91Ó°ÊÓ

The density of region is constant.

Region above sphere is given by equation $\mathrm{ÒÏ}=R$and below the cone is given by equation$\mathrm{Ï•}=\mathrm{Î\pm }$.

For the given information, Center of ,mass if given by

$\stackrel{¯}{x}=\frac{M_{yz}}{M}=\frac{{∭}_{E}x\mathrm{ÒÏ}dxdydz}{∭\mathrm{ÒÏ}dxdydz}$

$\stackrel{→}{y}=\frac{M_x}{M}=\frac{{∭}_{E}y\mathrm{ÒÏ}dxdydz}{{∭}_E\mathrm{ÒÏ}dxdydz}$

As density of region is constant $\mathrm{ÒÏ}=k$, the axis of center lies above $+z$axis and base lies in $xy$plane.

Hence,$\stackrel{¯}{x}=0,\stackrel{¯}{y}=0$

It can be given by $\stackrel{¯}{z}=\frac{M_{xy}}{M}\frac{{∭}_{E}z\mathrm{ÒÏ}dxdydz}{{∭}_E\mathrm{ÒÏ}dxdydz}$

$\stackrel{¯}{z}=\frac{M_{xy}}{M}=\frac{{∭}_{E}z\mathrm{ÒÏ}dV}{{∭}_E\mathrm{ÒÏ}dV}$

$=\frac{{∫}_{\mathrm{Ï•}=0}^{\mathrm{Î\pm }}{∫}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}{∫}_{\mathrm{ÒÏ}=0}^R\left(\mathrm{ÒÏ}\cos \mathrm{Ï•}\right)k{\mathrm{ÒÏ}}^2\sin \mathrm{Ï•}d\mathrm{ÒÏ}d\mathrm{θ}d\mathrm{Ï•}}{{∫}^a{∫}^{2\mathrm{Ï€}}{∫}^{R}k{\mathrm{ÒÏ}}^2\sin \mathrm{Ï•}d\mathrm{ÒÏ}d\mathrm{θ}d\mathrm{Ï•}}$

$=\frac{{\left.{\left.{\left.\left(\frac{{\sin }^2\mathrm{Ï•}}{2}\right)\right|}_{\mathrm{Ï•}=0}^{\mathrm{Ï•}=\mathrm{Î\pm }}k\left(\frac{{\mathrm{ÒÏ}}^4}{4}\right)\right|}_{\mathrm{ÒÏ}=0}^R\left(\mathrm{θ}\right)\right|}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}}{{\left.{\left.{\left.k(-\cos \mathrm{Ï•})\right|}_{\mathrm{Ï•}=0}^{\mathrm{Î\pm }}\left(\mathrm{θ}\right)\right|}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}\left(\frac{{\mathrm{ÒÏ}}^3}{3}\right)\right|}_{\mathrm{ÒÏ}=0}^R}$

Putting limits yields

$\stackrel{¯}{z}=\frac{\left(\frac{{\sin }^2\mathrm{Î\pm }}{2}\right)k\left(\frac{R^4}{4}\right)\left(2\mathrm{Ï€}\right)}{k(1-\cos \mathrm{Î\pm })\left(2\mathrm{Ï€}\right)\left(\frac{R^3}{3}\right)}$

$=\frac{{\sin }^2\mathrm{Î\pm }×R^4\left(2k\mathrm{Ï€}\right)}{8}×\frac{3}{\left(2k\mathrm{Ï€}\right)R^3(1-\cos \mathrm{Î\pm })}$

$=\left(\frac{3R}{8}\right)(1+\cos \mathrm{Î\pm })$

Hence, required Center of Mass is given by$(\stackrel{¯}{x},\stackrel{¯}{y},\stackrel{¯}{z})=\left(0,0,\left(\frac{3R}{8}\right)(1+\cos \mathrm{Î\pm })\right)$