Question

Find the specified quantities for the solids described below:

The center of mass of the region from Exercise $49$, assuming that the density at every point is proportional to the point’s distance from the $z$-axis.

Short answer

Center of mass is$(\overline{x},\overline{y},\overline{z})=\left(0,0,\frac{94}{15\sqrt{3}+40\pi }\right)$.

Step-by-step solution

Given Information

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

Evaluating of Center of Mass for x, y coordinates

Center of Mass as per given conditions is given by

$\overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{E}x\rho dxdydz}{\iiint \rho dxdydz}$

$\overrightarrow{y}=\frac{M_{xz}}{M}=\frac{{\iiint }_{E}y\rho dxdydz}{{\iiint }_E\rho dxdydz}$

$\overline{z}=\frac{M_{xy}}{M}\frac{{\iiint }_{E}z\rho dxdydz}{{\iiint }_E\rho dxdydz}$

Also

$\overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{E}z\rho dV}{{\iiint }_E\rho dV}$where $\rho =k\sqrt{\left(x^2+y^2\right)}=kr$

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

Hence, $\overline{x}=0,\overline{y}=0$

Evaluation of Center of Mass of z coordinate

It is given by

$\overline{z}=\frac{{\iiint }_{E}z\rho dV}{{\iiint }_E\rho dV}$

$=\frac{{\int }_{\theta =0}^{2\pi }{\int }_{r=1}^2{\int }_{z=0}^{z=\sqrt{4-r^2}}z\left(kr\right)rdrd\theta dz}{{\int }_{\theta =0}^{2\pi }{\int }_{r=1}^2{\int }_{z=0}^{\sqrt{4-r^2}}\left(kr\right)rdrd\theta dz}$

$=\frac{{\left(\frac{1}{2}\right)}_{\theta =0}^{2\pi }{\int }_{r=1}^{r=2}\left(k{r}^2\right)\left(4-r^2\right)drd\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{r=1}^2\left(k{r}^2\right)\left(\sqrt{4-r^2}\right)drd\theta }$

Use $\mathrm{r}=2\sin \alpha \Rightarrow \mathrm{dr}=2\cos \alpha \mathrm{d}\alpha$in denominator

If $r=1\Rightarrow \alpha =\frac{\pi }{6}$

If $r=2\Rightarrow \alpha =\frac{\pi }{2}$

Integral becomes

$=\frac{{\left.{\left(\frac{k}{2}\right)}_{\theta =0}^{2\pi }\left(\frac{4r^3}{3}-\frac{r^5}{5}\right)\right|}_{r=1}^{r=2}d\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{\alpha =\pi /6}^{\pi /2}k\left(2\sin \alpha {)}^2\right(2\cos \alpha {)}^{2}d\alpha }$

$\overline{z}=\frac{\left(\frac{k}{2}\right){\int }_{\theta =0}^{2\pi }\left(\frac{32}{3}-\frac{32}{5}\right)-\left(\frac{4}{3}-\frac{1}{5}\right)d\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{\alpha =\pi /6}^{\pi /2}\left(4k\right){\sin }^{2}2\alpha d\alpha }$

$=\frac{{\left.\left(\frac{47k}{30}\right)\left(\theta \right)\right|}_{\theta =0}^{\theta =2\pi }}{{\left.\left(4k\right){\int }_{\theta =0}^{2\pi }d\theta \left(\frac{1}{2}\right)\left(\alpha -\frac{\sin 4\alpha }{4}\right)\right|}_{\alpha =\pi /6}^{\pi /2}}$

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

$\overline{z}=\frac{\left(\frac{94k\pi }{30}\right)}{\left(4k\right)\left(\frac{1}{2}\right)\left(2\pi \right)\left(\frac{\pi }{3}+\frac{\sqrt{3}}{8}\right)}$

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

$=\left(\frac{94k\pi }{30}\right)\times \left(\frac{24}{4k\pi (8\pi +3\sqrt{3})}\right)$

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

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