91Ó°ÊÓ

The equation of sphere with radius $2$is $x^2+y^2+z^2=4$and for outside the cylinder$x^2+y^2=1$.

Center of Mass as per given conditions is given by

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

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

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

Also

$\stackrel{¯}{z}=\frac{M_{xy}}{M}=\frac{{∭}_{E}z\mathrm{ÒÏ}dV}{{∭}_E\mathrm{ÒÏ}dV}$where $\mathrm{ÒÏ}=k\sqrt{\left(x^2+y^2\right)}=kr$

It is given that density is proportional to point distance from $z$axis

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

It is given by

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

$=\frac{{∫}_{\mathrm{θ}=0}^{2\mathrm{π}}{∫}_{r=1}^2{∫}_{z=0}^{z=\sqrt{4-r^2}}z\left(kr\right)rdrd\mathrm{θ}dz}{{∫}_{\mathrm{θ}=0}^{2\mathrm{π}}{∫}_{r=1}^2{∫}_{z=0}^{\sqrt{4-r^2}}\left(kr\right)rdrd\mathrm{θ}dz}$

$=\frac{{\left(\frac{1}{2}\right)}_{\mathrm{θ}=0}^{2\mathrm{π}}{∫}_{r=1}^{r=2}\left(k{r}^2\right)\left(4-r^2\right)drd\mathrm{θ}}{{∫}_{\mathrm{θ}=0}^{2\mathrm{π}}{∫}_{r=1}^2\left(k{r}^2\right)\left(\sqrt{4-r^2}\right)drd\mathrm{θ}}$

Use $\mathrm{r}=2\sin \mathrm{Î\pm }⇒\mathrm{dr}=2\cos \mathrm{Î\pm }\mathrm{d}\mathrm{Î\pm }$in denominator

If $r=1⇒\mathrm{Î\pm }=\frac{\mathrm{Ï€}}{6}$

If $r=2⇒\mathrm{Î\pm }=\frac{\mathrm{Ï€}}{2}$

Integral becomes

$\stackrel{¯}{z}=\frac{\left(\frac{k}{2}\right){∫}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}\left(\frac{32}{3}-\frac{32}{5}\right)-\left(\frac{4}{3}-\frac{1}{5}\right)d\mathrm{θ}}{{∫}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}{∫}_{\mathrm{Î\pm }=\mathrm{Ï€}/6}^{\mathrm{Ï€}/2}\left(4k\right){\sin }^{2}2\mathrm{Î\pm }d\mathrm{Î\pm }}$

$=\frac{{\left.\left(\frac{47k}{30}\right)\left(\mathrm{θ}\right)\right|}_{\mathrm{θ}=0}^{\mathrm{θ}=2\mathrm{Ï€}}}{{\left.\left(4k\right){∫}_{\mathrm{θ}=0}^{2\mathrm{Ï€}}d\mathrm{θ}\left(\frac{1}{2}\right)\left(\mathrm{Î\pm }-\frac{\sin 4\mathrm{Î\pm }}{4}\right)\right|}_{\mathrm{Î\pm }=\mathrm{Ï€}/6}^{\mathrm{Ï€}/2}}$

$=\frac{\left(\frac{94k\mathrm{Ï€}}{30}\right)}{\left(4k\right)\left(2\mathrm{Ï€}\right)\left(\left(\frac{\mathrm{Ï€}}{2}-\frac{\mathrm{Ï€}}{6}\right)-\left(0-\frac{\sin (2\mathrm{Ï€}/3)}{4}\right)\right)}$

$\stackrel{¯}{z}=\frac{\left(\frac{94k\mathrm{π}}{30}\right)}{\left(4k\right)\left(\frac{1}{2}\right)\left(2\mathrm{π}\right)\left(\frac{\mathrm{π}}{3}+\frac{\sqrt{3}}{8}\right)}$

$=\frac{\left(\frac{94k\mathrm{Ï€}}{30}\right)}{\frac{\left(4k\mathrm{Ï€}\right)}{24}(8\mathrm{Ï€}+3\sqrt{3})}$

$=\left(\frac{94k\mathrm{π}}{30}\right)×\left(\frac{24}{4k\mathrm{π}(8\mathrm{π}+3\sqrt{3})}\right)$

$=\left(\frac{94}{15\sqrt{3}+40\mathrm{Ï€}}\right)$

Hence, Center of Mass is$(\stackrel{¯}{x},\stackrel{¯}{y},\stackrel{¯}{z})=\left(0,0,\frac{94}{15\sqrt{3}+40\mathrm{π}}\right)$