ is the surface mass density. This makes the moment of inertia $I \; = \; \tfrac43 MR^2 \frac{2 + \cos\alpha}{1 - \cos\alpha} \sin^4\tfrac{\alpha}{2}$ and so the answer is $4+3+2+1+1+1+4+2 = \boxed{18}$ .

The answer is in a slightly odd form, since a factor of $1 - \cos\alpha$ can be cancelled, so that $I \; = \; \tfrac13MR^2(1-\cos\alpha)(2 + \cos\alpha) \; = \; \tfrac23MR^2(2 + \cos\alpha)\sin^2\tfrac{\alpha}{2}$