Okay, I'll finish this, then:
Since z=r\cos\phi, it follows that on the plane where z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex], we have that the radius follows the the curve:<br />
r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}<br />
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Also, remember that the truncated sphere is axially symmetric, so that the x-y-coordinates must be zero.<br />
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Generally, assuming unit density, the z-coordinate of the center of mass of some region R will be:<br />
z_{c.m}=\frac{1}{V}\int_{R}zdV<br />
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We decompose our integral into two parts, then (with V as the ugly expression given above):<br />
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)<br />
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With Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}<br />
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We now calculate the two integrals:<br />
\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}<br />
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\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\frac{Z^{4}}{4}\frac{\sin\phi}{\cos^{3}\phi}d\phi=\frac{Z^{4}}{8}\frac{1}{\cos^{2}\phi}\mid_{0}^{\sin^{-1}(\frac{\alpha}{2R})}=\frac{Z^{4}}{8}(\frac{4R^{2}}{4R^{2}-\alpha^{2}}-1)=\frac{Z^{4}}{8}\frac{\alpha^{2}}{4R^{2}-\alpha^{2}}<br />
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Adding these, you should get something like:<br />
z_{c.m}=-\frac{\pi\alpha^{4}}{64V}
