Find the specified quantities for the solids described below:

The mass of the region from Exercise $47$assuming that the density at every point is proportional to the square of the point’s distance from the $z$-axis.

The region inside both the spheres is determined by equation $x^2+y^2+z^2=4$and cylinder with equation$x^2+(y-1{)}^2=1$.

The cylindrical and rectangular coordinates are related as

$r=\sqrt{x^2+y^2},\tan \theta =\frac{y}{x},z=z$

and $x=r\cos \theta ,y=r\sin \theta ,z=z$

Rectangular coordinates are $x^2+y^2+z^2=4$and $x^2+(y-1{)}^2=1$

Cylindrical coordinates are $z=\pm \sqrt{4-r^2}$and $r=2\sin \theta$

Hence, cylindrical limits are

$-\sqrt{4-r^2}\le z\le \sqrt{4-r^2},0\le r\le 2\sin \theta ,0\le \theta \le \pi$

At every point, density is proportional to the square of point distance from $z$axis.

$\Rightarrow \rho =k\left(x^2+y^2\right)=k{r}^2$

Required Mass,

$m={\iiint }_E\rho dxdydz$

$m={\iiint }_{E}k\left(x^2+y^2\right)dxdydz$

$m={\int }_{\theta =0}^{\pi }{\int }_{r=0}^{2\sin \theta }{\int }_{z=-\sqrt{4-r^2}}^{\sqrt{4-r^2}}\left(k{r}^2\right)rdzdrd\theta$

$m={\int }_{\theta =0}^{\pi }{\int }_{r=0}^{2\sin \theta }\left(k{r}^3\right)\left(2\sqrt{4-r^2}\right)drd\theta$

localid="1653041357266" $m=2k{\int }_{\theta =0}^{\pi }{\int }_{r=0}^{2\sin \theta }{r}^3\sqrt{4-r^2}drd\theta$

Simplify further

$m={\left.2k{\int }_{\theta =0}^{\pi }\left(\frac{1}{2}\left(\frac{2}{5}{\left(4-r^2\right)}^{5/2}-\frac{8}{3}{\left(4-r^2\right)}^{3/2}\right)\right)\right|}_{r=0}^{2\sin \theta }d\theta$

$m=2k{\int }_{\theta =0}^{\pi }\left(\frac{32}{15}\right)\left(\left(3{\cos }^2\theta -5\right)\left|{\cos }^3\theta \right|+2\right)d\theta$

For splitting integral, we can use

$\left|\cos \theta \right|=\left\{\begin{array}{cc}\cos \theta & 0\le \theta \le \pi /2\\ -\cos \theta & \pi /2\le \theta \le \pi \end{array}\right.$

$m=2k\left\{{\int }_{\theta =0}^{\pi /2}\left(\frac{32}{15}\right)\left(\left(3{\cos }^2\theta -5\right){\cos }^3\theta +2\right)d\theta +{\int }_{\theta =\pi /2}^{\pi }\left(\frac{32}{15}\right)\left(\left(-3{\cos }^2\theta +5\right){\cos }^3\theta +2\right)d\theta \right\}$$=2k\left\{\left(\frac{32}{225}\right)(15\pi -26)+\left(\frac{32}{225}\right)(15\pi -26)\right\}$

$=2k\left\{\frac{960\pi -1664}{225}\right\}$

Hence,$m=k\left(\frac{1920\pi -3328}{225}\right)$