Now, I'm not exactly sure what you mean by saying "h and R" is changing.
We have the relation:
[tex]h=R+\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]
For any particular truncation, (with [itex]\phi[/itex] being the angle between the upwards vertical and the position vector from the origin,) we may decompose our volume as a sphere with the appropriate cone cut out, plus that cone.
The volume of the cone is clearly: [tex]V_{cone}=\frac{1}{3}\pi(\frac{\alpha}{2})^{2}*\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]
wheras the volume of the "cone-less" sphere must be:
[tex]V=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{2\pi}\int_{0}^{R}r^{2}\sin\phi{dR}d\theta{d}\phi=\frac{2\pi{R}^{3}}{3}(1+\cos(\sin^{-1}(\frac{\alpha}{2R})))=\frac{2\pi{R}^{3}}{3}(1+\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{R})[/tex]
Since [itex]z=r\cos\phi[/itex], it follows that on the plane where [itex]z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex], we have that the radius follows the the curve:
[tex]r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}[/tex]
Also, remember that the truncated sphere is axially symmetric, so that the x-y-coordinates must be zero.
Generally, assuming unit density, the z-coordinate of the center of mass of some region R will be:
[tex]z_{c.m}=\frac{1}{V}\int_{R}zdV[/tex]
We decompose our integral into two parts, then (with V as the ugly expression given above):
[tex]z_{c.m}=\frac{2\pi}{V}(\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}zr^{2}\sin\phi{d\phi}dr+\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr)[/tex]
With [tex]Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]
We now calculate the two integrals:
[tex]\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\frac{R^{4}}{4}\cos\phi\sin(\phi)d\phi=\frac{R^{4}}{8}\sin^{2}\phi\mid_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}=-\frac{R^{2}\alpha^{2}}{32}[/tex]